M.C.Q

MCQ 11 Mark

The graph of the linear equation 2x + 3y = 6 meets the y-axis at the point.

A
(0, 3)
B
(2, 0)
C
(3, 0)
D
(0, 2)

Answer

(0, 2)
Solution:
If the graph of the linear equation 2x + 3y = 6 meets the y-axis, then x = 0.
Substituting the value of x = 0 in equation 2x + 3y = 6, we get
2(0) + 3y = 6
⇒ 3y = 6
$\Rightarrow\text{y}=\frac{6}{3}$
⇒ y = 2
So, the point of meeting is (0, 2).

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MCQ 21 Mark

The graph of the equation x + y = 4.

A
Intersects both the axis.
B
Parallel to the x-axis.
C
Intersects x-axis only.
D
Intersects y-axis only.

Answer

Intersects both the axis.
Solution:
The graph of the equation x + y = 4,
Put x = 0 cut y axis at y = 4,
Put y = 0 cut x axis at x = 4.

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MCQ 31 Mark

If the point (3, 4) lies on the graph of 3y = ax + 6, then the value of 'a' is:

Answer

2
Solution:
The point (3, 4) lies on the graph of 3y = ax + 6
So, it will satisfy the equation
3y = ax + 6
3(y) = ax + 6
12 = 3a + b
12 - 6 = 3a
3a = 6
$\text{a}=\frac{6}{3}$
$\text{a}=2$

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MCQ 41 Mark

The point of the form (a, –a) always lies on the line:

A
x = a
B
y = –a
C
y = x
D
x + y = 0

Answer

x + y = 0
Solution:
Taking option (d), x + y = a + (-a) = a – a = 0 [since, give point is of the form (a, -a)] Hence, the point (a, – a) always lies on the line x + y = 0.

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MCQ 51 Mark

The graph of the linear equation 4x + 2y = 12, cuts the x-axis at the point:

A
(3, 0)
B
(0, -2)
C
(-2, 0)
D
(0, 3)

Answer

(3, 0)
Solution:
The graph of the linear equation 4x + 2y = 12, cuts the x-axis at the point when line cut x-axis the co-ordinate of y becomes zero.
So we put y = 0 in given equation to find the co-ordinate,
4x + 2y = 124x + 2(0) = 124x = 12
$\text{x}=\frac{12}{4}$
x = 3
So the required coordinate is (3, 0).

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MCQ 61 Mark

x = 5, y = 2 is a solution of the linear equation:

A
x + 2y = 7
B
5x + 2y = 7
C
x + y = 7
D
5x + y = 7

Answer

x + y = 7
Solution:
Substituting x = 5 and y = 2 in L.H.S. of equation x + y = 7,
We get:
LHS
= 5 + 2
7 = RHS
Hence, x = 5 and y = 2 is a solution of the linear equation x + y = 7.

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MCQ 71 Mark

The equation x = 7 in two variables can be written as:

A
0.x + 0.y = 7
B
1.x + 0.y = 7
C
1.x + 1.y = 7
D
0.x + 1.y = 7

Answer

1.x + 0.y = 7
Solution:
The equation x = 7 in two variables can be written as exactly 1.x + 0.y = 7 because it contain two variable x and y and coefficient of y is zero as there is no term containing y in equation x = 7.

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MCQ 81 Mark

The graph of the linear equation x + y = 0 passes through the point.

A
(1, -1)
B
(1, 1)
C
(1, 0)
D
(0, 1)

Answer

(1, -1)
Solution:
The graph of the linear equation x + y = 0 passes through the point (1, -1) because the co-ordinate of x and y axis satisfy the given equation.
x + y = 0
1 - 1 = 0
So we can say (1, -1) is a solution of above equation

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MCQ 91 Mark

If a linear equation has solutions (1, 2), (-1, -16) and (0, -7), then it is of the form:

A
y = 9x - 7
B
9x - y + 7 = 0
C
x - 9y = 7
D
x = 9y - 7

Answer

y = 9x - 7
Solution:
Since all the given co- ordinate (1, 2), (-1, -16) and (0, -7) satisfy the given line y = 9x - 7
For point (1, 2)
y = 9x - 7
2 = 9(1) - 7
2 = 9 - 7
2 = 2
Hence (2, 1) is a solution.
For point (-1, -16)
y = 9x - 7
-16 = 9(-1) - 7
-16 = -9 - 7
-16 = -16
Hence (-1, -16) is a solution.
For point (0, -7)
y = 9x - 7
-7 = 9(0) -7
-7 = -7
Hence (0, -7) is a solution.

