2 Marks Questions

Question 12 Marks

Which value(s) of $\lambda,$ do the pair of linear equations $\lambda\text{x}+\text{y}=\lambda^2$ and $\text{x}+\lambda\text{y}=1$have:
Infinitely many solutions?

Answer

For infinitely many solutions,
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\Rightarrow\frac{\lambda}{1}=\frac{1}{\lambda}=\frac{\lambda^2}{1}$
$\Rightarrow\frac{\lambda}{1}=\frac{\lambda^2}{1}$
$\Rightarrow\lambda(\lambda-1)=0$
When $\lambda\neq0,$ then $\lambda=1$

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Question 22 Marks

On comparing the ratios $\frac{\text{a}_1}{\text{a}_2},\frac{\text{b}_1}{\text{b}_2}$ and $\frac{\text{c}_1}{\text{c}_2},$ and without drawing them, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincide.

$5x - 4y + 8 = 0$
$7x + 6y - 9 = 0$

Answer

Given equation are:
$5x - 4y + 8 = 0$
$7x + 6y - 9 = 0$
Where, $a_1=5, b_1=-4, c_1=8$
$a_2=7, b_2=6, c_3=-9$
We get $\frac{\text{a}_1}{\text{a}_2}=\frac{5}{7},\frac{\text{b}_1}{\text{b}_2}=\frac{-4}{6}=\frac{-2}{3}$
and $\frac{\text{c}_1}{\text{c}_2}=\frac{8}{-9}$
$\Rightarrow\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}$
Thus the pair of linear equation is intersecting.

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Question 32 Marks

Which value(s) of $\lambda,$ do the pair of linear equations $\lambda\text{x}+\text{y}=\lambda^2$ and $\text{x}+\lambda\text{y}=1$have:
a unique solutions?

Answer

For a Unique solution,
$\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}$
$\Rightarrow\frac{\lambda}{1}\neq\frac{1}{\lambda}$
$\Rightarrow\lambda^2\neq1$
$\Rightarrow\lambda\neq\pm1$
So, all real values of $\lambda$ except $\pm1$

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Question 42 Marks

On comparing the ratios $\frac{\text{a}_1}{\text{a}_2},\frac{\text{b}_1}{\text{b}_2}$ and $\frac{\text{c}_1}{\text{c}_2},$ and without drawing them, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincide.

$6x - 3y + 10 = 0$

$2x - y + 9 = 0$

Answer

Given equation are:

$6x - 3y + 10 = 0$

$2x - y + 9 = 0$

Where, $a_1=6, b_1=-3, c_1=10$
$a_2=2, b_2=-1, c_3=9$
We get $\frac{\text{a}_1}{\text{a}_2}=\frac{6}{2},\frac{\text{b}_1}{\text{b}_2}=\frac{-3}{-1}=\frac{\text{c}_1}{\text{c}_2}=\frac{10}{9}$
$\Rightarrow\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=3$
Thus the pair of line is parallel lines.

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Question 52 Marks

On comparing the ratios $\frac{\text{a}_1}{\text{a}_2},\frac{\text{b}_1}{\text{b}_2}$ and $\frac{\text{c}_1}{\text{c}_2},$ and without drawing them, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincide.

$9x + 3y + 12 = 0$

$18x + 6y + 24 = 0$

Answer

Given equation are:
$9x + 3y + 12 = 0$
$18x + 6y + 24 = 0$
Where, $a_1=9, b_1=3, c_1=12$
$a_2=18, b_2=6, c_3=24$
We get $\frac{\text{a}_1}{\text{a}_2}=\frac{9}{18},\frac{\text{b}_1}{\text{b}_2}=\frac{3}{6}$
and $\frac{\text{c}_1}{\text{c}_2}=\frac{12}{24}$
$\Rightarrow\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}=\frac{1}{2}$
Thus the pair of linear equation are coincide.

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Question 62 Marks

Solve the following systems of equations:
$\frac{4}{\text{x}}+3\text{y}=8,$
$\frac{6}{\text{x}}-4\text{y}=-5.$

Answer

$\frac{4}{\text{x}}+3\text{y}=8\ ......(\text{i})$
$\frac{6}{\text{x}}-4\text{y}=-5\ ......(\text{ii})$
equations $(i) × 3$
$\Rightarrow\frac{12}{\text{x}} + 9\text{y} = 24\ ...(\text{iii})$
and euations $(ii) × 2$
$\Rightarrow\frac{12}{\text{x}} - 8\text{y} - 10\ ...(\text{iv})$
Substracting $(iv)$ frpm $(iii)$
$\frac{12}{\text{x}}+9\text{y}-\frac{12}{\text{x}}+8\text{y}=24+10$
$17\text{y}=34$
$\text{y}=2$
Putting $y$ in $(iii)$ equation
$\frac{12}{\text{x}}+9\times2=24$
$\frac{12}{\text{x}}=24-18$
$\frac{12}{\text{x}}=6$
$\text{x}=2$
So that solution is $x = 2, y = 2.$

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