4 Marks Questions - Maths STD 10 Questions - Vidyadip

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Question 14 Marks

$ABCD$ is a cyclic quadrilateral such that $\angle\text{A}=(4\text{y}+20)^\circ,\angle\text{B}=(3\text{y}-5)^\circ$ $\angle\text{C}=(-4\text{x})^\circ$ and $\angle\text{D}=(7\text{x}+5)^\circ$ Find the four angles.

Answer

We know that the sum of the opposite angles of cyclic quadrilateral is $180^\circ $ in the cyclic quadrilateral $ABCD$ angle $A$ and $C$ angles $B$ and $D$ pairs of opposite angles
$\therefore\angle\text{A}+\angle\text{C}=180^\circ$ and $\angle\text{B}+\angle\text{D}=180^\circ$
$\angle\text{A}+\angle\text{C}=180^\circ$
By Substituting $\angle\text{A}=(4\text{y}+20)^\circ$ and $\angle\text{C}=(-4\text{x})^\circ$ we get
$4y + 20^\circ - 4x = 180^\circ $
$-4x + 4y + 20^\circ = 180^\circ $
$-4x + 4y = 180^\circ - 20^\circ $
$-4x + 4y = 160^\circ $
$4x - 4y = -160^\circ $
Divide both sides of equation by $4$ we get
$x - y = -40^\circ $
$x - y + 40^\circ = 0 .....(i)$
$\angle\text{B}+\angle\text{D}=180^\circ$
BY substituting $\angle\text{B}=(3\text{y}-5)^\circ$ and $\angle\text{D}=(7\text{x}+5)^\circ$ we get
$3y - 5^\circ + 7x + 5^\circ = 180^\circ $
$7x + 3y = 180^\circ $
$7x + 3y - 180^\circ = 0 ......(ii)$
By multiplying equation $(i)$ by $3$ we get
$3x - 3y + 120° = 0 .....(iii)$
By adding equation $(iii)$ from $(ii)$ we get
$3\text{x}\ -\ 3\text{y}\ +\ 120\ =\ 0\\7\text{x}\ +\ 3\text{y}\ -\ 180\ =\ 0\over10\text{x}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ 60$
$\text{x}=\frac{60}{10}$
$\text{x}=6^\circ$
By substituting $x = 6^\circ $ in equation $(i)$ we get
$x - y + 40^\circ = 0$
$6^\circ - y + 40^\circ = 0$
$-1y = -40^\circ - 6^\circ $
$-1y = -46^\circ $
$\text{y}=\frac{-46^\circ}{-1}$
$\text{y}=46^\circ$
The angles of a cyclic quadrilateral are
$\angle\text{A}=4\text{y}+20$
$= (4 \times 46)^\circ + 20^\circ $
$= 184 + 20$
$= 204^\circ $
$\angle\text{B}=3\text{y}-5$
$= 3 \times 46 - 5$
$= 138 - 5 = 133^\circ $
$\angle\text{C}=-4\text{x}^\circ$
$= -[4(6)]^\circ $
$= -24^\circ $
$\angle\text{D}=\big(7\text{x}+5\big)^\circ$
$= (7 \times 6)^\circ + 5^\circ $
$= (42 + 5)^\circ $
$= 47^\circ $
Hence the angles of quadrilateral are $\angle\text{A}=204^\circ,\angle\text{B}=133^\circ\angle\text{C}=-24^\circ,\angle\text{D}=47^\circ$

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Question 24 Marks

Solve the following systems of equations:
$\frac{2}{3\text{x}+2\text{y}}+\frac{3}{3\text{x}-2\text{y}}=\frac{17}{5},$
$\frac{5}{3\text{x}+2\text{y}}+\frac{1}{3\text{x}-2\text{y}}=2.$

Answer

The given equations are
$\frac{2}{3\text{x}+2\text{y}}+\frac{3}{3\text{x}-2\text{y}}=\frac{17}{5}$
$\frac{5}{3\text{x}+2\text{y}}+\frac{1}{3\text{x}-2\text{y}}=2$
Let $\frac{1}{3\text{x}-2\text{y}}=\text{u}$ and $\frac{1}{3\text{x}-2\text{y}}=\text{v}$ then equations are
$2\text{u}+3\text{v}=\frac{17}{5}\ ......(\text{i})$
$5\text{u}+\text{v}=2\ ......(\text{ii})$
Multiply equation $(ii)$ by $3$ and subtract $(ii)$ from $(i)$ we get
$2\text{u}\ \ +3\text{v}=\frac{17}{5}\\15\text{u}+3\text{v}=6\\\overline{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\\-13\text{u}=-\frac{13}{5}$
$\Rightarrow\text{u}=\frac{1}{5}$
Put the value of $u$ in equation $(i)$ we get
$2\times\frac{1}{5}+3\text{v}=\frac{17}{5}$
$\Rightarrow3\text{v}=3$
$\Rightarrow\text{v}=1$
Then
$\frac{1}{3\text{x}+2\text{y}}=\frac{1}{5}$
$\Rightarrow3\text{x}+2\text{y}=5\ ......(\text{iii})$
$\frac{1}{3\text{x}-2\text{y}}=1$
$\Rightarrow{3\text{x}-2\text{y}}=1\ ......(\text{iv})$
Add both equations we get
${3\text{x}\ +\ 2\text{y}}\ =\ 5\\{3\text{x}\ -\ 2\text{y}}\ =\ 1\over6\text{x}\ \ \ \ \ \ \ \ \ \ =\ 6$
$\Rightarrow\text{x}=1$
Put the value of $x$ in equation $(iii)$ we get
$3\times1+2\text{y}=5$
$\Rightarrow2\text{y}=2$
$\Rightarrow\text{y}=1$
Hence the value of $x = 1$ and $y = 1.$

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Question 34 Marks

Solve graphically the following system of linear equation. Also find the coordinates of the points where the lines meet axis of y.

$3x + 2y = 12$

$5x - 2y = 4$

Answer

We have,
$3x + 2y = 12$
$5x - 2y = 4$
Now, $3x + 2y = 12$
$⇒ 3x = 12 - 2y$
$\Rightarrow\text{x}=\frac{12-2\text{y}}{3}$
When $y = 3,$ we have,
$\text{x}=\frac{12-2\times3}{3}=2$
When $y = -3,$ we have,
$\text{x}=\frac{12-2\times(-3)}{3}=6$
Thus, we have the following table giving points on the line $3x + 2y = 12$

$x$	$2$	$6$
$y$	$3$	$-3$

Now, $5x - 2y = 4$
$⇒ 5x = 4 + 2y$
$\Rightarrow\text{x}=\frac{4+2\text{y}}{5}$
When $y = 3,$ we have,
$\text{x}=\frac{4+2\times3}{5}=2$
When $y = -7,$ we have,
$\text{x}=\frac{4+2\times(-7)}{5}=-2$
Thus, we have the following table points on the line $5x + 2y = 4$

$x$	$2$	$-2$
$y$	$3$	$-7$

Graph of the given equations,

Clearly, two intersect at $P(2, 3).$
Hence, $x = 2, y = 3$ is the solution of the given system of equations.
We also observe that lines represented by $3x + 2y = 12$ and $5x - 2y = 4$ meet y-axis at $A(0, 6)$ and $B(0, -2)$ respectively.

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Question 44 Marks

Represent the following pair of equations graphically and write the coordinates of points where the lines intersects $y-$axis.

$x + 3y = 6,$

$2x - 3y = 12.$

Answer

The given equations are
$x + 3y = 6 ......(i)$
$2x - 3y = 12 ........(ii)$
Putting $x = 0$ in equation $(i)$ we get,
$⇒ 0 + 3y = 6$
$⇒ y = 2$
$⇒ x = 0, y = 2$
Putting $y = 0$ in equation $(i)$ we get,
$⇒ x + 3 × 0 = 6$
$⇒ x = 6$
$⇒ x = 6, y = 0$
Use the following table to draw the graph.

$x$	$0$	$6$
$y$	$2$	$0$

The graph of $(i)$ can be obtained by plotting the two points $A(0, 2), B(6, 0).$
$2x - 3y = 12 ......(ii)$
Putting $x = 0$ in equation $(ii)$ we get,
$⇒ 2 × 0 - 3y = 12$
$⇒ y = -4$
$⇒ x = 0, y = -4$
Putting $y = 0 $ in equation $(ii)$ we get,
$⇒ 2x - 3 × 0 = 12$
$⇒ x = 6$
$⇒ x = 6, y = 0$
Use the following table to draw the graph.

$x$	$0$	$6$
$y$	$-4$	$0$

Draw the graph by plotting two points $C(0, -4), D(6, 0)$ from table.

Graph of lines represented by the equations $x + 3y = 6, 2x - 3y = 12$ meet y-axis at $A(0, 2), C(0, -4)$ respectively.

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Question 54 Marks

Akhila went to a fair in her village. She wanted to enjoy rides in the Giant Wheel and play Hoopla (a game in which you throw a rig on the items kept in the stall, and if the ring covers any object completely you get it). The number of times she played Hoopla is half the number of rides she had on the Giant Wheel. Each ride costs $Rs. 3,$ and a game of Hoopla costs $Rs. 4.$ If she spent $Rs. 20$ in the fair, represent this situation algebraically and graphically.

Answer

The pair of equations formed is:
$\text{y}=\frac{1}{2}\text{x}$
i.e.,$ x - 2y = 0$
$3x + 4y = 20$
Let us represent these equations graphically. For this, we need at least two solutions for each equation.
We give these solutions in Table:

$x$	$0$	$2$
$\text{y}-\frac{\text{x}}{2}$	$0$	$1$

$x$	$0$	$2$	$4$
$\text{y}=\frac{20-3\text{x}}{4}$	$5$	$0$	$2$

Recall from Class IX that there are infinitely many solutions of each linear equation. So each of you choose any two values, which may not be the ones we have chosen. Can you guess why we have chosen $x = 0$ in the first equation and in the second equation? When one of the variables is zero, the equation reduces to a linear equation is one variable, which can be solved easily. For instance, putting $x = 0$ in Equation $(2),$ we get $4y = 20$ i.e. $y = 5$ Similarly, putting $y = 0$ in equation $(2),$ we get $3x = 20$ i.e., $\text{x}=\frac{20}{3}.$ but as $\frac{20}{3}$ is not an integer, it will not be easy to plot exactly on the graph paper. So, we choose $y = 2$ Which gives $x = 4,$ an integral value.

Plot the points $A(0, 0), B(2, 1)$ and $P(0, 5), Q(4, 12),$ corresponding to the draw the lines $AB$ and $PQ,$ representing the equations $x - 2y = 0$ and $3x + 4y = 20,$ as shown. observe that the two lines representing the two equations are intersecting at the point $(4, 2).$

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Question 64 Marks

Aftab tells his daughter, "Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be". Is not this interesting? Represent this situation algebraically and graphically.

