NCERT Exemplar Solutions Class 10 Maths Pair of Linear Equations — Exercises 3.1 and 3.2

NCERT Exemplar Problems, Class 10 Mathematics — Pair of Linear Equations in Two Variables, Exercises 3.1 and 3.2. All 19 questions solved.

Exercise 3.1 is thirteen multiple-choice questions and Exercise 3.2 is six reasoning questions carrying thirteen sub-parts, and between them they test one table. For the pair

$$a_1x + b_1y + c_1 = 0, \qquad a_2x + b_2y + c_2 = 0$$

Ratios	Lines	Solutions	Called
$\dfrac{a_1}{a_2} \ne \dfrac{b_1}{b_2}$	intersecting	exactly one	consistent
$\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$	coincident	infinitely many	consistent (dependent)
$\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2}$	parallel	none	inconsistent

Two habits make the whole exercise mechanical. Put both equations in the form $ax+by+c=0$ before comparing anything — a constant left on the right-hand side is the single commonest source of a wrong $c$ ratio. And check the $c$ ratio last, because the first two ratios already separate “intersecting” from “the other two cases”.

Key insight. “Consistent” means has at least one solution, so it covers both intersecting and coincident lines. Reading it as “exactly one solution” turns three of these questions into the wrong answer.

Choose the correct answer from the given four options:

Exercise 3.1, Question 1

Graphically, the pair of equations $6x – 3y + 10 = 0$ and $2x – y + 9 = 0$ represents two lines which are

Solution. Both are already in the standard form, so read the ratios straight off:

$$\frac{a_1}{a_2} = \frac{6}{2} = 3, \qquad \frac{b_1}{b_2} = \frac{-3}{-1} = 3, \qquad \frac{c_1}{c_2} = \frac{10}{9}$$

The first two agree and the third does not, so the lines are parallel.

Option (B) is impossible for any pair of straight lines — two distinct lines meet at most once — and is there to be discarded on sight.

Exercise 3.1, Question 2

The pair of equations $x + 2y + 5 = 0$ and $-3x – 6y + 1 = 0$ have

Solution.

$$\frac{a_1}{a_2} = \frac{1}{-3}, \qquad \frac{b_1}{b_2} = \frac{2}{-6} = \frac{1}{-3}, \qquad \frac{c_1}{c_2} = \frac{5}{1} = 5$$

First two equal, third different: parallel lines, hence no solution.

A linear system can have exactly none, exactly one, or infinitely many solutions — never exactly two — so (B) is impossible in principle, just as in Question 1.

Exercise 3.1, Question 3

If a pair of linear equations is consistent, then the lines will be

Solution. Consistent means the system has a solution — one or many, the definition does not say which. Intersecting lines give one; coincident lines give infinitely many. Both qualify.

Parallel lines are the only inconsistent case, so (A) is exactly wrong. (B) and (D) each pick one of the two consistent cases and exclude the other; the word “always” in both is what makes them false.

Exercise 3.1, Question 4

The pair of equations $y = 0$ and $y = -7$ has

Solution. $y = 0$ is the $x$-axis and $y = -7$ is the horizontal line seven units below it. Two distinct horizontal lines never meet, so there is no point on both.

Algebraically the same thing: no value of $y$ can be $0$ and $-7$ at once, so the system is inconsistent whatever $x$ does.

Exercise 3.1, Question 5

The pair of equations $x = a$ and $y = b$ graphically represents lines which are

Solution. $x = a$ is a vertical line and $y = b$ a horizontal one, so they are perpendicular and certainly meet. The meeting point has $x$-coordinate $a$ and $y$-coordinate $b$, which is the point $(a, b)$.

Option (B) is the coordinates written in the wrong order — the standard trap, and worth a deliberate pause every time a question offers both orderings.

Exercise 3.1, Question 6

For what value of $k$ do the equations $3x – y + 8 = 0$ and $6x – ky = -16$ represent coincident lines?

Solution. Rewrite the second equation in standard form first: $6x – ky + 16 = 0$. Coincident lines need all three ratios equal:

$$\frac{3}{6} = \frac{-1}{-k} = \frac{8}{16}$$

The outer two are both $\frac12$, which is consistent, and the middle gives

$$\frac{1}{k} = \frac{1}{2} \implies k = 2$$

Had the $-16$ been left on the right, the constant ratio would have come out as $\frac{8}{-16} = -\frac12$ and the question would have looked impossible.

