LINEAR EQUATIONS IN TWO VARIABLES - MCQ - EM worksheet | Live ... - Free Printable

Let's solve each question one by one with detailed explanations.

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1) If linear equation in two variables x + 2y = 3 and 2x + 4y = k coincides then the value of 'k' is,

For two equations to coincide, they must represent the same line. That means one equation is a scalar multiple of the other.

Given:
- Equation 1: $ x + 2y = 3 $
- Equation 2: $ 2x + 4y = k $

Multiply Equation 1 by 2:
$ 2(x + 2y) = 2(3) \Rightarrow 2x + 4y = 6 $

So for the lines to coincide, $ k = 6 $

✔ Answer: B) 6

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2) 2x + 3y - 9 = 0 and 4x + 6y - 18 = 0 This pair of linear equations have ---- solutions.

Compare the ratios:

- $ \frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2} $
- $ \frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2} $
- $ \frac{c_1}{c_2} = \frac{-9}{-18} = \frac{1}{2} $

All ratios are equal → Coincident lines → Infinitely many solutions

✔ Answer: D) Infinite

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3) In the equation x + y = 7, if x = 3 then the value of 'y' is,

Substitute $ x = 3 $:
$ 3 + y = 7 \Rightarrow y = 4 $

✔ Answer: B) 4

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4) If a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are the pair of coinciding lines then the ratios of their coefficients are

For coincident lines, the ratios of coefficients must be equal:

$$
\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}
$$

✔ Answer: C) $ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} $

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5) If the pair of equations 3x + 2ky = 2 and 2x + 5y + 1 = 0 are parallel then the value of 'k' is

For parallel lines, the ratios of coefficients of x and y must be equal, but constant term ratio ≠ that.

So:

$$
\frac{3}{2} = \frac{2k}{5} \quad \text{(but not equal to } \frac{2}{1} \text{)}
$$

Solve:
$$
\frac{3}{2} = \frac{2k}{5} \Rightarrow 3 \cdot 5 = 2 \cdot 2k \Rightarrow 15 = 4k \Rightarrow k = \frac{15}{4}
$$

✔ Answer: A) 15/4

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6) The pair of equations 2x – 5y + 4 = 0 and 2x + y – 8 = 0 has

Check consistency:

- $ a_1 = 2, b_1 = -5, c_1 = 4 $
- $ a_2 = 2, b_2 = 1, c_2 = -8 $

Check ratios:
- $ \frac{a_1}{a_2} = \frac{2}{2} = 1 $
- $ \frac{b_1}{b_2} = \frac{-5}{1} = -5 $
- Not equal → Intersecting lines → Unique solution

✔ Answer: C) A unique solution

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7) The values of 'x' and 'y' when a point lies on the linear equation 2x – 3y = 12

We need to check which option satisfies $ 2x - 3y = 12 $

A) $ x=0, y=-3 $: $ 2(0) - 3(-3) = 0 + 9 = 9 $ ✘
B) $ x=2, y=3 $: $ 4 - 9 = -5 $ ✘
C) $ x=3, y=-2 $: $ 6 - (-6) = 6 + 6 = 12 $ ✔
D) $ x=-2, y=3 $: $ -4 - 9 = -13 $ ✘

✔ Answer: C) x = 3, y = -2

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8) Identify the wrong statement with respect to a pair of linear equations

A) If lines are parallel → no solution → ✔ Correct
B) If lines are perpendicular → no solution → ✘ Wrong! Perpendicular lines intersect, so one solution exists
C) Many solutions if lines coincide → ✔ Correct
D) Unique solution if they intersect → ✔ Correct (unless coincident)

So B is incorrect

✔ Answer: B) If the lines are perpendicular to each other, there is no solution

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9) In the equation 2x – y = 5 if y = 1 then the value of 'x' is,

Substitute $ y = 1 $:
$ 2x - 1 = 5 \Rightarrow 2x = 6 \Rightarrow x = 3 $

✔ Answer: A) 3

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10) The cost of 5 pens and 7 pencils is Rs 50. Which of the following equation describes the above statement?

Let pen = x, pencil = y
Then: $ 5x + 7y = 50 $

✔ Answer: B) 5x + 7y = 50

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11) The solution of the equation x – y = 2 and x + y = 4 is

Add both equations:
$ (x - y) + (x + y) = 2 + 4 \Rightarrow 2x = 6 \Rightarrow x = 3 $

Substitute into $ x + y = 4 $:
$ 3 + y = 4 \Rightarrow y = 1 $

Solution: $ (3,1) $

✔ Answer: C) 3, 1

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12) The line represented by 6x – 8y + 8 = 0 and 6x – 6y + 16 = 0

Check ratios:

- $ \frac{a_1}{a_2} = \frac{6}{6} = 1 $
- $ \frac{b_1}{b_2} = \frac{-8}{-6} = \frac{4}{3} $
- Not equal → So not parallel or coincident → Intersect

But let’s confirm: Since slopes differ, they intersect

Slope of first: $ 6x - 8y + 8 = 0 \Rightarrow y = \frac{6}{8}x + 1 = \frac{3}{4}x + 1 $

Second: $ 6x - 6y + 16 = 0 \Rightarrow y = x + \frac{8}{3} $

Different slopes → Intersect

✔ Answer: A) Intersects

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13) If the equations 4x + Py + 8 = 0 and 2x + 2y + 2 = 0 have unique solution then the value of 'P' is,

For unique solution, the lines must be neither parallel nor coincident → i.e., ratios of coefficients not all equal.

First simplify second equation:
$ 2x + 2y + 2 = 0 \Rightarrow x + y + 1 = 0 $

Now compare:
- Eq1: $ 4x + Py + 8 = 0 $
- Eq2: $ 2x + 2y + 2 = 0 $

For no unique solution, if $ \frac{4}{2} = \frac{P}{2} = \frac{8}{2} \Rightarrow 2 = P/2 = 4 $ → only possible if $ P = 4 $, but then $ P/2 = 2 $, $ 8/2 = 4 $ → $ 2 = 2 $, but $ 2 ≠ 4 $ → so not equal.

Wait — let's use condition:

For unique solution:
$$
\frac{a_1}{a_2} \neq \frac{b_1}{b_2}
$$

Here:
- $ \frac{4}{2} = 2 $
- $ \frac{P}{2} $

So for unique solution: $ 2 \neq \frac{P}{2} \Rightarrow P \neq 4 $

So P ≠ 4

Thus, except 4

✔ Answer: B) Except 4

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14) x + y = 20 and x – y = 4 solve for x and y, and substitute the values of x and y in y = mx + 3 then the value of 'm' is,

Solve:
Add: $ (x + y) + (x - y) = 20 + 4 \Rightarrow 2x = 24 \Rightarrow x = 12 $

Then $ 12 + y = 20 \Rightarrow y = 8 $

Now plug into $ y = mx + 3 $:
$ 8 = m(12) + 3 \Rightarrow 12m = 5 \Rightarrow m = \frac{5}{12} $

✔ Answer: A) 5 / 12

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✔ Final Answers:

| Q | Answer |
|---|--------|
| 1 | B) 6 |
| 2 | D) Infinite |
| 3 | B) 4 |
| 4 | C) $ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} $ |
| 5 | A) 15/4 |
| 6 | C) A unique solution |
| 7 | C) x = 3, y = -2 |
| 8 | B) If the lines are perpendicular... |
| 9 | A) 3 |
|10 | B) 5x + 7y = 50 |
|11 | C) 3, 1 |
|12 | A) Intersects |
|13 | B) Except 4 |
|14 | A) 5/12 |

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