Answer

Let present age of aftab be $x$ years and his daughter be $y$ years.
Seven years ago:
Age of aftab $= (x - 7)$ years
Age of his daughter $= (y - 7)$ years
According to question:
$(x - 7) = 7(y - 7)$
$⇒ x - 7 = 7y - 49$
$⇒ x - 7y + 42 = 0 .......(i)$
Three year from now:
Age of abtab $= (x + 3)$ years
Age of his daughter $= (y + 3)$ years
According to question:
$(x + 3) = 3(y + 3)$
$⇒ (x + 3) = 3y + 9$
$⇒ x - 3y - 6 = 0 ........(ii)$
Thus, equation $(i)$ and $(ii)$ show the situation algelraically.
Graphically represention:
In equation $(i),$ putting $x = 0,$ we get $y = 6$
Putting $x = 7,$ we get $y = 7$
Putting $x = 7,$ we get $y = 5$

$x$	$0$	$7$	$-7$
$y$	$6$	$7$	$5$

In equation $(ii),$ putting $x = 0,$ we get $y = -2$
putting $x = 3,$ we get $y = -1$
putting $x = 6, $ we get $y = 0$

$x$	$0$	$3$	$6$
$y$	$-2$	$-1$	$0$

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Question 74 Marks

Solve the following systems of equations:
$\frac{10}{\text{x}+\text{y}}+\frac{2}{\text{x}-\text{y}}=4,$
$\frac{15}{\text{x}+\text{y}}-\frac{5}{\text{x}-\text{y}}=-2.$

Answer

$\frac{10}{\text{x}+\text{y}}+\frac{2}{\text{x}-\text{y}}=4$
$\frac{15}{\text{x}+\text{y}}-\frac{5}{\text{x}-\text{y}}=-2$
Putting $\frac{1}{\text{x}+\text{y}}=\text{p} $ and $\frac{1}{\text{x}-\text{y}}=\text{q} $ in the given equations, we get:
$10p + 2q = 4$
$⇒10p + 2q - 4 = 0 ......(i)$
$15p - 2q = -2$
$⇒15p - 5q + 2 = 0 ........(ii)$
Using cross multiplication, we get
$\frac{\text{p}}{4-20}=\frac{\text{q}}{-60-(-20)}=\frac{1}{-50-30}$
$\frac{\text{p}}{-16}=\frac{\text{q}}{-80}=\frac{1}{-80}$
$\frac{\text{p}}{-16}=\frac{1}{-80}$ and $\frac{\text{q}}{-80}=\frac{1}{-80}$
$\text{p}=\frac{1}{5}$ and $\text{q}=1$
$\text{p}=\frac{1}{\text{x}+\text{y}}=\frac{1}{5}$ and $\text{q}=\frac{1}{\text{x}-\text{y}}=1$
$x + y = 5 .....(iii)$
and $x - y = 1 .....(iv)$
Adding equation $(iii)$ and $(iv),$ we get
$2x = 6$
$x = 3 .... (v)$
Putting value of $x$ in equation $(iii)$, we get
$y = 2$
Hence, $x = 3$ and $y = 2$.

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Question 84 Marks

Solve graphically the system of linear equations:

$4x - 3y + 4 = 0,$

$4x + 3y - 20 = 0.$

Find the area bounded by these lines and $x-$axis.

Answer

The given equations are,
$4x - 3y + 4 = 0 ......(i)$
$4x + 3y - 20 = 0 ........(ii)$
From $(i), \text{y}=\frac{4\text{x}+4}{3}$
Putting $x = 2,$ we get $y = 4$
Putting $x = 5,$ we get $y = 8$
Putting $x = -1,$ we get $y = 0$
Thus, plot the points $A(2, 4), B(5, 8)$ and $C(-1, 0)$ On the graph paper.
From $(ii), \text{y}=\frac{20-4\text{x}}{3}$
Putting $x = -1$, we get $y = 8$
Putting $x = 2$, we get $y = 4$
Putting $x = 5,$ we get $y = 0$
Thus Plot the points $P(-1, 8), Q(2, 4)$ and $R(5, 0)$ on the graph paper.

Thus, theb are boundad by these lines and $x-$axis in from of $\triangle\text{OQR}.$
Area of $\triangle\text{OQR} =\frac{1}{2}\times\text{OR}\times\text{QD}$
$=\frac{1}{2}\times6\times4$
$= 3 × 4$
$= 12 sq.$ units.
When, we solve these equations we get $x = 2$ and $y = 4$.

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Question 94 Marks

The boat goes $30\ km$ upstream and $44\ km$ downstream in $10$ hours. In $13$ hours, it can go $40\ km$ upstream and $55\ km$ downstream. Determine the speed of stream and that of the boat in still water.

Answer

Let the speed of the boat in still water be $x\ km/hr$ and the speed of the stream be $y\ km/hr$
Speed upstream $= (x - y)\ km/hr$
Speed down stream $= (x + y)\ km/hr$
Now,
Time taken to cover $30\ km$ upstream $=\frac{30}{\text{x}-\text{y}}\text{hrs.}$
Time taken to cover $44\ km$ down stream $=\frac{44}{\text{x}+\text{y}}\text{hrs.}$
But total time of journey is $10$ hours
$\frac{30}{\text{x}-\text{y}}+\frac{44}{\text{x}+\text{y}}=10\ ....(\text{i})$
Time taken to cover $40\ km$ upstream $=\frac{40}{\text{x}-\text{y}}\text{hrs.}$
Time taken to cover $55\ km$ down stream $=\frac{55}{\text{x}+\text{y}}\text{hrs.}$
In this case total time of journey is given to be $13$ hours
$\therefore\frac{40}{\text{x}-\text{y}}+\frac{55}{\text{x}+\text{y}}=13\ ....(\text{ii})$
Putting $\frac{1}{\text{x}-\text{y}}=\text{u}$ and $\frac{1}{\text{x}-\text{y}}=\text{v}$ in equation $(i)$ and $(ii)$ we get
$30\text{u}+44\text{v}=10$
$40\text{u}+55\text{v}=10$
$30\text{u}+44\text{v}-10=0\ ....(\text{iii})$
$40\text{u}+55\text{v}-13=0\ .....(\text{iv})$
Solving these equations by cross multiplication we get
$\Rightarrow\frac{\text{u}}{44\times-13-55\times-10}=\frac{-\text{v}}{30\times-13-40\times-10}\\=\frac{1}{30\times55-40\times44}$
$\Rightarrow\frac{\text{u}}{-572+550}=\frac{-\text{v}}{-390+400}=\frac{1}{1650-1760}$
$\Rightarrow\frac{\text{u}}{-22}=\frac{-\text{v}}{10}=\frac{1}{-110}$
$\Rightarrow\text{u}=\frac{-22}{-110}$
$\Rightarrow\text{v}=\frac{-10}{-110}$
$\text{u}=\frac{1}{5}$ and $\text{v}=\frac{1}{11}$
Now, $\text{u}=\frac{2}{10}$
$\frac{1}{\text{x}-\text{y}}=\frac{2}{10}$
$1\times10=2(\text{x}-\text{y})$
$10=2\text{x}-2\text{y}\div2$
$5=\text{x}-\text{y}\ ....(\text{v})$
$\text{v}=\frac{1}{11}$
$\frac{1}{\text{x}+\text{y}}=\frac{1}{11}$
$1\times11=1(\text{x}+\text{y})$
$11=\text{x}+\text{y}\ ....(\text{vi})$
By solving equation $(v)$ and $(vi)$ we get
$\text{x}\ -\ \text{y}\ =\ 15\\\text{x}\ +\ \text{y}\ =\ 16\over2\text{x}\ \ \ \ \ \ =\ 16$
$\text{x}=\frac{16}{2}$
$\text{x}=8$
Substituting $x = 8$ in equation $(vi)$ we get
$\text{x}+\text{y}=11$
$8+\text{y}=11$
$\text{y}=11-8$
$\text{y}=3$
Hence, speed of the boat in still water is $8\ km/hr.$ Speed of the stream is $3\ km/hr.$

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Question 104 Marks

Ritu can row downstream $20\ km$ in $2$ hours, and upstream $4\ km$ in $2$ hours. Find her speed of rowing in still water and the speed of the current.

Answer

Let the speed of rowing in still water be $x\ km/hr$ and the speed of the current be $y\ km/hr$ Let the speed of rowing in still water be $x\ km/ hr$ and the speed of the current be $y\ km/ hr.$
Speed upstream $= (x - y)\ km/ hr$
Speed downstream $= (x + y)\ km/ hr$
Now,
Time taken to cover $20\ km$ down steam $=\frac{20}{\text{x}+\text{y}}\text{ hrs.}$
Time taken to cover $4\ km$ upstream $=\frac{4}{\text{x}+\text{y}}\text{ hrs.}$
But, time taken to cover 20km downstream in $2$ hours
$\frac{20}{\text{x}+{\text{y}}}=2$
$20=2(\text{x}+\text{y})$
$20=2\text{x}+2\text{y}\ ......(\text{i})$
Time taken to cover 4km upstream in $2$ hours
$\frac{4}{\text{x}-\text{y}}=2$
$4=2(\text{x}-\text{y})$
$4=2\text{x}-2\text{y}\ .....(\text{ii})$
By solving these equation $(i)$ and $(ii)$ we get
$2\text{x}\ +\ 2\text{y}\ =\ 20\\2\text{x}\ -\ 2\text{y}\ =\ \ \ 4\over4\text{x}\ \ \ \ \ \ \ \ \ \ \ =\ 24$
$\text{x}=\frac{24}{4}$
$\text{x}=6$
Substitute $x = 6$ in equation $(i)$ we get
$2\text{x}+2\text{y}=20$
$2+2\text{y}=20$
$2\text{y}=20-12$
$\text{y}=\frac{8}{2}$
$\text{y}=4$
Hence, the speed of rowing in still water is $6\ km/ hr.$
The speed of current is $4\ km/ hr.$

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Question 114 Marks

Solve the following system of equations by the method of cross-multiplication:
$\text{bx}+\text{cy}=\text{a}+\text{b}$
$\text{ax}\Big(\frac{1}{\text{a}-\text{b}}-\frac{1}{\text{a}+\text{b}}\Big)+\text{cy}\Big(\frac{1}{\text{b}-\text{a}}-\frac{1}{\text{b}+\text{a}}\Big)=\frac{2\text{a}}{{\text{a}+\text{b}}}$

Answer

The given system of equation is,
$\text{bx}+\text{cy}=\text{a}+\text{b}\ ....(\text{i})$
$\text{a}\text{x}\Big(\frac{1}{\text{a}-\text{b}}-\frac{1}{\text{a}+\text{b}}\Big)+\text{cy}\Big(\frac{1}{\text{b}-\text{a}}-\frac{1}{\text{b}+\text{a}}\Big)=\frac{2\text{a}}{\text{a}+\text{b}}\ ...(\text{ii})$
From equation (i) we get
$\text{bx}+\text{cy}-(\text{a}+\text{b})=0\ ...(\text{iii})$
From equation (ii) we get
$\text{a}\text{x}\Big[\frac{\text{a}+\text{b}-(\text{a}-\text{b})}{(\text{a}-\text{b})(\text{a}+\text{b})}\Big]+\text{cy}\Big(\frac{\text{b}+\text{a}-(\text{b}-\text{a})}{(\text{b}-\text{a})(\text{b}+\text{a})}\Big)-\frac{2\text{a}}{\text{a}+\text{b}}=0$
$ \Rightarrow\text{a}\text{x}\Big[\frac{\text{a}+\text{b}-\text{a}+\text{b}}{(\text{a}-\text{b})(\text{a}+\text{b})}\Big]+\text{cy}\Big(\frac{\text{b}+\text{a}-\text{b}+\text{a}}{(\text{b}-\text{a})(\text{b}+\text{a})}\Big)-\frac{2\text{a}}{\text{a}+\text{b}}$
$\Rightarrow\text{a}\text{x}\Big[\frac{2\text{b}}{(\text{a}-\text{b})(\text{a}+\text{b})}\Big]+\text{cy}\Big(\frac{2\text{a}}{(\text{b}-\text{a})(\text{b}+\text{a})}\Big)-\frac{2\text{a}}{\text{a}+\text{b}}=0$
$ \Rightarrow\text{x}\Big[\frac{2\text{a}\text{b}}{(\text{a}-\text{b})(\text{a}+\text{b})}\Big]+\text{y}\Big(\frac{2\text{a}\text{c}}{-(\text{a}-\text{b})(\text{a}+\text{b})}\Big)-\frac{2\text{a}}{\text{a}+\text{b}}=0$
$ \Rightarrow\text{x}\Big[\frac{2\text{a}\text{b}}{(\text{a}-\text{b})(\text{a}+\text{b})}\Big]+\text{y}\Big(\frac{-2\text{a}\text{c}}{(\text{a}-\text{b})(\text{a}+\text{b})}\Big)-\frac{2\text{a}}{\text{a}+\text{b}}=0$
$\Rightarrow\frac{1}{\text{a}+\text{b}}\Big[\frac{2\text{abx}}{\text{a}-\text{b}}-\frac{2\text{acy}}{\text{a}-\text{b}}-2\text{a}\Big]=0$
$\Rightarrow\frac{2\text{abx}}{\text{a}-\text{b}}-\frac{2\text{acy}}{\text{a}-\text{b}}-2\text{a}=0$
$ \Rightarrow\frac{2\text{abx}-2\text{acy}-2\text{a}(\text{a}-\text{b})}{\text{a}-\text{b}}=0$
$\Rightarrow2\text{abx}-2\text{acy}-2\text{a}(\text{a}-\text{b})=0\ ...(\text{iv})$
From equation (i) and equation (ii) we get
$ a_1=b, b_1=c, c_1=-(a+b) $
$ a_2=2 a b, b 2=-2 a c \text { and } c_2=-2 a(a-b)$
By cross-multiplication we get,
$\Rightarrow\frac{\text{x}}{-2\text{ac}(\text{a}-\text{b})-\big[-(\text{a}+\text{b})\big][-2\text{ac}]}\\=\frac{-\text{y}}{-2\text{ab}(\text{a}-\text{b})-\big[-(\text{a}+\text{b})\big][2\text{ab}]}\\=\frac{1}{-2\text{abc}-2\text{abc}}$
$\Rightarrow\frac{\text{x}}{-2\text{a}^2\text{c}+2\text{ab}-\big[2\text{a}^2\text{c}+2\text{abc}\big]}\\=\frac{-\text{y}}{-2\text{a}^2\text{b}+2\text{ab}^2+\big[2\text{a}^2\text{b}+2\text{ab}^2\big]}\\=\frac{1}{-4\text{abc}}$
$\Rightarrow\frac{\text{x}}{-2\text{a}^2\text{c}+2\text{ab}-2\text{a}^2\text{c}-2\text{abc}}\\=\frac{-\text{y}}{-2\text{a}^2\text{b}+2\text{ab}^2+2\text{a}^2\text{b}+2\text{ab}^2}\\=\frac{1}{-4\text{abc}}$
$\Rightarrow\frac{\text{x}}{-4\text{a}^2\text{c}}=\frac{-\text{y}}{4\text{a}\text{b}^2}=\frac{-1}{4\text{abc}}$
Now,
$\frac{\text{x}}{-4\text{a}^2\text{c}}=\frac{-1}{4\text{abc}}$
$\Rightarrow\text{x}=\frac{4\text{a}^2\text{c}}{4\text{abc}}=\frac{\text{a}}{\text{b}}$
and,
$\Rightarrow\text{y}=\frac{4\text{a}^2\text{c}}{4\text{abc}}=\frac{\text{b}}{\text{c}}$
Hence, $\text{x}=\frac{\text{a}}{\text{b}},\text{y}=\frac{\text{b}}{\text{c}}$ is the solution of the given system of the equations.