Exercise 3.1, Question 7

If the lines given by $3x + 2ky = 2$ and $2x + 5y + 1 = 0$ are parallel, then the value of $k$ is

Solution. Parallel needs the first two ratios equal:

$$\frac{3}{2} = \frac{2k}{5} \implies 4k = 15 \implies k = \frac{15}{4}$$

It is worth confirming that this really gives parallel rather than coincident lines. With $k = \frac{15}{4}$ the equations are $3x + \frac{15}{2}y – 2 = 0$ and $2x + 5y + 1 = 0$, and the constant ratio is $\frac{-2}{1} = -2$, which differs from $\frac32$. Parallel, as required.

Exercise 3.1, Question 8

The value of $c$ for which the pair of equations $cx – y = 2$ and $6x – 2y = 3$ will have infinitely many solutions is

Solution. In standard form: $cx – y – 2 = 0$ and $6x – 2y – 3 = 0$. Infinitely many solutions needs

$$\frac{c}{6} = \frac{-1}{-2} = \frac{-2}{-3}$$

The two known ratios are $\frac12$ and $\frac23$, and these are not equal. So the chain can never hold, whatever $c$ is — the $y$ and constant terms already rule it out before $c$ is even considered.

This is why (A) is the most dangerous option: solving $\frac{c}{6} = \frac12$ alone gives $c = 3$, which satisfies one of the two conditions and neither of the others. Always check the full chain.

Exercise 3.1, Question 9

One equation of a pair of dependent linear equations is $-5x + 7y = 2$. The second equation can be

Solution. “Dependent” means the two equations describe the same line, so the second must be a non-zero constant multiple of the first. Write the first as $-5x + 7y – 2 = 0$ and multiply throughout by $-2$:

$$10x – 14y + 4 = 0$$

Option (D) reads $10x – 14y = -4$, which is exactly this. The other three each change a sign somewhere and so describe a different line — (C), for instance, is $-10x+14y+4=0$, which is $-2$ times the first two terms but $+4$ instead of $+4$ on the wrong side, giving the constant ratio $\frac{-2}{4} = -\frac12$ against a coefficient ratio of $\frac12$.

Exercise 3.1, Question 10

A pair of linear equations which has a unique solution $x = 2$, $y = -3$ is

Solution. Two things have to hold: the point must satisfy both equations, and the solution must be unique. Test in that order.

• (A) $2 + (-3) = -1$ ✓, but $2(2) – 3(-3) = 4 + 9 = 13 \ne -5$ ✗.
• (B) $4 – 15 = -11$ ✓ and $8 – 30 = -22$ ✓ — the point fits. But the second equation is exactly twice the first, so the lines are coincident and there are infinitely many solutions, not a unique one. ✗
• (C) $2(2) – (-3) = 7 \ne 1$ ✗.
• (D) $2 – 4(-3) – 14 = 2 + 12 – 14 = 0$ ✓ and $5(2) – (-3) – 13 = 10 + 3 – 13 = 0$ ✓. Uniqueness: $\frac{1}{5} \ne \frac{-4}{-1}$, so the lines intersect at exactly one point. ✓

Option (B) is the whole point of the question — it satisfies the arithmetic and fails the word “unique”.

Exercise 3.1, Question 11

If $x = a$, $y = b$ is the solution of the equations $x – y = 2$ and $x + y = 4$, then the values of $a$ and $b$ are, respectively

Solution. Add the two equations, which eliminates $y$ immediately:

$$(x – y) + (x + y) = 2 + 4 \implies 2x = 6 \implies x = 3$$

Then $y = 4 – 3 = 1$. So $a = 3$ and $b = 1$.

Exercise 3.1, Question 12

Aruna has only ₹1 and ₹2 coins with her. If the total number of coins that she has is $50$ and the amount of money with her is ₹75, then the number of ₹1 and ₹2 coins are, respectively

Solution. Let $x$ be the number of ₹1 coins and $y$ the number of ₹2 coins. Counting coins and counting rupees gives two different equations:

$$x + y = 50 \qquad\text{(coins)}, \qquad x + 2y = 75 \qquad\text{(rupees)}$$

Subtracting the first from the second, $y = 25$, and then $x = 25$.

Option (B) can be dismissed without any algebra: $35 + 20 = 55 \ne 50$.

Exercise 3.1, Question 13

The father’s age is six times his son’s age. Four years hence, the age of the father will be four times his son’s age. The present ages, in years, of the son and the father are, respectively

Solution. Let the son be $s$ and the father $f$ years old now. “Six times” gives $f = 6s$. Four years hence both ages increase by four:

$$f + 4 = 4(s + 4)$$

Substituting $f = 6s$:

$$6s + 4 = 4s + 16 \implies 2s = 12 \implies s = 6, \quad f = 36$$

Option (D) fails the very first condition — $24$ is eight times $3$, not six times — so it can go immediately.

