NCERT Class 10 Mathematics — Pair of Linear Equations in Two Variables, Exercise 3.2. All 3 questions solved.

This exercise practises the substitution method. It has three steps:

1. From one equation, express one variable in terms of the other — preferably one whose coefficient is $1$ or $-1$, so no fractions appear.
2. Substitute that expression into the other equation. This leaves a linear equation in one variable; solve it.
3. Put the value back into the expression from step 1 to find the second variable.

Step 2 can also end with no variable left at all. A true statement such as $9 = 9$ means the two equations are the same line and there are infinitely many solutions; a false statement such as $-4 = 0$ means the lines are parallel and there is no solution.

Key insight. Substitution does more than produce the answer — it tells you which of the three cases you are in. If the variable survives, the solution is unique; if it cancels, what remains decides between “every point” and “no point”. Question 1(iii) is there to show the cancelling case.

Question 1

Solve the following pair of linear equations by the substitution method.

(i) $x + y = 14$, $\;x – y = 4$

(ii) $s – t = 3$, $\;\dfrac{s}{3} + \dfrac{t}{2} = 6$

(iii) $3x – y = 3$, $\;9x – 3y = 9$

(iv) $0.2x + 0.3y = 1.3$, $\;0.4x + 0.5y = 2.3$

(v) $\sqrt2\,x + \sqrt3\,y = 0$, $\;\sqrt3\,x – \sqrt8\,y = 0$

(vi) $\dfrac{3x}{2} – \dfrac{5y}{3} = -2$, $\;\dfrac{x}{3} + \dfrac{y}{2} = \dfrac{13}{6}$

Solution.

(i) The second equation gives $x$ directly: $x = y + 4$. Substitute into the first:

$$(y + 4) + y = 14 \;\Longrightarrow\; 2y = 10 \;\Longrightarrow\; y = 5, \qquad x = 5 + 4 = 9$$

(ii) From the first equation, $s = t + 3$. Substitute into the second, then clear fractions by multiplying by $6$ (the LCM of $3$ and $2$):

$$\frac{t + 3}{3} + \frac{t}{2} = 6 \;\Longrightarrow\; 2(t + 3) + 3t = 36 \;\Longrightarrow\; 5t = 30 \;\Longrightarrow\; t = 6$$

Then $s = 6 + 3 = 9$.

(iii) From the first equation, $y = 3x – 3$. Substitute into the second:

$$9x – 3(3x – 3) = 9 \;\Longrightarrow\; 9x – 9x + 9 = 9 \;\Longrightarrow\; 9 = 9$$

The variable has vanished and what is left is always true. So every pair $(x, y)$ satisfying the first equation also satisfies the second — the second is just the first multiplied by $3$. There are infinitely many solutions: $y = 3x – 3$, with $x$ taking any real value.

(iv) Multiply both equations by $10$ to remove the decimals — every term, including the right-hand side:

$$2x + 3y = 13, \qquad 4x + 5y = 23$$

From the first, $x = \dfrac{13 – 3y}{2}$. Substitute into the second:

$$4 \cdot \frac{13 – 3y}{2} + 5y = 23 \;\Longrightarrow\; 26 – 6y + 5y = 23 \;\Longrightarrow\; y = 3$$

Then $x = \dfrac{13 – 9}{2} = 2$.

(v) From the first equation, $x = -\dfrac{\sqrt3}{\sqrt2}\,y$. Substitute into the second, remembering that $\sqrt8 = 2\sqrt2$:

$$\sqrt3\left(-\frac{\sqrt3}{\sqrt2}\,y\right) – 2\sqrt2\,y = 0 \;\Longrightarrow\; -\frac{3}{\sqrt2}\,y – 2\sqrt2\,y = 0$$

Multiplying by $\sqrt2$ gives $-3y – 4y = 0$, so $-7y = 0$ and $y = 0$. Then $x = -\dfrac{\sqrt3}{\sqrt2} \times 0 = 0$.

Both equations have zero on the right, so $(0, 0)$ was bound to satisfy them; the working shows it is the only solution, because the $y$ did not cancel.

(vi) Clear the fractions first. Multiplying each equation by $6$:

$$9x – 10y = -12, \qquad 2x + 3y = 13$$

From the second, $x = \dfrac{13 – 3y}{2}$. Substitute into the first and multiply by $2$:

$$9 \cdot \frac{13 – 3y}{2} – 10y = -12 \;\Longrightarrow\; 117 – 27y – 20y = -24 \;\Longrightarrow\; 47y = 141 \;\Longrightarrow\; y = 3$$

Then $x = \dfrac{13 – 9}{2} = 2$.

Question 2

Solve $2x + 3y = 11$ and $2x – 4y = -24$ and hence find the value of ‘$m$’ for which $y = mx + 3$.

Solution. Both equations contain $2x$, so take $2x = 11 – 3y$ from the first and substitute it into the second:

$$(11 – 3y) – 4y = -24 \;\Longrightarrow\; -7y = -35 \;\Longrightarrow\; y = 5$$

Then $2x = 11 – 15 = -4$, so $x = -2$.

“Hence” means the line $y = mx + 3$ must pass through this point. Substituting $x = -2$, $y = 5$:

$$5 = m(-2) + 3 \;\Longrightarrow\; -2m = 2 \;\Longrightarrow\; m = -1$$

Question 3

Form the pair of linear equations for the following problems and find their solution by substitution method.