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MCQ 101 Mark

Write the linear equation such that each point on its graph has an ordinate 5 times it's abscissa.

A
5x + y = 2
B
None of these
C
y = 5x
D
x = 5y

Answer

y = 5x
Solution:
y = 5x
At x = 1
y = 5.1 = 5
y = 5
(1, 5)
At x = 2
y = 5.2 = 10
y = 10
(2, 10)
At x = 3
y = 5.3 = 15
y = 15
(3, 15).

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MCQ 111 Mark

x = 2, y = - 1 is a solution of the linear equation:

A
x + 2y = 4
B
2x + y = 5
C
2x + y = 0
D
x + 2y = 0

Answer

x + 2y = 0
Solution:
2 + 2(-1) = 2 - 2 = 0.

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MCQ 121 Mark

If the point (3, 4) lies on the graph of 3y = ax + 6, then the value of ‘a’ is:

Answer

2
Solution:
The point (3, 4) lies on the graph of 3y = ax + 6
So it will satisfy the equation
3y = ax + 6
3(y) = ax + 6
12 = 3a + 6
12 - 6 = 3a
3a = 6
$\text{a}=\frac{6}{3}$
a = 2

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MCQ 131 Mark

If we multiply both sides of a linear equation with a non-zero number, then the solution of the linear equation:

A
Gets multiplied by the number.
B
Remains the same.
C
Changes.
D
None of these.

Answer

Remains the same.
Solution:
If for any c. where c is any natural number.
Like addition and subtraction, we can multiply and divide both sides of an equation by a number, c, without changing the equation, where c is any natural number

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MCQ 141 Mark

Express y in terms of x in the equation 5x - 2y = 7.

A
$\text{y}=\frac{5\text{x}+7}{2}$
B
$\text{y}=\frac{7\text{x}+5}{2}$
C
$\text{y}=\frac{5\text{x}-7}{2}$
D
$\text{y}=\frac{7-5\text{x}}{2}$

Answer

$\text{y}=\frac{5\text{x}-7}{2}$
Solution:
5x - 2y = 7
-2y = 7 - 5x
2y = 5x - 7
$\text{y}=\frac{5\text{x}-7}{2}.$

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MCQ 151 Mark

The graph of the linear equation 3x - 5y = 15, cuts the y-axis at the point:

A
(2, 0)
B
(-2, 0)
C
(0, 3)
D
(0, -3)

Answer

(0, -3)
Solution:
The graph of the linear equation 3x - 5y = 15, cuts the y-axis at the point when line cut y-axis the co-ordinate of x becomes zero.
So we put x = 0 in given equation to find the co-ordinate.
3x - 5y = 15
3(0) - 5y = 15
-5y = 15
$\text{y} = −\frac{15}{5}$
y = -3
So the required cordinate is (0, -3).

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MCQ 161 Mark

Which of the following is a linear equation in two variables?

A
2x - 5y = 0
B
x + y = 8
C
x² = 5x + 3
D
5x = y² + 3

Answer

2x - 5y = 0
Solution:
In linear equation power of variable x and y should be 1 and here, the given linear equation has two variable x and y.

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MCQ 171 Mark

The graph of y = 5 is a line.

A
Parallel to the x-axis at a distance of 6 units from the origin.
B
Making an intercept 5 on the x-axis.
C
Parallel to the y-axis at a distance of 5 units from the origin.
D
Making an intercept 5 on the y-axis.

Answer

Making an intercept 5 on the y-axis.
Solution:
As, the graph of y = 5 is a line parallel to x-axis i.e. y = 0.
⇒ The line represented by the equation y = 5 is parallel to x-axis and intersects y-axis at y = 5.
So, the graph of y = 5 is a line parallel to the x-axis at a distance of 5 units from the origin making an intercept 5 on the y-axis.

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MCQ 181 Mark

The graph of the linear equation y = 3x passes through the point.

A
$\Big(0,-\frac{2}{3}\Big)$
B
$\Big(-\frac{2}{3},0\Big)$
C
$\Big(0,\frac{2}{3}\Big)$
D
$\Big(\frac{2}{3},2\Big)$

Answer

$\Big(\frac{2}{3},2\Big)$
Solution:
$\text{y}=3\text{x}$
$\frac{\text{y}}{3}=\text{x}$
For, y = 2, the value of x will be $\frac{2}{3}$
So, $\Big(\frac{2}{3},2\Big)$

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MCQ 191 Mark

Any point on the y - axis is of the form.