Exercise 3.2, Question 1

Do the following pairs of linear equations have no solution? Justify your answer.

(i) $2x + 4y = 3$ and $12y + 6x = 6$

Yes. Rearrange the second into the usual order: $6x + 12y = 6$. Then

$$\frac{a_1}{a_2} = \frac{2}{6} = \frac13, \qquad \frac{b_1}{b_2} = \frac{4}{12} = \frac13, \qquad \frac{c_1}{c_2} = \frac{3}{6} = \frac12$$

The first two agree and the third differs, so the lines are parallel and there is no solution.

(ii) $x = 2y$ and $y = 2x$

No. Write both in standard form: $x – 2y = 0$ and $2x – y = 0$. Now

$$\frac{a_1}{a_2} = \frac12, \qquad \frac{b_1}{b_2} = \frac{-2}{-1} = 2$$

These differ, so the lines intersect and there is exactly one solution — namely $x = y = 0$, which both equations obviously satisfy.

(iii) $3x + y – 3 = 0$ and $2x + \dfrac{2}{3}y = 2$

No. The second is $2x + \frac23 y – 2 = 0$, so

$$\frac{a_1}{a_2} = \frac32, \qquad \frac{b_1}{b_2} = \frac{1}{\,2/3\,} = \frac32, \qquad \frac{c_1}{c_2} = \frac{-3}{-2} = \frac32$$

All three are equal: the lines are coincident, so there are infinitely many solutions. Having many solutions is emphatically not the same as having none.

Exercise 3.2, Question 2

Do the following equations represent a pair of coincident lines? Justify your answer.

(i) $3x + \dfrac{1}{7}y = 3$ and $7x + 3y = 7$

No. $\dfrac{a_1}{a_2} = \dfrac{3}{7}$ but $\dfrac{b_1}{b_2} = \dfrac{1/7}{3} = \dfrac{1}{21}$. The first two ratios already differ, so the lines intersect at a single point — not coincident.

(ii) $-2x – 3y = 1$ and $6y + 4x = -2$

Yes. Standard form: $-2x – 3y – 1 = 0$ and $4x + 6y + 2 = 0$. Then

$$\frac{a_1}{a_2} = \frac{-2}{4} = -\frac12, \qquad \frac{b_1}{b_2} = \frac{-3}{6} = -\frac12, \qquad \frac{c_1}{c_2} = \frac{-1}{2} = -\frac12$$

All three equal, so the equations describe the same line. Indeed the second is $-2$ times the first.

(iii) $\dfrac{x}{2} + y + \dfrac{2}{5} = 0$ and $4x + 8y + \dfrac{5}{16} = 0$

No.

$$\frac{a_1}{a_2} = \frac{1/2}{4} = \frac18, \qquad \frac{b_1}{b_2} = \frac{1}{8}, \qquad \frac{c_1}{c_2} = \frac{2/5}{5/16} = \frac{2}{5} \times \frac{16}{5} = \frac{32}{25}$$

The coefficient ratios match but the constant ratio does not, so these are parallel, not coincident. The question is a good reminder that “$\frac{a_1}{a_2} = \frac{b_1}{b_2}$” on its own settles nothing between the last two rows of the table.

Exercise 3.2, Question 3

Are the following pairs of linear equations consistent? Justify your answer.

(i) $-3x – 4y = 12$ and $4y + 3x = 12$

No. In standard form: $-3x – 4y – 12 = 0$ and $3x + 4y – 12 = 0$.

$$\frac{a_1}{a_2} = -1, \qquad \frac{b_1}{b_2} = -1, \qquad \frac{c_1}{c_2} = \frac{-12}{-12} = 1$$

Coefficient ratios equal, constant ratio different: parallel lines, no solution, inconsistent.

(ii) $\dfrac{3}{5}x – y = \dfrac{1}{2}$ and $\dfrac{1}{5}x – 3y = \dfrac{1}{6}$

Yes.

$$\frac{a_1}{a_2} = \frac{3/5}{1/5} = 3, \qquad \frac{b_1}{b_2} = \frac{-1}{-3} = \frac13$$

These differ, so the lines intersect at exactly one point. A system with a solution is consistent, and here there is no need to find the solution to answer the question.

(iii) $2ax + by = a$ and $4ax + 2by – 2a = 0$, where $a, b \ne 0$

Yes. Write the first as $2ax + by – a = 0$. Then

$$\frac{a_1}{a_2} = \frac{2a}{4a} = \frac12, \qquad \frac{b_1}{b_2} = \frac{b}{2b} = \frac12, \qquad \frac{c_1}{c_2} = \frac{-a}{-2a} = \frac12$$

(all three divisions are legitimate because $a, b \ne 0$ — that is what the condition is for). The three ratios agree, so the lines are coincident: infinitely many solutions, and therefore consistent.