Solution. In each part, name the two unknowns first, turn each sentence into one equation, and solve by substitution. Checking the answer against the words of the problem, not just the equations, catches a wrongly formed equation.

Question 3 (i)

The difference between two numbers is 26 and one number is three times the other. Find them.

Let the numbers be $x$ and $y$ with $x > y$. Then

$$x – y = 26, \qquad x = 3y.$$

The second equation is already solved for $x$. Substituting: $3y – y = 26$, so $2y = 26$, $y = 13$, and $x = 39$. Check: $39 – 13 = 26$ and $39 = 3 \times 13$.

Question 3 (ii)

The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

Let the larger angle be $x^\circ$ and the smaller $y^\circ$. Supplementary angles add to $180^\circ$:

$$x + y = 180, \qquad x – y = 18.$$

From the second, $x = y + 18$. Substituting: $2y + 18 = 180$, so $y = 81$ and $x = 99$.

Question 3 (iii)

The coach of a cricket team buys 7 bats and 6 balls for ₹3800. Later, she buys 3 bats and 5 balls for ₹1750. Find the cost of each bat and each ball.

Let one bat cost ₹$x$ and one ball ₹$y$. Then

$$7x + 6y = 3800, \qquad 3x + 5y = 1750.$$

Neither variable has coefficient $1$, so take $x = \dfrac{1750 – 5y}{3}$ from the second and substitute into the first, multiplying through by $3$:

$$7(1750 – 5y) + 18y = 11400 \;\Longrightarrow\; 12250 – 17y = 11400 \;\Longrightarrow\; 17y = 850 \;\Longrightarrow\; y = 50$$

Then $x = \dfrac{1750 – 250}{3} = 500$. Check: $7 \times 500 + 6 \times 50 = 3800$ and $3 \times 500 + 5 \times 50 = 1750$.

Question 3 (iv)

The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ₹105 and for a journey of 15 km, the charge paid is ₹155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?

Let the fixed charge be ₹$x$ and the charge per km ₹$y$. The fare is the fixed charge plus the distance times the rate:

$$x + 10y = 105, \qquad x + 15y = 155.$$

From the first, $x = 105 – 10y$. Substituting: $105 – 10y + 15y = 155$, so $5y = 50$, $y = 10$, and $x = 5$.

For 25 km the fare is $x + 25y = 5 + 250 = 255$.

Question 3 (v)

A fraction becomes $\dfrac{9}{11}$, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes $\dfrac56$. Find the fraction.

Let the fraction be $\dfrac{x}{y}$. Cross-multiplying each condition:

$$\frac{x + 2}{y + 2} = \frac{9}{11} \;\Longrightarrow\; 11x + 22 = 9y + 18 \;\Longrightarrow\; 11x – 9y + 4 = 0$$

$$\frac{x + 3}{y + 3} = \frac56 \;\Longrightarrow\; 6x + 18 = 5y + 15 \;\Longrightarrow\; 6x – 5y + 3 = 0$$

From the second, $x = \dfrac{5y – 3}{6}$. Substituting into the first and multiplying by $6$:

$$11(5y – 3) – 54y + 24 = 0 \;\Longrightarrow\; 55y – 33 – 54y + 24 = 0 \;\Longrightarrow\; y = 9$$

Then $x = \dfrac{45 – 3}{6} = 7$. Check: $\dfrac{9}{11}$ ✓ and $\dfrac{10}{12} = \dfrac56$ ✓.

Question 3 (vi)

Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

Let Jacob’s present age be $x$ years and his son’s $y$ years. In five years both are five years older; five years ago both were five years younger:

$$x + 5 = 3(y + 5) \;\Longrightarrow\; x – 3y – 10 = 0$$

$$x – 5 = 7(y – 5) \;\Longrightarrow\; x – 7y + 30 = 0$$

From the first, $x = 3y + 10$. Substituting: $3y + 10 – 7y + 30 = 0$, so $4y = 40$, $y = 10$, and $x = 40$. Check: in five years $45 = 3 \times 15$; five years ago $35 = 7 \times 5$.

Common mistakes

• Question 1(iii), reading $9 = 9$ as an error or as “no solution”. A true statement with no variable left means the equations describe the same line: infinitely many solutions. It is a false statement that means no solution.
• Question 1(iv), multiplying only the left side by $10$. $0.2x + 0.3y = 1.3$ becomes $2x + 3y = 13$; forgetting to scale the $1.3$ gives a completely different line.
• Question 1(v), mishandling $\sqrt8$. Simplify it to $2\sqrt2$ first. Treating it as $8$ or as $2\sqrt8$ makes the $y$-terms fail to combine, and the unique solution $(0, 0)$ gets lost in the algebra.
• Question 2, stopping at $x$ and $y$. The question continues with “hence find $m$”. The point $(-2, 5)$ has to be substituted into $y = mx + 3$, with $x$ and $y$ in the right places.
• Question 3(iv), leaving out the fixed charge. The 25 km fare is $5 + 25 \times 10$, not $25 \times 10 = 250$. The fixed charge is paid on every journey.
• Question 3(vi), shifting only one age. “Five years ago” applies to Jacob and his son, so both $x$ and $y$ lose $5$. Writing $x – 5 = 7y$ gives the wrong ages.

Practise next

• Exercise 3.3 — the elimination method, and the same kind of word problems solved a second way.
• Exercise 3.1 — the graphical picture behind question 1(iii): coincident lines are what “infinitely many solutions” looks like.