A
(y, y)
B
(0, y)
C
(x, y)
D
(x, 0)

Answer

(0, y)
Solution:
Any point on the y - axis is of the form (0, y).
On the y - axis, y can take any values and x should be equal to 0.

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MCQ 201 Mark

Write the correct answer in the following:
The positive solutions of the equation ax + by + c = 0 always lie in the,

A
I^st quadrant.
B
II^nd quadrant.
C
III^rd quadrant.
D
IV^th quadrant.

Answer

I^st quadrant.
Solution:
We know that, if a line passes through the Ist quadrant, then all solution lying on the line in first quadrant must be positive because the coordinate of all points in the Ist quadrant are positive.

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MCQ 211 Mark

Any point on the line y = 3x is of the form.

A
$(\text{a}, 3\text{a})$
B
$(3\text{a}, \text{a})$
C
$(\text{a}, \frac{\text{a}}{3})$
D
$(\frac{\text{a}}{3}, \text{-a})$

Answer

$(\text{a}, 3\text{a})$

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MCQ 221 Mark

Point (3, 4) lies on the graph of the equation 3y = kx + 7. The value of k is:

A
$\frac{4}{3}$
B
$\frac{5}{3}$
C
$3$
D
$\frac{7}{3}$

Answer

$\frac{5}{3}$
Solution:
3y = kx + 7
Here, x = 3 and y = 4
Hence,
(3 × 4) = (kx3) + 7
12 = 3k + 7
3k = 12 - 7
3k = 5
$\text{k}=\frac{5}{3}$

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MCQ 231 Mark

Write the correct answer in the following:
The equation x = 7, in two variables can be written as,

A
1 - x + 1.y = 7
B
1 - x + 0.y = 7
C
0 - x + 1.y = 7
D
0 - x + 0.y = 7

Answer

1 - x + 0.y = 7
Solution:
The equation x = 7 in two variables can be expressed as 1.x + 0.y = 7.

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MCQ 241 Mark

If (2k - 1, k) is a solution of the equation 10x - 9y = 12, then k =

Answer

2
Solution:
If (2k - 1, k) is solution of equation 10x - 9y = 12, then it must satisfy this equation.
Thus, we have
10(2k - 1) - 9k = 12
20k - 10 - 9k = 12
11k = 22
k = 2
Hence, correct option is (b).

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MCQ 251 Mark

If the graph of the equation 4x + 3y = 12 cuts the coordinate axes at A and B, then hypotenuse of right triangle AOB is of length.

A
3 units.
B
4 units.
C
5 units.
D
None of these.

Answer

5 units.
Solution:
According to the given question, triangle so formed has sides of units 3 and 4, using pythagoras theorem, the largest side is of 5 units.

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MCQ 261 Mark

How many lines pass through one point?

A
Two.
B
Three.
C
Many.
D
One.

Answer

Many.
Solution:
Because one point can be solution of many equations. So many equations can be pass from one point.

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MCQ 271 Mark

The equation y = 2x - 7 has:

A
Many solutions.
B
No solution.
C
Two solutions.
D
One solution.

Answer

Many solutions.
Solution:
y = 2x - 7
Has many solutions because for different value of x we have different value of y for example.
At x = 1
y = 2 (1) - 7
y = 2 - 7
y = -5
At x = 2
y = 2(2) - 7
y = 4 - 7
y = -3
So we can say for many value of x there is many value of y.

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MCQ 281 Mark

lf the graph of the equation 4x + 3y = 12 cuts the coordinate axes at A and B, then hypotenuse of right triangle AOB is of length:

A
4 units.
B
3 units.
C
5 units.
D
None of these.

Answer

5 units.
Solution:

4x + 3y = 12
At x = 0, 3y = 12 ⇒ y = 4 units
At y = 0, 4x = 12 ⇒ x = 3 units
The triangle formed is $\triangle\text{AOB},$ where
OB = 4 units
OA = 3 units
Hypotenuse $=\text{AB}=\sqrt{\text{OB}^2+\text{OA}^2}=\sqrt{16+9}=5\text{ units}$
Hence, correct option is (c).

