NCERT Class 10 Mathematics — Pair of Linear Equations in Two Variables, Exercise 3.3. Both questions solved, all nine parts.

This exercise introduces the elimination method:

1. Multiply one or both equations by suitable non-zero numbers so that one variable has the same coefficient (or the same coefficient with opposite sign) in both.
2. Subtract the equations if those coefficients are equal, add them if they are opposite. That variable disappears, leaving an equation in one variable.
3. Solve it, and substitute back into either original equation for the other variable.

Question 1 asks for each pair to be solved both ways, by elimination and by substitution. Question 2 turns five word problems into equations and solves them by elimination.

Key insight. Elimination and substitution are two routes to the same point, so the two methods in question 1 must give the same answer — which makes the second one a free check on the first. For elimination, choose multipliers that make the coefficients of one variable equal with the least work (usually their LCM); for substitution, isolate whichever variable has coefficient $1$.

Question 1

Solve the following pair of linear equations by the elimination method and the substitution method:

(i) $x + y = 5$ and $2x – 3y = 4$

(ii) $3x + 4y = 10$ and $2x – 2y = 2$

(iii) $3x – 5y – 4 = 0$ and $9x = 2y + 7$

(iv) $\dfrac{x}{2} + \dfrac{2y}{3} = -1$ and $x – \dfrac{y}{3} = 3$

Solution.

Question 1 (i)

Elimination. Multiply the first equation by $2$ so that both have $2x$:

$$2x + 2y = 10, \qquad 2x – 3y = 4$$

Subtracting the second from the first removes $x$: $2y – (-3y) = 10 – 4$, so $5y = 6$ and $y = \dfrac65$. Then $x = 5 – \dfrac65 = \dfrac{19}{5}$.

Substitution. From the first equation $x = 5 – y$. Then $2(5 – y) – 3y = 4$, so $10 – 5y = 4$, giving $y = \dfrac65$ and $x = \dfrac{19}{5}$ — the same point.

Question 1 (ii)

Elimination. The $y$-coefficients are $4$ and $-2$. Multiply the second equation by $2$ to make them $4$ and $-4$, then add, since they are opposite:

$$(3x + 4y) + (4x – 4y) = 10 + 4 \;\Longrightarrow\; 7x = 14 \;\Longrightarrow\; x = 2$$

From $2x – 2y = 2$: $4 – 2y = 2$, so $y = 1$.

Substitution. Dividing the second equation by $2$ gives $x – y = 1$, so $x = y + 1$. Then $3(y + 1) + 4y = 10$, so $7y = 7$, $y = 1$ and $x = 2$.

Question 1 (iii)

Write both in the form $ax + by = c$ first, so that like terms line up:

$$3x – 5y = 4, \qquad 9x – 2y = 7$$

Elimination. Multiply the first by $3$ to match the $9x$:

$$9x – 15y = 12, \qquad 9x – 2y = 7$$

Subtracting the first from the second: $-2y – (-15y) = 7 – 12$, so $13y = -5$ and $y = -\dfrac{5}{13}$. Then $3x = 4 + 5y = 4 – \dfrac{25}{13} = \dfrac{27}{13}$, so $x = \dfrac{9}{13}$.

Substitution. From the first equation, $x = \dfrac{4 + 5y}{3}$. Then

$$9 \cdot \frac{4 + 5y}{3} – 2y = 7 \;\Longrightarrow\; 12 + 15y – 2y = 7 \;\Longrightarrow\; 13y = -5,$$

giving the same $y = -\dfrac{5}{13}$ and $x = \dfrac{9}{13}$.

Question 1 (iv)

Clear the fractions first. Multiplying the first equation by $6$ and the second by $3$ — every term, the right-hand sides included:

$$3x + 4y = -6, \qquad 3x – y = 9$$

Elimination. The $x$-coefficients already match, so subtract: $4y – (-y) = -6 – 9$, so $5y = -15$ and $y = -3$. Then $3x = 9 + y = 6$, so $x = 2$.

Substitution. From the second, $y = 3x – 9$. Then $3x + 4(3x – 9) = -6$, so $15x = 30$, $x = 2$ and $y = -3$.

Question 2

Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:

Solution. “If they exist” is a reminder that a pair of equations need not have a solution. Here all five do, and each has exactly one.

Question 2 (i)

If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes $\dfrac12$ if we only add 1 to the denominator. What is the fraction?