(iv) $x + 3y = 11$ and $2(2x + 6y) = 22$

No. Expand the second: $4x + 12y = 22$. Then

$$\frac{a_1}{a_2} = \frac14, \qquad \frac{b_1}{b_2} = \frac{3}{12} = \frac14, \qquad \frac{c_1}{c_2} = \frac{11}{22} = \frac12$$

Parallel, hence inconsistent. The bracket is doing the deceiving here: $2(2x+6y)$ looks like $4$ times $x + 3y$, and it is — but $22$ is only $2$ times $11$, not $4$ times. Expanding before comparing catches it.

Exercise 3.2, Question 4

For the pair of equations $\lambda x + 3y = -7$ and $2x + 6y = 14$ to have infinitely many solutions, the value of $\lambda$ should be $1$. Is the statement true? Give reasons.

Solution. No, the statement is false — and not because $1$ is the wrong number, but because no value of $\lambda$ works at all.

Infinitely many solutions requires all three ratios to be equal. Two of them involve no $\lambda$:

$$\frac{b_1}{b_2} = \frac{3}{6} = \frac12, \qquad \frac{c_1}{c_2} = \frac{-7}{14} = -\frac12$$

These are already unequal, so the chain $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ is broken before $\lambda$ enters. Whatever $\lambda$ is, the pair can never have infinitely many solutions.

(With $\lambda = 1$ the ratios are $\frac12$, $\frac12$, $-\frac12$ — parallel lines and no solution, the opposite of what the statement claims.)

Exercise 3.2, Question 5

For all real values of $c$, the pair of equations $x – 2y = 8$ and $5x – 10y = c$ have a unique solution. Justify whether it is true or false.

Solution. False. A unique solution requires $\frac{a_1}{a_2} \ne \frac{b_1}{b_2}$, but here

$$\frac{a_1}{a_2} = \frac15, \qquad \frac{b_1}{b_2} = \frac{-2}{-10} = \frac15$$

The two are equal for every $c$, because $c$ appears in neither. So the lines are never intersecting: they are parallel when $c \ne 40$ and coincident when $c = 40$. There is no value of $c$ giving a unique solution — the statement is as wrong as it could be.

Exercise 3.2, Question 6

The line represented by $x = 7$ is parallel to the $x$-axis. Justify whether the statement is true or not.

Solution. Not true. The equation $x = 7$ says that the $x$-coordinate is $7$ whatever $y$ is, so the graph is the set of points $(7, y)$ — a vertical line, seven units to the right of the $y$-axis. It is parallel to the $y$-axis and perpendicular to the $x$-axis.

The line parallel to the $x$-axis is the one of the form $y = k$, where the $y$-coordinate is fixed instead.

Common mistakes

• Comparing ratios before putting both equations in $ax+by+c=0$ form. Question 6 of Exercise 3.1 flips sign if the $-16$ is left on the right; Question 3(iv) of Exercise 3.2 hides the mismatch inside a bracket. Rearrange first, every time.
• Solving only the first ratio equation. In Exercise 3.1 Question 8, $\frac{c}{6} = \frac12$ gives $c = 3$ and option (A) is waiting. The other two ratios are $\frac12$ and $\frac23$, which already disagree, so no $c$ works. The same trap is Exercise 3.2 Question 4.
• Reading “consistent” as “unique solution”. It means at least one, so coincident lines are consistent too. This decides Exercise 3.1 Question 3 and Exercise 3.2 Question 3(iii).
• Confusing “no solution” with “infinitely many”. Exercise 3.2 Question 1(iii) is coincident, so the answer to “do they have no solution?” is No — even though a quick glance at three equal ratios might suggest something exceptional is going on.
• Swapping the coordinates in Exercise 3.1 Question 5. $x = a$, $y = b$ meet at $(a, b)$; option (B) offers $(b, a)$ for exactly this reason.
• Forgetting that “unique” is part of the question in Exercise 3.1 Question 10. Option (B) is satisfied by the given point and still wrong, because its two equations are proportional.
• Taking $x = 7$ for a horizontal line. The variable that is fixed names the axis the line is perpendicular to, not parallel to.

Practise next

• Exemplar Exercise 3.3 — the same ratio conditions used to solve for a parameter, plus the substitution and elimination work the objective band only hints at.
• Exemplar Exercise 3.4 for the long word problems: boats, fares, investments and digits.
• NCERT textbook Exercise 3.1 — the graphical method and the same consistency table applied to worded situations.
• NCERT textbook Exercise 3.2 and Exercise 3.3 for substitution and elimination as the book sets them.