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Question 124 Marks

Solve the following systems of equations:

$x - y + z = 4,$

$x - 2y - 2z = 9,$

$2x + y + 3z = 1.$

Answer

We have,
$x - y + z = 4 ........(i)$
$x - 2y - 2z = 9 .......(ii)$
$2x + y + 3z = 1 .........(iii)$
From equation $(i)$ we get
$z = 4 - x + y$
$⇒ z = -x + y + 4$
Substituting the value of $z$ in equation $(ii)$ we get
$x - 2y - 2(-x + y + 4) = 9$
$⇒ x - 2y + 2x - 2y - 8 = 9$
$⇒ 3x - 4y = 9 + 8$
$⇒ 3x - 4y = 17 .......(iv)$
Substituting the value of $z$ in equation $(iii)$ we get
$2x + y + 3(-x + y + 4) = 1$
$⇒ 2x + y - 3x + 3y + 12 = 1$
$⇒ -x + 4y = 1 - 12$
$⇒ -x + 4y = -11 .......(v)$
Adding equations $(iv)$ and $(v)$ we get
$3x - x - 4y + 4y = 17 - 11$
$⇒ 2x = 6$
$\Rightarrow\text{x}=\frac{6}{2}=3$
Puuting $x = 3$ in equation $(iv)$ we get
$3 × 3 - 4y = 17$
$⇒ 9 - 4y = 17$
$⇒ -4y = 17 - 9$
$⇒ -4y = 8$
$\Rightarrow\text{y}=\frac{8}{-4}=-2$
Putting $x = 3$ and $y = -2$ in $z = -x + y + 4$ we get
$z = -3 - 2 + 4$
$⇒ z = -5 + 4$
$⇒ z = -1$
Hence solution of the giving system of equation is $x = 3, y = -2, z = -1.$

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Question 134 Marks

Solve the following system of equations graphically:
Shade the region between the lines and the $y-$axis.

$4x - y = 4,$

$3x + 2y = 14.$

Answer

The given system of equations is
$4x - y = 4$
$3x + 2y = 14$
Now, $4x - y = 4$
$⇒ 4x - 4 = y$
$⇒ y = 4x - 4$
When $x = 0,$ we have
$y = 4 × 0 - 4 = -4$
When $x = -1,$ we have
$y = 4 × (-1) - 4 = -8$
Thus, we have the following table.

$x$	$0$	$-1$
$y$	$-4$	$-8$

We have,
$3x + 2y = 14$
$⇒ 2y = 14 - 3x$
$\Rightarrow\text{y}=\frac{14-3\text{x}}{2}$
When $x = 0,$ we have
$\text{y}=\frac{14-3\times0}{2}=7$
When $x = 4,$ we have
$\text{y}=\frac{14-3\times4}{2}=1$
Thus, we have the following table.

$x$	$0$	$4$
$y$	$7$	$1$

Graph of the given system of equations

Clearly, the two lines intersect at $A(2, 4).$ Hence, $x = 2, y = 4$ is the solution of the given system of equations.
We also observe $\triangle\text{ABC}$ is the required shaded region.

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Question 144 Marks

Solve graphically the following system of linear equation. Also find the coordinates of the points where the lines meet axis of y.

$2x - y - 5 = 0,$

$x - y - 3 = 0.$

Answer

$2x - y - 5 = 0 .....(i)$
$x - y - 3 = 0 ..........(ii)$
The two points satisfying $(i)$ can be listed in a table as,

$x$	$1$	$3$
$y$	$-3$	$1$

The two points satisfying $(ii)$ can be listed in a table as,

$x$	$1$	$5$
$y$	$-2$	$2$

Now, graph of equations $(i)$ and $(ii)$ can be drawn as,

It is seen that the solution of the given system of equations is given by $x = 2, y = -1.$ Also, it is observed that the lines $(i)$ and $(ii)$ meet the y-axis at the points. $(0, -3)$ and $(0, -5)$ respectively.

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Question 154 Marks

Solve the following system of linear equation graphically and shade the region between the two lines and $x-$axis:

$3x + 2y -11 = 0,$

$2x - 3y + 10 = 0.$

Answer

The system of given equations are,
$3x + 2y - 11 = 0 ......(i)$
$2x - 3y + 10 = 0 ........(ii)$
From $(i)$, $\text{y}=\frac{11-3\text{x}}{2}$
Putting $x = 1,$ we get $y = 4$
Putting $x = 3,$ we get $y = 1$
Putting $x = 5,$ we get $y = -2$
Thus, plot $A(1, 4), B(3, 1)$ and $C(5, -2)$ on graph paper.
From $(ii), \text{y}=\frac{2\text{x}+10}{3}$
Putting $x = 1,$ we get $y = 4$
Putting $x = 4,$ we get $y = 6$
Putting $x = 7,$ we get $y = 8$
Thus, plot $P(1, 4), Q(4, 6)$ and $R(7, 8)$ on graph paper.

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Question 164 Marks

Solve the following system of linear equation graphically and shade the region between the two lines and x-axis:

$2x + 3y = 12,$

$x - y = 1.$

Answer

The system of given equations is,
$2x + 3y = 12$
$x - y = 1$
Now, $2x + 3y = 12$
$⇒ 2x = 12 - 3y$
When $y = 2,$ we have,
$\text{x}=\frac{12-3\times2}{2}=3$
When y = 4, we have,
$\text{x}=\frac{12-3\times4}{2}=0$
Thus, we have the following table,

$x$	$0$	$3$
$y$	$4$	$2$

We have, $x - y = 1$
$⇒ x = 1 + y$
When $y = 0,$ we have,
$x = 1$
When $y = 1,$ we have,
$x = 1 + 1 = 2$
Thus, we have the following table,

$x$	$1$	$2$
$y$	$0$	$1$

Graph of the given system of equations.

Clearly, the two lines intersect at $P(3, 2).$
Hence, $x = 3, y = 2$ is the solution of the given system of equations.

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Question 174 Marks

Solve the following systems of equations graphically:

$2x + 3y + 5 = 0$

$3x - 2y - 12 = 0$

Answer

The given equations are,
$2x + 3y + 5 = 0 .......(i)$
$3x - 2y - 12 = 0 ..........(ii)$
From $(i),$ $\text{y}=\frac{-5-2\text{x}}{3}\ .......(\text{iii})$
Putting $x = -1$ in $(iii)$, we get $y = -1$
Putting $x = 2$ in $(iii),$ we get $y = -3$
Putting $x = 5$ in $(iii),$ we get $y = -5$

$x$	$-1$	$2$	$5$
$y$	$-1$	$-3$	$-5$

From $(ii)$, $\text{y}=\frac{3\text{x}-12}{2}\ ......(\text{iv})$
Putting $x = 0$ in $(iv)$, we get $y = -6$
Putting $x = 2$ in $(iv),$ we get $y = -3$
Putting $x = 4$ in $(iv),$ we get $y = 0$

$x$	$0$	$2$	$4$
$y$	$-6$	$-3$	$0$

Clearly, from the above graph solution of above system of equations is $x = 2$ and $y = -3.$

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Question 184 Marks

A boat goes $12\ km$ upstream and $40\ km$ downstream in $8$ hours. I can go $16\ km$ upstream and $32\ km$ downstream in the same time. Find the speed of the boat in still water and the speed of the stream.

Answer

Let the speed of boat in still water and speed of stream be $x$ and $y\ km/ hr.$ respectively.
Speed of upstream $= (x - y)\ km/ hr.$
Speed of downstream $= (x + y)\ km/ hr.$
Time taken to cover $12\ km$. upstream $=\frac{12}{(\text{x}-\text{y})}\text{ hour.}$
Times taken to cover $40\ km.$ downstream $\frac{40}{(\text{x}+\text{y})}\text{ hour.}$
It takes 8 hours in journey so,
$\frac{12}{\text{x}-\text{y}}+\frac{40}{\text{x}+\text{y}}=8\ ....(\text{i})$
and, $\frac{16}{\text{x}-\text{y}}+\frac{32}{\text{x}+\text{y}}=8\ ....(\text{ii})$
let $\frac{1}{\text{x}-\text{y}}=\text{u}$ and $\frac{1}{\text{x}+\text{y}}=\text{v}$ then
$12u + 40v = 8$
$⇒ 3u + 10v = 2 ....(iii)$
$⇒ 16u + 32v = 8$
$⇒ 2u + 4v = 1 .....(iv)$
Multiplying $(iii)$ by $2$ and $(iv)$ by $3,$ we get
$⇒ 6u + 20v = 4 .....(v)$
$⇒ 6u + 12v = 3 .......(vi)$
Subtracting $(vi)$ from $(v)$ we get
$⇒ 8v = 1$
$\Rightarrow\text{v}=\frac{1}{8}$
Putting $\text{v}=\frac{1}{8}$ in $(iv)$ we get
$\Rightarrow2\text{u}+4\times\frac{1}{8}=1$
$\Rightarrow2\text{u}+\frac{1}{2}=1$
$\Rightarrow2\text{u}=\frac{1}{2}$
$\Rightarrow\text{u}=\frac{1}{4}$
$\because\frac{1}{\text{x}-\text{y}}=\text{u}$
$\text{x}-\text{y}=4\ ....(\text{vii})$
and $\frac{1}{\text{x}+\text{y}}=\text{v}$
$\text{x}+\text{y}=8\ .....(\text{viii})$
Adding $(vii)$ and $(viii)$ we get
$⇒ 2x = 12$
$⇒ x = 6\ km/ hr$.
Putting $x = 6$ in $(viii)$ we get
$⇒ 6 + y = 8$
$⇒ y = 2\ km/ hr.$
Thus, speed of boat in still water $= 6\ km/ hr.$ and speed of stream $= 2\ km/ hr.$

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Question 194 Marks

Solve the following systems of equations:
$\frac{15}{\text{u}}+\frac{2}{\text{v}}=17$
$\frac{1}{\text{u}}+\frac{1}{\text{v}}=\frac{36}{5}$

Answer

$\frac{15}{\text{u}}+\frac{2}{\text{v}}=17$
$\frac{1}{\text{u}}+\frac{1}{\text{v}}=\frac{36}{5}$
Let us consider $\frac{1}{\text{u}}=\text{x}$ and $\frac{1}{\text{v}}=\text{y}$
$15\text{x}+2\text{y}=17\ ...(\text{i})$
$\text{x}+\text{y}=\frac{36}{5}\ ...(\text{ii})$
Now multiplying equation $2^{nd}$ by $2$ and substract from $(i)$
$15\text{x}+2\text{x}=17-\frac{72}{5}$
$17\text{x}=\frac{85-72}{5}$
$\text{x}=\frac{13}{85}$
And $\text{y}=\frac{36}{5}-\frac{13}{85}$
$\Rightarrow\text{y}=\frac{612-13}{85}$
$\text{y}=\frac{599}{85}$
Thus $\text{u}=\frac{85}{13}$ and $\text{v}=\frac{85}{599}$

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Question 204 Marks

Abdul travelled $300\ km$ by train and $200\ km$ by taxi, it took him $5$ hours $30$ minutes. But if he travels $260\ km$ by train and $240\ km$ by taxi he takes $6$ minutes longer. Find the speed of the train and that of the taxi.