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

$$\frac{x + 1}{y – 1} = 1 \;\Longrightarrow\; x + 1 = y – 1 \;\Longrightarrow\; x – y + 2 = 0$$

$$\frac{x}{y + 1} = \frac12 \;\Longrightarrow\; 2x = y + 1 \;\Longrightarrow\; 2x – y – 1 = 0$$

The $y$-coefficients are both $-1$, so subtract the first from the second: $(2x – y – 1) – (x – y + 2) = 0$, which gives $x – 3 = 0$, so $x = 3$. Then $y = x + 2 = 5$. Check: $\dfrac{4}{4} = 1$ and $\dfrac{3}{6} = \dfrac12$.

Question 2 (ii)

Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

Let Nuri’s present age be $x$ years and Sonu’s $y$ years.

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

$$x + 10 = 2(y + 10) \;\Longrightarrow\; x – 2y – 10 = 0$$

Both have $x$ with coefficient $1$, so subtract the first from the second: $(-2y – 10) – (-3y + 10) = 0$, which gives $y – 20 = 0$, so $y = 20$. Then $x = 2y + 10 = 50$. Check: five years ago they were $45$ and $15$, and $45 = 3 \times 15$; in ten years they will be $60$ and $30$, and $60 = 2 \times 30$.

Question 2 (iii)

The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Let the tens digit be $x$ and the units digit $y$. The number is then $10x + y$, and with the digits reversed it is $10y + x$ — the digits carry place value, so the number is not simply “$xy$”.

$$x + y = 9$$

$$9(10x + y) = 2(10y + x) \;\Longrightarrow\; 90x + 9y = 20y + 2x \;\Longrightarrow\; 88x – 11y = 0 \;\Longrightarrow\; 8x – y = 0$$

The $y$-coefficients are $1$ and $-1$, so add the equations: $9x = 9$, $x = 1$, and $y = 8$. The number is $18$. Check: $1 + 8 = 9$, and $9 \times 18 = 162 = 2 \times 81$.

Question 2 (iv)

Meena went to a bank to withdraw ₹2000. She asked the cashier to give her ₹50 and ₹100 notes only. Meena got 25 notes in all. Find how many notes of ₹50 and ₹100 she received.

Let her receive $x$ notes of ₹50 and $y$ notes of ₹100. One equation counts the notes, the other counts the money:

$$x + y = 25, \qquad 50x + 100y = 2000 \;\Longrightarrow\; x + 2y = 40$$

Subtract the first from the second: $y = 15$, and then $x = 10$. Check: $10 \times 50 + 15 \times 100 = 500 + 1500 = 2000$.

Question 2 (v)

A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹27 for a book kept for seven days, while Susy paid ₹21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Let the fixed charge be ₹$x$ and the charge for each extra day ₹$y$. The fixed charge covers the first three days, so seven days means $7 – 3 = 4$ extra days and five days means $2$:

$$x + 4y = 27, \qquad x + 2y = 21$$

Subtract: $2y = 6$, so $y = 3$, and $x = 21 – 6 = 15$.

Common mistakes

• Question 1(i), a sign slip when subtracting. $2y – (-3y)$ is $5y$, not $-y$. Subtracting an equation means changing the sign of every term in it; writing the subtraction out in brackets prevents the slip.
• Question 1(iii), eliminating before rearranging. $9x = 2y + 7$ has its $y$-term on the wrong side. Put both equations in the form $ax + by = c$ first, or the terms you line up are not like terms.
• Question 1(iv), clearing fractions on one side only. Multiplying $\frac{x}{2} + \frac{2y}{3} = -1$ by $6$ gives $3x + 4y = -6$; leaving the right side as $-1$ changes the line.
• Question 2(iii), writing the reversed number as $yx$. That is a product. A number with tens digit $y$ and units digit $x$ is $10y + x$.
• Question 2(v), counting all seven days as extra. The fixed charge already pays for the first three days, so Saritha pays for only $4$ extra days. Using $7$ and $5$ gives $y = 3$ but a fixed charge of ₹6, which fails the “first three days” condition.
• Question 2(ii), ageing only one person. “Ten years later” adds $10$ to Nuri’s age and Sonu’s. Writing $x + 10 = 2y$ changes the answer.

Practise next

• Chapter 4, Exercise 4.2 — solving quadratic equations by factorisation, the next kind of equation in the book.
• Exercise 3.2 — the substitution method on its own, with a second set of word problems.