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MCQ 291 Mark

The graph of the linear equation x - y = 0 passes through the point:

A
$(-1,1)$
B
$\Big(\frac{1}{1},\frac{1}{2}\Big)$
C
$\Big(\frac{1}{1},-\frac{1}{2}\Big)$
D
$(0,1)$

Answer

$\Big(\frac{1}{1},\frac{1}{2}\Big)$
Solution:
The graph of the linear equation x - y = 0 passes through the point $\Big(\frac{1}{1},\frac{1}{2}\Big)$ because the co-ordinate of x and y axis satisfy the given equation x - y = 0.
$\frac{1}{1}-\frac{1}{2}=0$
0 = 0
So we can say $\Big(\frac{1}{1},\frac{1}{2}\Big)$ is a solution of above equation.
So we can say the value of x co-ordinate must be equal to y co-ordinate.

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MCQ 301 Mark

The graph of a linear equation $\text{y}=\frac{9}{5}\text{x}+32$ cuts the y-axis at the point:

A
(0, 32)
B
(-32, 0)
C
(0, -32)
D
(32, 0)

Answer

(0, 32)
Solution:
When the graph cut at y axis in that case the value of x- coordinate is 0.
$\text{y}=\frac{9}{5}\text{x}+32$
$\text{y}=\frac{9}{5}.0+\text{32}$
$\text{y}=32$
So, the co-ordinates are (32, 0)

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MCQ 311 Mark

The area of the triangle formed by the line 3x + 4y = 12 and the co-ordinate axis is:

A
6 sq. units.
B
12 sq. units.
C
4 sq. units.
D
3 sq. units.

Answer

6 sq. units.
Solution:

To find the area of the triangle AOB formed by the line 3x + 4y = 12 and co-ordinate axis we put x = 0 in given equation to find the point on y axies.
So, at x = 0
3(0) + 4y = 12
4y = 12
We get y = 3
At y = 0
3x + 4(0) = 12
3x = 12
We get x = 4
So the line cut y axis at 3 and x axis at 4
So the hight of triangle AOB is OB = 3 unit and base OA = 4 unit
Area ot triangle AOB = 12(base × height)
= 12 × 4 × 3
= 6 unit square.

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MCQ 321 Mark

x = 2, y = 5 is a solution of the linear equation.

A
x + 2y = 7
B
x + y = 7
C
5x + y = 7
D
5x + 2y = 7

Answer

x + y = 7
Solution:
x = 2 and y = 5 satisfy the given equation.

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MCQ 331 Mark

The point of the form (a, -a), where a lies on:

A
The line x = y.
B
The x-axis.
C
The line y + x = 0.
D
The y-axis.

Answer

The line y + x = 0.
Solution:
The point (a, -a) lies on line x + y = 0
Here, is the verification
Put x = a in equation
x + y = 0
a + y = 0
y = -a
Hence, it is prove that (a, -a) is a solution of x + y = 0.

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MCQ 341 Mark

The graph of the linear equation 3x - 2y = 6, cuts the x-axis at the point:

A
(2, 0)
B
(0, 2)
C
(0, -2)
D
(-2, 0)

Answer

(2, 0)
Solution:
The linear equation 3x - 2y = 6, cuts the x-axis when y co-ordinate is 0.
So we put y = 0 in given equation 3x - 2y = 6
3x - 2.0 = 6
3x = 6
$\text{x}=\frac{6}{3}$
x = 2
So the co-ordinates are (2, 0).

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MCQ 351 Mark

Write the correct answer in the following:
Any point on the X-axis is of the form,

A
(x, y)
B
(0, y)
C
(x, 0)
D
(x, x)

Answer

(x, 0)
Solution:
Every point on the X-axis has its y-coordinate equal to zero. i.e., y = 0.

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MCQ 361 Mark

How many lines pass through two points?

A
Two.
B
Only one.
C
Many.
D
Three.

Answer

Only one.
Solution:
Only one because if a line is passing through two points then that two points are solution of a single linear equation so only one line passes over two given points.

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MCQ 371 Mark

Write the correct answer in the following:
The graph of the linear equation 2x + 3y = 6 is a line which meets the X-axis at the point.

A
(0, 2)
B
(2, 0)
C
(3, 0)
D
(0, 3)

Answer

(3, 0)
Solution:
Since, the graph of linear equation 2x + 3y = 6 meets the X-axis.
So, we put y = 0 in $2\text{x} + 3\text{y} = 6 \Rightarrow 2\text{x} + 3(0) = 6$
$\Rightarrow2\text{x}+0=6$
$\Rightarrow\text{x}=\frac{6}{2}\Rightarrow\text{x}=3$

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MCQ 381 Mark

The taxi fare in a city is as follows: For the first kilometer, the fare is ₹ 8 and for the subsequent distance it is ₹ 5 per kilometer. Taking the distance covered as x km and total fare as ₹ y, write a linear equation for this information.