Answer

Let the speed of the train be $x\ km/ hr.$ and that of that of the taxi be $y\ km/ hr.$ we have the following cases:
Case I: When abdul travels $300\ km$ by train and $200\ km$ by taxi:
In this case we have
Time taken by abdul to travel $300\ km$ by train $=\frac{300}{\text{x}}\text{hrs.}$
Time taken by abdul to travel $200\ km$ by taxi $=\frac{200}{\text{y}}\text{hrs.}$
$\therefore$ Total time taken by abdul $=\frac{300}{\text{x}}+\frac{200}{\text{y}}$
It is given that the total time taken is $5$ hours $30$ minutes.
$\therefore\frac{300}{\text{x}}+\frac{200}{\text{y}}=5\text{ hours 30 minutes}$
$\Rightarrow\frac{300}{\text{x}}+\frac{200}{\text{y}}=5\frac{1}{2}$
$\Rightarrow\frac{300}{\text{x}}+\frac{200}{\text{y}}=\frac{11}{2}$
$\Rightarrow\frac{600}{\text{x}}+\frac{400}{\text{y}}=11\ ....(\text{i})$
Case II: When abdul travels $260\ km$ by train and $240\ km$ by taxi:
In this case we have
Time taken by abdul to travel $200\ km$ by train $=\frac{260}{\text{x}}\text{hrs.}$
Time taken by abdul to travel $240\ km$ by taxi $=\frac{240}{\text{y}}\text{hrs.}$
In this case total time of time of the journey is $(5$ hours $30$ minutes $+ 6$ minutes$)$
$=5\frac{1}{2}+\frac{1}{10}$
$=\frac{11}{2}+\frac{1}{10}$
$=\frac{55+1}{10}$
$=\frac{56}{10}$
$=\frac{28}{5}\text{hrs.}$
$\therefore\frac{260}{\text{x}}+\frac{240}{\text{y}}=\frac{28}{5}$
$\Rightarrow4\Big(\frac{65}{\text{x}}+\frac{60}{\text{y}}\Big)=\frac{28}{5}$
$\Rightarrow\frac{65}{\text{x}}+\frac{60}{\text{y}}=\frac{7}{5}$
$\Rightarrow\frac{65\times5}{\text{x}}+\frac{60\times5}{\text{y}}=7$
$\Rightarrow\frac{325}{\text{x}}+\frac{300}{\text{y}}=7\ .....(\text{ii})$
Putting $\frac{1}{\text{x}}=\text{u}$ and $\frac{1}{\text{y}}=\text{v}$ in equation $(i)$ and $(ii)$ we get
$600\text{u}+400\text{v}=11$
$\Rightarrow600\text{u}+400\text{v}-11=0\ .....(\text{iii})$
And, $325\text{u}+300\text{v}-7=0\ ....(\text{iv})$
By cross-multiplying we have
$\Rightarrow\frac{\text{u}}{400\times(-7)-(-11)\times300}=\frac{-\text{v}}{600\times(-7)-(-11)\times325}\\=\frac{1}{600\times300-400\times325}$
$\Rightarrow\frac{\text{u}}{-2800+3300}=\frac{-\text{v}}{-4200+3575}=\frac{1}{180000-130000}$
$\Rightarrow\frac{\text{u}}{500}=\frac{-\text{v}}{-625}=\frac{1}{50000}$
$\Rightarrow\frac{\text{u}}{500}=\frac{\text{v}}{625}=\frac{1}{50000}$
$\Rightarrow\frac{\text{u}}{500}=\frac{1}{50000}$ and $\frac{\text{v}}{625}=\frac{1}{50000}$
$\Rightarrow\text{u}=\frac{1}{100}$ and $\text{v}=\frac{1}{80}$
Now, $\text{u}=\frac{1}{100}$
$\Rightarrow\frac{1}{\text{x}}=\frac{1}{100}$
$\Rightarrow\text{x}=100$
And, $\text{v}=\frac{1}{80}$
$\Rightarrow\frac{1}{\text{y}}=\frac{1}{80}$
$\Rightarrow\text{y}=80$
Hence, speed of the train $= 100\ km/ hr.$ Speed of the taxi $= 80\ km/ hr.$

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Question 214 Marks

Roohi travels $300\ km$ to her home partly by train and partly by bus. She takes $4$ hours if she travels $60\ km$ by train and the remaining by bus. If she travels $100\ km$ by train and the remaining by bus, she takes $10$ minutes longer. Find the speed of the train and the bus separately.

Answer

Let the speed of the train be $x\ km/ hr$. and that of the bus be $y\ km/ hr$. we have the following cases:
Case I: When Roohi travels $60\ km$ by train and the rest by bus in this case, we have
Time taken by Roohi to travel $60\ km$ by train $=\frac{60}{\text{x}}\text{ hrs.}$
Time taken by Roohi to travel $(300 - 60) = 240\ km$ by bus $=\frac{240}{\text{y}}\text{ hrs.}$
$\therefore$ Total times taken by Roohi to cover 300km $=\frac{60}{\text{x}}+\frac{240}{\text{y}}$
It is given that the total times taken is $4$ hours.
$\therefore\frac{60}{\text{x}}+\frac{240}{\text{y}}=4$
$\Rightarrow4\Big[\frac{15}{\text{x}}+\frac{60}{\text{y}}\Big]=4$
$\Rightarrow\frac{15}{\text{x}}+\frac{60}{\text{y}}=1\ ....(\text{i})$
Case II: When Roohi travels $100\ km$ by train and the rest by bus in this case we have
Time taken by Roohi to travel $100\ km$ by train $=\frac{100}{\text{x}}\text{hrs.}$
Time taken by Roohi to travel $(300 - 100) = 200\ km$ by bus $=\frac{200}{\text{y}}\text{hrs.}$
In this case total time of the journey is $4$ hrs $10$ minutes
$\therefore\frac{100}{\text{x}}+\frac{200}{\text{y}}=4\text{ hrs.}10\text{ minutes}$
$\Rightarrow\frac{100}{\text{x}}+\frac{200}{\text{y}}=4\frac{1}{6}$
$\Rightarrow\frac{100}{\text{x}}+\frac{200}{\text{y}}=\frac{25}{6}$
$\Rightarrow25\Big(\frac{4}{\text{x}}+\frac{8}{\text{y}}\Big)=\frac{25}{6}$
$\Rightarrow\frac{4}{\text{x}}+\frac{8}{\text{y}}=\frac{1}{6}$
$\Rightarrow6\Big(\frac{4}{\text{x}}+\frac{8}{\text{y}}\Big)=1$
$\Rightarrow\frac{24}{\text{x}}+\frac{48}{\text{y}}=1\ ....(\text{ii})$
Putting $\frac{1}{\text{x}}=\text{u}$ and $\frac{1}{\text{y}}=\text{v},$ in equation $(i)$ and $(ii)$ we get
$15\text{u}+60\text{v}=1\ ....(\text{iii})$
$24\text{u}+48\text{v}=1\ ....(\text{iv})$
By cross-multiplication we have
$\Rightarrow\frac{\text{u}}{60\times(-1)-48\times(-1)}=\frac{-\text{v}}{15\times(-1)-24\times(-1)}\\=\frac{1}{15\times48-60\times24}$
$\Rightarrow\frac{\text{u}}{-60+48}=\frac{-\text{v}}{-15+24}=\frac{1}{720-1440}$
$\Rightarrow\frac{\text{u}}{-12}=\frac{-\text{v}}{9}=\frac{1}{-720}$
$\Rightarrow\frac{\text{u}}{-12}=\frac{1}{-720}$ and $\frac{-\text{v}}{9}=\frac{1}{-720}$
$\Rightarrow\text{u}=\frac{-12}{-720}$ and $\text{v}=\frac{-9}{-720}$
$\Rightarrow\text{u}=\frac{1}{60}$ and $\text{v}=\frac{1}{80}$
Now, $\text{u}=\frac{1}{60}$
$\Rightarrow\frac{1}{\text{x}}=\frac{1}{60}$
$\Rightarrow\text{x}=60\text{km/ hr.}$
And $\text{v}=\frac{1}{80}$
$\Rightarrow\frac{1}{\text{y}}=\frac{1}{80}$
$\Rightarrow\text{y}=80\text{km/ hr.}$
Hence, speed of the train $= 60\ km/ hr.$ speed of the car $= 80\ km/ hr.$

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Question 224 Marks

Solve the following system of equations by the method of cross-multiplication:

$\frac{\text{x}+\text{y}}{\text{xy}}=2,$

$\frac{\text{x}-\text{y}}{\text{xy}}=6.$

Answer

The given system of equations is

$\frac{\text{x}+\text{y}}{\text{xy}}=2$

$\Rightarrow\frac{\text{x}}{\text{xy}}+\frac{\text{y}}{\text{xy}}=2$

$\Rightarrow\frac{1}{\text{y}}+\frac{1}{\text{x}}=2$

$\Rightarrow\frac{1}{\text{x}}+\frac{1}{\text{y}}=2\ .....(\text{i})$

and $\frac{\text{x}-\text{y}}{\text{xy}}=6$

$\Rightarrow\frac{\text{x}}{\text{xy}}-\frac{\text{y}}{\text{xy}}=6$

$\Rightarrow\ \frac{1}{\text{y}}-\frac{1}{\text{x}}=6$

$\Rightarrow\ \frac{1}{\text{x}}-\frac{1}{\text{y}}=6\ .....(\text{ii})$

Taking $\text{u}=\frac{1}{\text{x}}$ and $\text{v}=\frac{1}{\text{y}},$ we get

$\text{u}+\text{v}=2$

$\Rightarrow\text{u}+\text{v}-2=0\ ...(\text{iii})$

and $\text{u}-\text{v}=-6$

$\Rightarrow\text{u}-\text{v}+6=0\ ....(\text{iv})$

Here, $\text{a}_1=1,\text{b}_1=1,\text{c}_1=-2$

$\text{a}_2=1,\text{b}_2=-1$ and $\text{c}_2=6$

By cross multiplication

$ \Rightarrow\ \frac{\text{u}}{1\times6-(-2)\times(-1)}=\frac{-\text{v}}{1\times6-(-2)\times1}\\=\frac{1}{1\times(-1)-1\times1}$

$\Rightarrow\ \frac{\text{u}}{6-2}=\frac{-\text{v}}{6+2}=\frac{1}{-1-1}$

$\Rightarrow\ \frac{\text{u}}{4}=\frac{-\text{v}}{8}=\frac{1}{-2}$

Now,

$\frac{\text{u}}{4}=\frac{1}{-2}$

$\Rightarrow\ \text{u}=\frac{4}{-2}=-2$

And, $\frac{-\text{v}}{8}=\frac{1}{-2}$

$\Rightarrow-\text{v}=\frac{8}{-2}=-4$

$\Rightarrow-\text{v}=-4$

$\Rightarrow\ \text{v}=4$

Now, $\text{x}=\frac{1}{\text{u}}=\frac{-1}{2}$ and $\text{y}=\frac{1}{\text{v}}=\frac{1}{4}$

Hence, $\text{x}=\frac{-1}{2},\ \text{y}=\frac{1}{4}$ is the solution of the given system of equations.

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Question 234 Marks

Solve the following system of linear equations graphically:

$3x + y - 11 = 0,$

$x - y - 1 = 0.$

Shade the region bounded by these lines and $y-$axis. Also, find the area of the region bounded by these lines and $y-$axis.

Answer

The given equations are,
$3x + y - 11 = 0 ......(i)$
$x - y - 1 = 0 ........(ii)$
Putting $x = 0$ in equations $(i),$ we get,
$⇒ 3 × 0 + y = 11$
$⇒ y = 11$
$⇒ x = 0, y = 11$
Putting $y = 0$ in equations $(i), $ we get,
$⇒ 3x + 0 = 11$
$\Rightarrow\text{x}=\frac{11}{3}$
$\Rightarrow\text{x}=\frac{11}{3},\ \text{y}=0$
Use the following table to draw the graph.

$x$	$0$	$\frac{11}{3}$
$y$	$11$	$0$

Draw the graph by plotting the two points from table.

$x - y = 1 ......(ii)$
Putting x = 0 in equation $(ii),$ we get,
$⇒ 0 - y = 1$
$⇒ y = -1$
$⇒ x = 0, y = -1$
Putting y = 0 in equation $(ii),$ we get,
$⇒ x - 0 = 1$
$⇒ x = 1$
$⇒ x = 1, y = 0$
Use the following table to draw the graph.

$x$	$0$	$1$
$y$	$-1$	$0$

Draw the graph by plotting the two points from table. The two lines intersect at $P(3, 2).$
Hence $x = 3, y = 2$ is the solution of the given equations. The area enclosed by the lines represented by the given equations and the y−axis is shaded region in the,
Now, Required area $=$ Area of shaded region
$⇒$ Required area $=$ Area of $PAC$
$⇒$ Required area $=\frac{1}{2} ($base $×$ height$)$
$⇒$ Required area $=\frac{1}{2}$ $(AC × PM)$
$⇒$ Required area $=\frac{1}{2}$ $(12 × 3)$ sq. units
$⇒$ Required area $=\frac{1}{2}$ (36)
$⇒$ Required area $= 18$
Hence the required area is $18$ sq. unit.

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Question 244 Marks

Solve the following systems of equations:
$\frac{1}{5\text{x}}+\frac{1}{6\text{y}}=12,$
$\frac{1}{3\text{x}}-\frac{3}{7\text{y}}=8,\text{x}\neq0,\text{y}\neq0.$

Answer

Taking $\frac{1}{\text{x}}=\text{u}$ and $\frac{1}{\text{y}}=\text{v},$ the given equations become
$\frac{\text{u}}{5}+\frac{\text{v}}{6}=12$
$\Rightarrow\frac{6\text{u}+5\text{v}}{30}=12$
$\Rightarrow6\text{u}+5\text{v}=360\ ....(\text{i})$
And, $\frac{\text{u}}{3}-\frac{3\text{v}}{7}=8$
$\Rightarrow\frac{7\text{u}-9\text{v}}{21}=8$
$\Rightarrow7\text{u}-9\text{v}=168\ .......(\text{ii})$
Let us eliminate $'v'$ from equation $(i)$ and $(ii)$, multiolying equation $(i)$ by $9$ and equation $(ii)$ by $5,$ we get
$54\text{u}+45\text{v}=3240\ .....(\text{iii})$
$35\text{u}-45\text{u}=840\ ......(\text{iv})$
Adding equation $(i)$ adding equation $(ii),$ we get
$54\text{u}+35\text{u}=3240+840$
$\Rightarrow89\text{u}=4080$
$\Rightarrow\text{u}=\frac{4080}{89}$
Putting $\text{u}=\frac{4080}{89}$ in equation $(i)$, we get
$6\times\frac{4080}{89}+5\text{v}=360$
$\Rightarrow\frac{24480}{89}+5\text{v}=360$
$\Rightarrow5\text{v}=360-\frac{24480}{89}$
$\Rightarrow5\text{v}=\frac{32040-24480}{89}$
$\Rightarrow5\text{v}=\frac{7560}{89}$
$\Rightarrow\text{v}=\frac{7560}{5\times89}$
$\Rightarrow\text{v}=\frac{1512}{89}$
Hence, $\text{x}=\frac{1}{\text{u}}=\frac{89}{4080}$ and $\text{y}=\frac{1}{\text{v}}=\frac{89}{1512}$
So, the solution of the given system of equation is $\text{x}=\frac{89}{4080},\ \text{y}=\frac{89}{1512}.$

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Question 254 Marks

Solve the following systems of equations:
$7\text{x}-\frac{2\text{y}}{\text{xy}}=5,$
$8\text{x}+\frac{7\text{y}}{\text{xy}}=15.$

Answer

$\frac{7\text{x}-2\text{y}}{\text{xy}}=5$
$\Rightarrow\ \frac{7\text{x}}{\text{xy}}-\frac{2\text{y}}{\text{xy}}=5$
$\Rightarrow\ \frac{7}{\text{y}}-\frac{2}{\text{x}}=5\dots(\text{i})$
$\frac{ 8\text{x}+7\text{y}}{\text{xy}}=15$
$\frac{8\text{x}+7\text{y}}{\text{xy}}=15$
$\Rightarrow\frac{8\text{x}}{\text{xy}}+\frac{7\text{y}}{\text{xy}}=15$
$\Rightarrow\ \frac{8}{\text{y}}+\frac{7}{\text{x}}=15\dots(\text{ii})$
Putting $\frac{1}{\text{x}} =\text{p}$ and $\frac{1}{\text{y}} =\text{q}$ in $(i)$ and $(ii)$ we get,
$7q - 2p = 5...(iii)$
$8q + 7p = 15 ...(iv)$
Multiplying equation $(iii)$ by $7$ and multiplying equation $(iv)$ by $2$ we get,
$49q - 14p = 35 ...(v)$
$16q + 14p = 30..(vi)$
Now. adding equation $(v)$ and $(vi)$ we get,
$49q - 14p + 16q + 14p = 35 + 30$
$\Rightarrow\ 65\text{q}=65$
$\Rightarrow\ \text{q}=1$
Putting the value of $q$ in equation $(iv)$
$8 + 7p = 15$
$7p = 15 - 8$
$\Rightarrow7\text{p}=7$
$\Rightarrow\ \text{p}=1$
Now,
$\text{p}=\frac{1}{\text{x}}=1$
$\Rightarrow\frac{1}{\text{x}}=1$
$\Rightarrow\text{x}=1$
also, $\text{q}=1 =\frac{1}{\text{y}}$
$\Rightarrow\ \frac{1}{\text{y}}=1$
$\Rightarrow\ \text{y}=1$
Hence, $x = 1$ and $y = 1$ is the solution.

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Question 264 Marks

Draw the graphs of $x - y + 1 = 0$ and $3x + 2y - 12 = 0$. Determine the coordinates of the vertices of the triangle formed by these lines and $x-$axis and shade the triangular area. Calculate the area bounded by these lines and $x-$axis.

Answer

The given system of equations is
$x - y + 1 = 0$
$3x + 2y - 12 = 0$
Now, $x - y + 1 = 0$
$⇒ x = y - 1$
When $y = 3,$ we have
$x = 3 - 1 = 2$
When $y = -1,$ we have
$x = -1 - 1 = -2$
Thus, we have the following table,

$x$	$2$	$-2$
$y$	$3$	$-1$

We have
$⇒ 3x + 2y - 12 = 0$
$⇒ 3x = 12 - 2y$
$\Rightarrow\text{x}=\frac{12-2\text{y}}{3}$
When $y = 6,$ we have
$\text{x}=\frac{12-2\times6}{3}=0$
When $y = 3,$ we have
$\text{x}=\frac{12-2\times3}{3}=2$
Thus, we have the following table,

$x$	$0$	$2$
$y$	$6$	$3$

Graph of the given system of equations.

Clearly, the two lines intersect at $A(2, 3).$ We also observe that lines represented by the equations. $x - y + 1 = 0$ and $3x + 2y - 12 = 0$ meet $x-$axis at $B(-1, 0)$ and $C(4, 0)$ respectively. Thus, $x = 2, y = 3$ is the solution of the given system of equations. Draw $AD$ per perpendicular from $A$ on $x-$ axis. Clearly, we have $AD = y-$coordinate of point $A(2, 3),$
$AD = 3$ and $BC = 4 - (-1) = 4 + 1 = 5.$

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Question 274 Marks

The ages of two friends Ani and Biju differ by $3$ years. Ani's father Dharma is twice as old as Ani and Biju as twice as old as his sister Cathy. The ages of Cathy and Dharam differ by $30$ years. Find the ages of Ani and Biju.

Answer

Let the present ages of Ani, Biju, Dharam and Cathy be $x, y, z$ and $t$ years respectively.
The ages of Ani and Biju differ by $3$ years. Thus, we have
$\text{x}-\text{y}=\pm3$
$\Rightarrow\text{x}=\text{y}\pm3$
Dharam is twice as old as Ani. Thus, we have $z = 2x$
Biju is twice as old as Cathy. Thus, we have $y = 2t$
The ages of Cathy and Dharam differ by $30$ years. Clearly, Dharam is older than Cathy.
Thus, we have $z - t = 30$
So, we have two systems of simultaneous equations
$x = y + 3 .....(i)$
$z = 2x$
$y = 2t$
$z - t = 30$
$x = y - 3 ......(ii)$
$z = 2x$
$y = 2t$
$z - t = 30$
Here $x, y, z$ and t are unknowns. We have to find the value of $x$ and $y.$
eq. $(i)$ by using the third equation, the first equation becomes $x = 2t + 3$
From the fourth equation, we have
$t = z - 30$
Hence, we have
$x = 2(z - 30) + 3$
$= 2z - 60 + 3$
$= 2z - 57$
Using the second equation, we have
$x = 2 × 2x - 57$
$⇒ x = 4x - 57$
$⇒ 4x - x = 57$
$⇒ 3x = 57$
$\Rightarrow\text{x}=\frac{57}{3}$
$⇒ x = 19$
From the first equation we have
$x = y + 3$
$⇒ y = x - 3$
$⇒ y = 19 - 3$
$⇒ y = 16$
Hence, the age of Ani is $19$ years and the age of Biju is $16$ years.
By using the third equation the first equation becomes $x = 2t - 3$
From the fourth equation we have
$t = z - 30$
Hence, we have
$x = 2(z - 30) - 3$
$= 2z - 60 - 3$
$= 2z - 60 - 3$
$= 2z - 63$
Using the second equation, we have
$x = 2 × 2x - 63$
$⇒ x = 4x - 63$
$⇒ 4x - x = 63$
$\Rightarrow\text{x}=\frac{63}{3}$
$⇒ x = 21$
From the first equation we have
$x = y - 3$
$⇒ y = x + 3$
$⇒ y = 21 + 3$
$⇒ y = 24$
Hence, the age of Ani is $21$ years and the age of Biju is $24$ years.
Note: that there are two possibilites.

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Question 284 Marks

Write a pair of linear equations which has the unique solution $x = -1, y = 3$. How many such pairs can you write$?$

Answer

Condition for the pair of system to have unique solution
$\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}$
Let the equations are,
$a_1 x+b_1 y+c_1=0$
and $a_2 x+b_2 y+c_2=0$
Since, $x=-1$ and $y=3$ is the unique solution of these two equations, then
$a_1(-1)+b_1(3)+c_1=0 $
$\Rightarrow-a_1+3 b_1+c_1=0$
$ \text { and } a_2(-1)+b_2(3)+c 2=0 $
$ \Rightarrow-a_2+3 b_2+c_2=0$

So, the different valume of $a_1, a_2, b_1, b_2, c_1$ and $c_2$ satisfy the eqs. $(i)$ and $(ii).$
Hence, infinitely many pairs of linear equations are possible.