A
x = 5y - 3
B
y = 5x + 3
C
x = 5y + 3
D
y = 5x - 3

Answer

y = 5x + 3
Solution:
Taxi fare for first kilometer = ₹ 8
Taxi fare for subsequent distance = ₹ 5
Total distance covered = x
Total fare = y
Since the fare for first kilometer = ₹ 8
According to problem, Fare for (x - 1) kilometer = 5(x - 1)
So, the total fare y = 5(x - 1) + 8
⇒ y = 5(x - 1) + 8
⇒ y = 5x - 5 + 8
⇒ y = 5x + 3
Hence, y = 5x + 3 is the required linear equation.

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MCQ 391 Mark

Express 'y' in terms of 'x' in the equation 5y - 3x - 10 = 0.

A
$\text{y}=\frac{3-10\text{x}}{5}$
B
$\text{y}=\frac{3+10\text{x}}{5}$
C
$\text{y}=\frac{3\text{x}-10\text{}}{5}$
D
$\text{y}=\frac{3\text{x}+10}{5}$

Answer

$\text{y}=\frac{3\text{x}+10}{5}$
Solution:
5y - 3, x - 10 = 0
5y - 3, x = 10
5y = 10 + 3x
$\text{y}=\frac{3\text{x}+10}{5}$

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MCQ 401 Mark

How many linear equation can be satisfied by x = 2 and y = 3?

A
Only one.
B
Only two.
C
Only three.
D
Infinitely many.

Answer

Infinitely many.
Solution:
Infinitely many linear equations can be satisfied by x = 2 and y = 3.

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MCQ 411 Mark

y = 0 is the equation of:

A
A line parallel to x - axis
B
A line parallel to y - axis
C
x - axis
D
y - axis

Answer

A line parallel to y - axis

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MCQ 421 Mark

Each of the points (-2, 2), (0, 0), (2, 2) satisfies the linear equation:

A
x - y = 0
B
x + y = 0
C
-x + 2y = 0
D
x - 2y = 0

Answer

x + y = 0
Solution:
Since given that each of the three points is a solution of the linear equation, all three points have to satisfy the linear equation.
We need to check for each of the four given equations.
Substituting x = -2 and y = 2 in option (b),
We get:
LHS
= x + y
= -2 + 2
0 = RHS
$\therefore\ $x = -2 and y = 2
Satisfy the given linear equation.
Substituting x = 0 and y = 0 in option (b),
We get:
LHS
= x + y
= 0 + 0
0 = RHS
$\therefore\ $x = 0 and y = 0
Satisfy the given linear equation.
Substituting x = -2 and y = 2 in option (b),
We get:
LHS
= x + y
= 2 - 2
0 = RHS
$\therefore\ $x = 2 and y = -2
Satisfy the given linear equation.
So, clearly all the three points satisfy the equation
x + y = 0.

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MCQ 431 Mark

If we multiply or divide both sides of a linear equation with a non-zero number, then the solution of the linear equation:

A
Changes.
B
Remains the same.
C
Only changes in case of multiplication.
D
Only changes in case of division.

Answer

Remains the same.
Solution:
If we multiply or divide both sides of a linear equation with a non-zero number, then the solution of the linear equation remains the same.

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MCQ 441 Mark

Write the correct answer in the following:
If (2, 0) is a solution of the linear equation 2x + 3y = k, then the value of k is:

Answer

4
Solution:
Since, (2, 0) is a solution of the given linear equation 2x + 3y = k, then put x = 2 and y = 0 in the equation.
⇒ 2(2) + 3(0) = k ⇒ k = 4

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MCQ 451 Mark

Which of the following is not a solution of 3x + 4y = 12?

A
(0, 3)
B
(2, 3)
C
(4, 0)
D
(8, -3)

Answer

(2, 3)
Solution:
The given co-ordinate is solution of a eqution if on puting the co-ordiates L.H.S = R.H.S
3x + 4y = 12
Put co-ordinate (2, 3) in given equation,
L.H.S
3.2 + 4.3
6 + 12 = 18
L.H.S ≠ R.H.S
So we can say (2, 3) is a not a solution of 3x + 4y = 12.

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MCQ 461 Mark

All linear equations in two variables have __________.