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Question 294 Marks

Solve the following systems of equations:
$\frac{\text{xy}}{\text{x}+\text{y}}=\frac{6}{5}$
$\frac{\text{xy}}{\text{y}-\text{x}}=6$ where $\text{x}+\text{y}\neq0,\text{y}-\text{x}\neq0.$

Answer

The given system of equation is
$\frac{\text{xy}}{\text{x}+\text{y}}=\frac{6}{5}$
$\Rightarrow5\text{xy}=6(\text{x}+\text{y})$
$\Rightarrow5\text{xy}=6\text{x}+6\text{y}\ ......(\text{i})$
And, $\frac{\text{xy}}{\text{y}-\text{x}}=6$
$\Rightarrow\text{xy}=6(\text{y}-\text{x})$
$\Rightarrow\text{xy}=6\text{y}-6\text{x}\ .....(\text{ii})$
Adding equation $(i)$ and equation $(ii)$ we get
$6\text{xy}=6\text{y}+6\text{y}$
$\Rightarrow6\text{xy}=12\text{y}$
$\Rightarrow\text{x}=\frac{12\text{y}}{6\text{y}}=2$
Putting $x = 2$ in equation $(i)$ we get
$5\times2\times\text{y}=6\times2+6\text{y}$
$\Rightarrow10\text{y}=12+6\text{y}$
$\Rightarrow10\text{y}-6\text{y}=12$
$\Rightarrow4\text{y}=12$
$\Rightarrow\text{y}=\frac{12}{4}=3$
Hence solution of the given system of equations is $x = 2, y = 3.$

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Question 304 Marks

In the following system of equation determine whether the system has a unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it:

$x - 2y = 8$

$5x - 10y = 10$

Answer

Given
$x - 2y = 8$
$5x - 10y = 10$
To find: To determine whether the system has a uniqfue solution, no solution or infinitely many solutions
We know that the system of equations,
$a_1 x+b_1 y+c_1=0 $
$ a_2 x+b_2 y+c_2=0 $
For unique solution
$\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}$
For no solution
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$
For infinitely many solution
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
Here,
$\frac{1}{5}=\frac{-2}{-10}=\frac{8}{10}$
$\frac{1}{5}=\frac{1}{5}\neq\frac{4}{5}$
Since, $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$ which means $\frac{1}{5}=\frac{1}{5}\neq\frac{4}{5}$ hence the system of equation has no solution.
Hence the system of equation has no solution.

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Question 314 Marks

Find the value of k for which each of the following system of equations have infinitely many solutions:

$2x + 3y = k$

$(k - 1)x + (k + 2)y = 3k$

Answer

The given system of equations may be written as
$2x + 3y - k = 0$
$(k - 1)x + (k + 2)y - 3k = 0$
The system of equations is of the form
$ a_1 x+b_1 y+c_1=0$
$ a_2 x+b_2 y+c_2=0$
Where, $a_1=2, b_1=3, c_1=-k$
and, $a_2=k-1, b_2=k+2, c_2=-3 k$
For infinitely many solutions, we must have
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\Rightarrow\frac{2}{\text{k}-1}=\frac{3}{\text{k}+2}=\frac{-\text{k}}{-3\text{k}}$
$\Rightarrow\frac{2}{\text{k}-1}=\frac{3}{\text{k}+2}$ and $\frac{3}{\text{k}+2}=\frac{-\text{k}}{-3\text{k}}$
$\Rightarrow 2(k + 2) = 3(k - 1)$ and $3 \times 3k = k(k + 2)$
$\Rightarrow 2k + 4 = 3k - 3$ and $ 9 = k + 2$
$\Rightarrow 4 + 3 = 3k - 2k$ and $9 - 2 = k$
$\Rightarrow 7 = k$ and $7 = k$
$\Rightarrow k = 7$ and $k = 7$
$\Rightarrow k = 7$ satisfies both the conditions
Hence, the given system of equations will have infinitely many solutions, if $k = 7$

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Question 324 Marks

Find the values of a and b for which the following system of equations has infinitely many solutions:

$2x + 3y = 7$

$(a - 1)x + (a + 2)y = 3a$

Answer

The given system of equations is
$ 2 x+3 y-7=0 $
$ (a-1) x+(a+2) y-3 a=0$
It is of the form
$a_1 x+b_1 y+c_1=0$
$ a_2 x+b_2 y+c_2=0$
Where, $a_1=2, b_1=3, c_1=-7$
And, $a_2=(a-1), b_2=(a+2)$ and $c_2=-3 a$
The given system of equations will be have infinite number of solution, if
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\Rightarrow\frac{2}{\text{a}-1}=\frac{3}{\text{a}+2}=\frac{-7}{-3\text{a}}$
$\Rightarrow\frac{2}{\text{a}-1}=\frac{3}{\text{a}+2}=\frac{7}{3\text{a}}$
$\Rightarrow\frac{2}{\text{a}-1}=\frac{3}{\text{a}+2}$ and $\frac{3}{\text{a}+2}=\frac{7}{3\text{a}}$
$\Rightarrow 2(a + 2) = 3(a - 1)$ and $3 \times 3a = 7(a + 2)$
$\Rightarrow 2a + 4 = 3a - 3$ and $9a = 7a + 14$
$\Rightarrow 4 + 3 = 3a - 2a$ and $9a - 7a = 14$
$\Rightarrow 7 = a$ and $2a = 14$
$\Rightarrow a = 7$ and $a = 7$
$\Rightarrow a = 7$
Hence, the given system of equations will have infinitely many solutions, if $a = 7$

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Question 334 Marks

Solve the following systems of equations graphically:

$x - 2y = 5$

$2x + 3y = 10$

Answer

The given equations are:
$x - 2y = 5 ......(i)$
$2x + 3y = 10 .....(ii)$
Puting $x = 0$ in equation $(i)$, we get,
$⇒ 0 - 2y = 5$
$\Rightarrow\text{y}=\frac{-5}{2}$
$\Rightarrow\text{x}=0,\ \text{y}=\frac{-5}{2}$
Puting $y = 0$ in equation $(i)$, we get,
$⇒ x + 2 × 0 = 5$
$⇒ x = 5$
$⇒ x = 5, y = 0$
Use the following table to draw the graph.

$x$	$0$	$5$
$y$	$\frac{-5}{2}$	$0$

Draw the graph by plotting the two points $\text{A}\Big(0,\frac{-5}{2}\Big)$ and $B(5, 0)$ from table.

Graph the equation$ (ii),$
$⇒ 2x + 3y = 10 ......(ii)$
Putting $x = 0 $in equation $(ii)$, we get,
$⇒ 2 × 0 + 3y = 10$
$\Rightarrow\text{y}=\frac{10}{3}$
$\Rightarrow\text{x}=0,\ \text{y}=\frac{10}{3}$
Putting $y = 0$ in equation $(ii),$ we get,
$⇒ 2x + 3 × 0 = 10$
$⇒ x = 5$
$x = 5, y = 0$
use the following table to draw the graph.

$x$	$0$	$5$
$y$	$\frac{10}{3}$	$0$

Draw the graph by plotting the two points $\text{C}\Big(0,\frac{10}{3}\Big)$ and $B(5, 0)$ from table.
The two lines intersects at points $B(5, 0).$
Hence $x = 5, y = 0$ is the solution.

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Question 344 Marks

Solve the following system of equations by the method of cross-multiplication:
$\text{x}\Big(\text{a}-\text{b}+\frac{\text{ab}}{\text{a}-\text{b}}\Big)=\text{y}\Big(\text{a}-\text{b}-\frac{\text{ab}}{\text{a}+\text{b}}\Big)$
$\text{x}+\text{y}=2\text{a}^2$

Answer

The given equations are,
$\text{x}\Big(\text{a}-\text{b}+\frac{\text{ab}}{\text{a}-\text{b}}\Big)-\text{y}\Big(\text{a}-\text{b}-\frac{\text{ab}}{\text{a}+\text{b}}\Big)=0\ .....(\text{i})$
$\text{x}+\text{y}=2\text{a}^2\ .....(\text{ii})$
By cross-multiplication we get,
$\Rightarrow\frac{\text{x}}{-\Big(\text{a}+\text{b}-\frac{\text{ab}}{\text{a}+\text{b}}\Big)\big(-2\text{a}^2\big)-0}=\frac{\text{y}}{0-\Big(\text{a}-\text{b}+\frac{\text{ab}}{\text{a}-\text{b}}\Big)\big(-2\text{a}^2\big)}\\=\frac{1}{\Big(\text{a}-\text{b}+\frac{\text{ab}}{\text{a}-\text{b}}\Big)+\Big(\text{a}+\text{b}-\frac{\text{ab}}{\text{a}+\text{b}}\Big)}$
$\Rightarrow\frac{\text{x}}{\bigg\{\frac{(\text{a}+\text{b})^2-\text{ab}}{(\text{a}+\text{b})}\bigg\}2\text{a}^2}=\frac{\text{y}}{\bigg\{\frac{(\text{a}-\text{b})^2+\text{ab}}{\text{a}-\text{b}}\bigg\}2\text{a}^2}\\=\frac{1}{\frac{\big\{(\text{a}-\text{b})^2+\text{ab}\big\}}{(\text{a}-\text{b})}+\frac{\big\{(\text{a}+\text{b}^2)-\text{ab}\big\}}{(\text{a}+\text{b})}}$
$\Rightarrow\frac{\text{x}}{\frac{(\text{a}^2+\text{b}^2+2\text{ab}-\text{ab})2\text{a}^2}{(\text{a}+\text{b})}}=\frac{\text{y}}{\frac{(\text{a}^2+\text{b}^2-2\text{ab}+\text{ab})2\text{a}^2}{(\text{a}-\text{b})}}\\=\frac{1}{\frac{(\text{a}^2+\text{b}^2-2\text{ab}+\text{ab})}{(\text{a}-\text{b})}+\frac{(\text{a}^2+\text{b}^2+2\text{ab}-\text{ab})}{(\text{a}+\text{b})}}$
$\Rightarrow\frac{\text{x}}{\frac{(\text{a}^2+\text{b}^2+\text{ab})2\text{a}^2}{(\text{a}+\text{b})}}=\frac{\text{y}}{\frac{(\text{a}^2+\text{b}^2-\text{ab})2\text{a}^2}{(\text{a}-\text{b})}}\\=\frac{1}{\frac{(\text{a}^2+\text{b}^2-\text{ab})}{(\text{a}-\text{b})}+\frac{(\text{a}^2+\text{b}^2+\text{ab})}{(\text{a}+\text{b})}}$
$\Rightarrow \frac{\text{x}}{\frac{(\text{a}-\text{b})(\text{a}^2+\text{b}^2+\text{ab})2\text{a}^2}{(\text{a}+\text{b})(\text{a}-\text{b})}}=\frac{\text{y}}{\frac{(\text{a}-\text{b})(\text{a}^2-\text{ab}+\text{b}^2)2\text{a}^2}{(\text{a}-\text{b})(\text{a}+\text{b})}}\\=\frac{(\text{a}-\text{b})(\text{a}+\text{b})}{\text{a}^3+\text{ab}^2-\text{a}^2\text{b}+\text{a}^2\text{b}+\text{b}^3-\text{ab}^2+\text{a}^2+\text{ab}^2+\text{a}^2\text{b}-\text{a}^2\text{b}-\text{b}^3-\text{ab}^2}$
$ \frac{\text{x}}{\frac{(\text{a}^3-\text{b}^3)2\text{a}^2}{\text{a}^2-\text{b}^2}}=\frac{\text{y}}{\frac{(\text{a}^3+\text{b}^3)2\text{a}^2}{\text{a}^2-\text{b}^2}}=\frac{\text{a}^2-\text{b}^2}{2\text{a}^3}$
$\text{x}=\frac{(\text{a}^2-\text{b}^2)}{2\text{a}^3}\times\frac{(\text{a}^3-\text{b}^3)2\text{a}^2}{(\text{a}^2-\text{b}^2)}$
$ \text{x}=\frac{\text{a}^3-\text{b}^3}{\text{a}}$
$\text{y}=\frac{\text{a}^2-\text{b}^2}{2\text{a}^3}\times\frac{(\text{a}^3+\text{b}^3)2\text{a}^2}{(\text{a}^2-\text{b}^2)}$
$\text{y}=\frac{\text{a}^3+\text{b}^3}{\text{a}}$
Thus, $\text{x}=\frac{\text{a}^3-\text{b}^3}{\text{a}}$ and $\text{y}=\frac{\text{a}^3+\text{b}^3}{\text{a}}$

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Question 354 Marks

Solve the following systems of equations:

$x - y + z = 4,$

$x + y + z = 2,$

$2x + y - 3z = 0.$

Answer

We have,
$x - y + z = 4 ........(i)$
$x + y + z = 2 .......(ii)$
$2x + y - 3z = 0 .........(iii)$
From equation $(i)$ we get
$z = 4 - x + y$
$⇒ z = -x + y + 4$
Substituting $z = -x + y + 4 $in equation $(ii)$ we get
$x + y + (-x + y + 4) = 2$
$⇒ x + y - x + y + 4 = 2$
$⇒ 2y + 4 = 2$
$⇒ 2y = 2 - 4 = -2$
$⇒ 2y = -2$
$\Rightarrow\text{y}=\frac{-2}{2}=-1$
Substituting the value of $z$ in equation $(iii)$ we get
$2x + y - 3(-x + y + 4) = 0$
$⇒ 2x + y + 3x - 3y - 12 = 0$
$⇒ 5x - 2y - 12 = 0$
$⇒ 5x - 2y = 12 .......(iv)$
Putting $y = -1$ in equation $(iv)$ we get
$5x - 2 × (-1) = 12$
$⇒ 5x + 2 = 12$
$⇒ 5x = 12 - 2 = 10$
$\Rightarrow\text{x}=\frac{10}{5}=2$
Puuting $x = 2 $and $y = -1$ in $z = -x + y + 4 $we get
$z = -2 + (-1) + 4$
$= -2 - 1 + 4$
$= -3 + 4$
$= 1$
Hence solution of the giving system of equation is $x = 2, y = -1, z = 1.$

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Question 364 Marks

Solve graphically that the following system of equation has infinitely many solutions:

$3x + y = 8$

$6x + 2y = 16$

Answer

The given equations are,
$3x + y = 8 .......(i)$
$6x + 2y = 16 ........(ii)$
From $(i), y = 8 - 3x .......(iii)$
Putting $x = 0$ in $(iii)$, we get $y = 8$
Putting $x = 1$ in $(iii)$, we get $y = 5$
Putting $x = 2$ in $(iii)$, we get $y = 2$

$x$	$0$	$1$	$2$
$y$	$8$	$5$	$2$

From $(ii)$, $\text{y}=\frac{16-6\text{x}}{2}\ .....(\text{iv})$
Putting $x = 1$ in $(iv)$, we get $y = 5$
Putting $x = 2$ in $(iv)$, we get $y = 2$
Putting $x = 3$ in $(iv)$, we get $y = -1$

$x$	$1$	$2$	$3$
$y$	$5$	$2$	$-1$

When we polt these points on graph paper we observe that all then points on a line so the given system of equations has infinitaly many solutions.

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Question 374 Marks

Form the pair of linear equations in the following problems, and find their solution graphically:
$5$ pencils and $7$ pens together cost Rs. $50$, whereas $7$ pencils and $5$ pens together cost Rs. $46.$ Find the cost of one pencil and a pen.

Answer

Let the number of pencils and pens be $x$ and $y$ respectively.
According to questions.
$5x + 7y = 50 .......(i)$
$7x + 5y = 46 ........(ii)$
From (i), $\text{y}=\frac{50-5\text{x}}{7}\ ......(\text{iii})$
Putting $x = 3$ in $(iii)$, we get $Y = 5$
Putting $x = -4$ in $(iii)$, we get $Y = 10$

$x$	$3$	$-4$
$Y$	$5$	$10$

From $(ii)$, $\text{y}=\frac{46-7\text{x}}{5}\ ......(\text{iv})$
Putting $x = 3$ in $(iv)$, we get $y = 5$
Putting $x = -2$ in $(iv)$, we get $y = 12$

$x$	$3$	$-2$
$y$	$5$	$12$

Thus, from graph cost of pencil $= Rs. 3$ and cost of pen$= Rs. 5.$

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Question 384 Marks

The area of a rectangle remains the same if the length is increased by $7$ meters and the breadth is decreased by $3$ meters. The area remains unaffected if the length is decreased by $7$ meters and breadth in increased by $5$ meters. Find the dimensions of the rectangle.

Answer

Let the length and breadth of the rectangle be xm and ym respectively. Then,
Area $= xym^2$
If length is increased by $7\ m$ and the breadth is decreased by $3\ m$, the area rem ains same
$\therefore xy = (x + 7)(y - 3)$
$\Rightarrow xy = xy - 3x + 7y - 21$
$\Rightarrow 3x - 7y = -21 ......(i)$
When length is decreased by $7\ m$ and breasth is increased by $5m$, then area rem ains unaffected
$\therefore xy = (x - 7)(y + 5)$
$\Rightarrow xy = xy + 5x - 7y - 35$
$\Rightarrow 35 = 5x - 7y$
$\Rightarrow 5x - 7y = 35 .....(ii)$
Subtracting equation $(i)$ from $(ii)$, we get
$5x - 3x = 35 - (-21)$
$\Rightarrow 2x = 35 + 21$
$\Rightarrow\text{x}=\frac{56}{2}=28$
Putting $x = 28$ inj equation $(ii)$ we get
$5 \times 28 - 7y = 35$
$\Rightarrow 140 - 7y = 35$
$\Rightarrow -7y = 35 - 140$
$\Rightarrow -7y = -105$
$\Rightarrow\text{y}=\frac{105}{7}=15$
Hence, length and breadth of the rectangle are $28\ m$ and $15\ m$ respectively.

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Question 394 Marks

Solve graphically each of the following systems of linear equations. Also, find the coordinates of the points where the lines meet the axis of x in each system.

$x + 2y = 5,$

$2x - 3y = -4.$

Answer

$x + 2y = 5 ......(i)$
$2x - 3y = -4 ......(ii)$
$x + 2y = 8 .....(i)$
$⇒ x = 5 - 2y$
Substituting some different values of $x$, we get corresponding values of $y$ as shown below.

$x$	$3$	$1$	$-1$
$y$	$1$	$2$	$3$

Now plot the points on the graph and join them similarly in equation
$2x - 3y = 4$
$⇒ 2x = 3y - 4$
$\text{x}=\frac{3\text{y}-4}{2}$

$x$	$-2$	$1$	$4$
$y$	$0$	$2$	$4$

polt these points and join them we see that these two lines intersect each other at $(1, 2) x = 1, y = 2$. and these two lines also meet x-axis at $(1, 0)$ and $(2, 0)$ respectively as shown in the,

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Question 404 Marks

$A$ and $B$ each has some money. If $A$ gives $R s .30$ to $B$, then $B$ will have twice the money left with $A$. But, if $B$ gives $Rs. 10$ to $A$, then $A$ will have thrice as much as is left with $B$. How much money does each have?

Answer

Let the money with $A$ be Rs. $x$ and the money with $B$ be Rs. $y$
It is given that if A give Rs. $30$ to $B$. Then $B$ will have twice the money left with A. Then According to the question we have
$\Rightarrow y + 30 = 2(x - 30)$
$\Rightarrow y + 30 = 2x - 60$
$\Rightarrow y - 2x = -60 - 30$
$\Rightarrow -2x + y + 90 = 0 .....(i)$
It is also given that if $B$ gives Rs. $10$ to $A$ then a will have thric as much as is left with $B$
$\Rightarrow x + 10 = 3(y - 10)$
$\Rightarrow x + 10 = 3y - 30$
$\Rightarrow x - 3y + 10 + 30 = 0$
$\Rightarrow x - 3y + 40 = 0 .......(ii)$
Multiplying eq. $(ii)$ by $2$ and we get
$\Rightarrow 2x - 6y + 80 = 0 .....(iii)$
Now, adding eq $(i)$ and $(iii)$ and we get
$\Rightarrow -2x + y + 90 + 2x - 6y + 80 = 0$
$\Rightarrow -5y + 170 = 0$
$\Rightarrow -5y = -170$
$\Rightarrow\text{y}=\frac{170}{5}$
$\Rightarrow y = 34$
Now, Putting the value os y in eq. $(i)$
$\Rightarrow -2x + y + 90 = 0$
$\Rightarrow -2x + 34 + 90 = 0$
$\Rightarrow -2x + 124 = 0$
$\Rightarrow -2x = -124$
$\Rightarrow\text{x}=\frac{124}{2}$
$\Rightarrow x = 62$
Hence, the money with A be Rs. $62$ and money with $B$ be Rs. $34$

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Question 414 Marks

One says, "Give me a hundred, friend! I shall then become twice as rich as you." The other replies, "If you give me ten, I shall be six times as rich as you."Tell me what is the amount of their respective capital?

Answer

Let the money with the first person and second person be Rs. $x$ and Rs. $y$ respectively.
According to the question,
$x + 100 = 2(y - 100)$
$x + 100 = 2y - 200$
$x - 2y = -300 .....(i)$
$6(x - 10) = (y + 10)$
$6x - 60 = y + 10$
$6x - y = 70 ......(ii)$
Multiplying equation $(ii)$ by $2$ we obtain
$12x - 2y = 140 ......(iii)$
Subtracting equation $(i)$ from equation $(iii)$ we obtain
$11x = 140 + 300$
$11x = 440$
$x = 40$
Putting the value of $x$ in equation $(i)$ we obtain
$40 - 2y = -300$
$40 + 300 = 2y$
$2y = 340$
$y = 170$
Thus, the two friends had Rs. $40$ and Rs. $170$ with them.

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Question 424 Marks

In a rectangle, if the length is increased by $3$ meters and breadth is decreased by $4$ meters, the area of the rectangle is reduced by $67$ square meters. If length is reduced by $1$ meter and breadth is increased by $4$ meters, the area is increased by $89 $Sq. meters. Find the dimensions of the rectangle.

Answer

Let the length and breadth of the rectangle be $x$ and $y$units respectively.
Then, area of rectangle $= xy$ square units
If the length is increased by $3$ meters and breath is reduced each by $4$ square meters the area is reduced by $67$ square units
$\therefore xy - 67 = (x + 3)(y - 4)$
$xy - 67 = xy + 3y - 4x - 12$
$xy - 67 = xy + 3y - 4x - 12$
$4x - 3y - 67 + 12 = 0$
$4x - 3y - 55 = 0 ......(i)$
Then the length is reduced by $1$ meter and breadth is increased by $4$ meter then the area is increased by $89$ square units
$xy + 89 = (x - 1)(y + 4)$
$xy + 89 = xy + 4x - y - 4$
$4x - y - 4 - 89 = 0$
$4x - y - 93 = 0 ......(ii)$
Thus we get the following system of linear equation
$4x - 3y - 55 = 0$
$4x - y - 93 = 0$
By using cross multiplication we have
$\Rightarrow\frac{\text{x}}{(-3\times-93)-(-1\times-55)}=\frac{\text{-y}}{(4\times-93)-(4\times-55)}$
$=\frac{1}{(4\times-1)-(4\times-3)}$
$\Rightarrow\frac{\text{x}}{279-55}=\frac{\text{-y}}{-372+220}=\frac{1}{-4+12}$
$\Rightarrow\frac{\text{x}}{224}=\frac{-\text{y}}{-152}=\frac{1}{8}$
$\text{x}=\frac{224}{8}$
$\text{x}=28$
and,
$\text{y}=\frac{152}{8}$
$\text{y}=19$
Hence the length of rectangle is $28$ meter, The breath of rectangle is $19$ meter.

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Question 434 Marks

Solve the following system of equations by the method of cross-multiplication:
$\text{ax}+\text{by}=\frac{\text{a}+\text{b}}{2}$
$3\text{x}+5\text{y}=4$

Answer

Given,
$\text{ax}+\text{by}=\frac{\text{a}+\text{b}}{2}$
$3\text{x}+5\text{y}=4$
To find: The solution of the systems of equation by the method of cross-multiplication:
Here we have the pair of simultaneous equation,
$\text{ax}+\text{by}-\frac{\text{a}+\text{b}}{2}=0$
$3\text{x}+5\text{y}-4=0$
By cross multiplication method we get,
$\Rightarrow\frac{\text{x}}{(-4\text{b})-\Big(-\frac{5(\text{a}+\text{b})}{2}\Big)}=\frac{-\text{y}}{(-4\text{a})-\Big(-\frac{3(\text{a}+\text{b})}{2}\Big)}\\=\frac{1}{5\text{a}-3\text{b}}$
$\Rightarrow\frac{\text{x}}{(-4\text{b})+\frac{5(\text{a}+\text{b})}{2}}=\frac{-\text{y}}{(-4\text{a})+\frac{3(\text{a}+\text{b})}{2}}=\frac{1}{5\text{a}-3\text{b}}$
$\Rightarrow\frac{\text{x}}{(-4\text{b})+\frac{5(\text{a}+\text{b})}{2}}=\frac{1}{5\text{a}-3\text{b}}$
$\Rightarrow\text{x}(5\text{a}-3\text{b})=-4\text{b}+\frac{5(\text{a}+\text{b})}{2}$
$\Rightarrow\text{x}(5\text{a}-3\text{b})=\frac{5\text{a}-3\text{b}}{2}$
$\Rightarrow\text{x}=\frac{1}{2}$
And, $\frac{-\text{y}}{(-4\text{a})+\frac{3(\text{a}+\text{b})}{2}}=\frac{1}{5\text{a}-3\text{b}}$
$\Rightarrow\frac{-\text{y}}{\frac{-8\text{a}+3\text{a}+3\text{b}}{2}}=\frac{1}{5\text{a}-3\text{b}}$
$\Rightarrow\frac{-\text{y}}{\frac{-5\text{a}+3\text{b}}{2}}=\frac{1}{5\text{a}-3\text{b}}$
$ \Rightarrow\text{y}(5\text{a}-3\text{b})=\frac{5\text{a}-3\text{b}}{2}$
$\Rightarrow\text{y}=\frac{1}{2}$
Hence we get the value of $\text{x}=\frac{1}{2}$ and $\text{y}=\frac{1}{2}$

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Question 444 Marks

Solve the following systems of equations:
$\frac{2}{\text{x}}+\frac{3}{\text{y}}=\frac{9}{\text{xy}},$
$\frac{4}{\text{x}}+\frac{9}{\text{y}}=\frac{21}{\text{xy}},\text{x}\neq0,\text{y}\neq0.$

Answer

The system og given equation is
$\frac{2}{\text{x}}+\frac{3}{\text{y}}=\frac{9}{\text{xy}}\ .......(\text{i})$
$\frac{4}{\text{x}}+\frac{9}{\text{y}}=\frac{21}{\text{xy}},\text{where}\text{ x}\neq0,\text{y}\neq0\ .......(\text{ii}) $
Multiplying equation (i) and equation (ii) by xy we get
$2\text{y}+3\text{x}=9\ .....(\text{iii})$
$4\text{y}+9\text{x}=21\ ......(\text{iv})$
From (iii) we get
$3\text{x}=9-2\text{y}$
$\Rightarrow\text{x}=\frac{9-2\text{y}}{3}$
Substituting $\text{x}=\frac{9-2\text{y}}{3}$ in equation (iv) we get
$4\text{x}+9\Big(\frac{9-2\text{y}}{3}\Big)=21$
$\Rightarrow4\text{y}+3(9-2\text{y})=21$
$\Rightarrow4\text{y}+27-6\text{y}=21$
$\Rightarrow-2\text{y}=21-27$
$\Rightarrow-2\text{y}=-6$
$\Rightarrow\text{y}=\frac{-6}{-2}=3$
Putting y = 3 in $\text{x}=\frac{9-2\text{y}}{3},$ we get
$\text{x}=\frac{9-2\times3}{3}$
$=\frac{9-6}{3}$
$=\frac{3}{3}=1$
Hence, solution of the system of equation$ x = 1, y = 3.$

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Question 454 Marks

Places $A$ and $B$ are $100 \ km$ apart on a highway. One car starts from $A$ and another from $B$ at the same time. If the cars travel in the same direction at different speeds, they meet in $5$ hours. If they travel towards each other, they meet in $1$ hour. What are the speeds of two cars.

Answer

Let $x km / hr$ and $y km / hr$ be the speed of cars $A$ and $B$ respectively. When they move in same direction, they meet in 5 hours.
Suppose they meet at C. Then,
Distance cover by $A = AC$
Distance cover by $B=B C$
Distance $=$ Speed $\times$ time
Distance cover by $A=5 x km$
Distance cover by $B =5 y km$
Clearly, $A C-B C=A B$
$5 x-5 y=100$
$x-y=20 \ldots \text { (i) }$
When they move towards each other they meet in one hours.
Distance cover by $A = AD = xkm$
Distance cover by $B=B D=y km$
Cleary, $A D+B D=A B$
$\Rightarrow x+y=100$
Adding (i) and (ii) we get
$\Rightarrow 2 x=120$
$\Rightarrow x=60 km / hr$
Putting $x=60$ in (ii) we get
$\Rightarrow 60+y=100$
$\Rightarrow y=40 km / hr$
Thus, speed of $A=60 km / hr$. and Speed of $B=40 km / hr$.

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Question 464 Marks

$5$ pens and $6$ pencils together cost Rs. $9$ and $3$ pens and $2$ pencils cost $Rs. 5$. Find the cost of $1$ pen and $1$ pencil.

Answer

Let the coist of a pen be Rs. $x$ and that of a pencil be Rs. $y$. Then,
$5x + 6y = 9 .....(i)$
and $3x + 2y = 5 .....(ii)$
Multiplying equations $(i)$ by $2$ and equation $(ii)$ by $6,$ we get
$10x + 12y = 18 .....(ii)$
$18x + 12y = 30.......(iv)$
Subtracting equation $(iii)$ by equation $(iv)$ we get
$18x - 10x + 12y - 12y = 30 - 18$
$\Rightarrow 8x = 12$
$\Rightarrow\text{x}=\frac{12}{8}=\frac{3}{2}=1.5$
Substituting $x = 1.5$ in equation $(i)$ we get
$5 \times 1.5 + 6y = 9$
$\Rightarrow 7.5 + 6y = 9$
$\Rightarrow 6y = 9 - 7.5$
$\Rightarrow 6y = 1.5$
$\Rightarrow\text{y}=\frac{1.5}{6}=\frac{1}{4}=0.25$
Hence, cost of one pen $= Rs. 1.50$ and cost of one pencil $= Rs.0.25$

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Question 474 Marks

Solve the following system of equations by the method of cross-multiplication:

$mx - ny = m^2 + n^2$

$x + y = 2m$

Answer

Given,

$mx - ny = m^2 + n^2$

$x + y = 2m$

To find: The solution of the systems of equation by the method of cross-multiplication:
Here we have the pair of simultaneous equation
$mx - ny - m^2 + n^2) = 0$
$x + y - 2m = 0$
By cross multiplication method we get,
$\Rightarrow\frac{\text{x}}{(-2\text{m})(-\text{n})-[-(\text{m}^2+\text{n}^2)]}=\frac{-\text{y}}{(-2\text{m})(\text{m})-[-(\text{m}^2+\text{n}^2)]}\\=\frac{1}{\text{m + n}}$
$\Rightarrow\frac{\text{x}}{(\text{m}+\text{n})^2}=\frac{-\text{y}}{(-2\text{m}^2)+(\text{m}^2+\text{n}^2)}=\frac{1}{\text{m + n}}$
$\Rightarrow \frac{\text{x}}{(\text{m}+\text{n})^2}=\frac{1}{\text{m + n}}$
$\Rightarrow\text{x}=\text{m + n}$
Now for y,
$\Rightarrow\frac{-\text{y}}{(-2\text{m}^2)+(\text{m}^2+\text{n}^2)}=\frac{1}{\text{m + n}}$
$\Rightarrow \frac{\text{y}}{(\text{m}^2-\text{n}^2)}=\frac{1}{\text{m + n}}$
$\Rightarrow\frac{\text{y}}{(\text{m}-\text{n})(\text{m}+\text{n})}=\frac{1}{\text{m + n}}$
Hence we get the value of $x = m + n$ and $y = m - n.$

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Question 484 Marks

Solve the following systems of equations:

$x + y = 5xy,$

$3x + 2y = 13xy$, $\text{x}\neq0,\text{y}\neq0.$

Answer

The given system of equation is
$x + y= 5xy ......(i)$
$3x + 2y = 13xy .......(ii)$
Multiplying equation $(i)$ by$ 2$ and equation $(ii)$ by $1$, we get
$2x + 2y = 10xy .......(iii)$
$3x + 2y = 13xy .........(iv)$
Subtracting equation $(iii)$ from equation $(iv)$ we get
$3x - 2x = 13xy - 10xy$
$\Rightarrow x = 3xy$
$\Rightarrow\frac{\text{x}}{3\text{x}}=\text{y}$
$\Rightarrow\text{y}=\frac{1}{3}$
Putting $\text{y}=\frac{1}{3}$ in equation (i) we get
$\text{x}+\frac{1}{3}=5\text{x}\times\text{x}\times\frac{1}{3}$
$\text{x}+\frac{1}{3}=\frac{5\text{x}}{3}$
$\Rightarrow\frac{1}{3}=\frac{5\text{x}}{3}-\text{x}$
$\Rightarrow\frac{1}{3}=\frac{5\text{x}-3\text{x}}{3}$
$\Rightarrow1=2\text{x}$
$\Rightarrow2\text{x}=1$
$\Rightarrow\text{x}=\frac{1}{2}$
Hence, solution of the given system of equation is $\text{x}=\frac{1}{2},\text{y}=\frac{1}{3}.$

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Question 494 Marks

Solve graphically the following system of linear equation. Also find the coordinates of the points where the lines meet axis of $y$.

$3x + y - 5 = 0,$

$2x - y - 5 = 0.$

Answer

The given equations are
$3x + y - 5 = 0 ........(i)$
$2x - y - 5 = 0 ...........(ii)$
From $(i), y = 5 - 3x ........(iii)$
Putting $x = 0$ in $(iii)$, we get $y = 5$
Putting $x = 1$ in $(iii)$, we get $y = 2$
Putting $x = 2$ in $(iii)$, we get $y = -1$

$x$	$0$	$1$	$2$
$y$	$5$	$2$	$-1$

From $(ii), y = 2x - 5 ......(iv)$
Putting $x = 0$ in $(iv)$, we get $y = -5$
Putting $x = 1$ in $(iv)$, we get $y = -3$
Putting $x = 2$ in $(iv)$, we get $y = -1$

$x$	$0$	$1$	$2$
$y$	$-5$	$-3$	$-1$

When we solve yhese equations $x =2$ and $y =-1$ this is the point where these lines intersect each other. $(0,5)$ and $(0$, $-5)$ are the points where lines meeet axis of $y$.

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Question 504 Marks

In the following system of equation determine whether the system has a unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it:

$x - 3y = 3$

$3x - 9y = 2$

Answer

The given system of equations may be written as,
$x - 3y - 3 = 0$
$3x - 9y - 2 = 0$
The given system of equations is of the form,
$ a_1 x+b_1 y+c_1=0$
$ a_2 x+b_2 y+c_2=0$
Where, $a_1=1, b_1=-3, c_1=-3$
And $a_2=3, b_2=-9, c_2=-2$
We have,
$\frac{\text{a}_1}{\text{a}_2}=\frac{1}{3}$
$\frac{\text{b}_1}{\text{b}_2}=\frac{-3}{-9}=\frac{1}{3}$
And $\frac{\text{c}_1}{\text{c}_2}=\frac{-3}{-2}=\frac{3}{2}$
Clearly, $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$
So, the given system of equation has no solutions.

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4 Marks Questions - Maths STD 10 Questions - Vidyadip