Pair of Linear Equations in Two Variables | Exercise 3.3

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

Understand the Question

To solve a system of two linear equations in two variables ( $x$ and $y$ ):

Elimination Method: Multiply one or both equations by suitable non-zero constants so that the coefficients of one variable become equal (or opposites). Then add or subtract the equations to eliminate that variable and solve for the remaining one.
Substitution Method: Express one variable in terms of the other from one equation, and substitute this expression into the second equation to obtain a single-variable linear equation.

(i) $x + y = 5$ and $2x - 3y = 4$

Step 1 · Elimination Method

Given equations: $x + y = 5 \quad \dots (1)$ $2x - 3y = 4 \quad \dots (2)$

Multiply equation $(1)$ by $2$ :

\begin{aligned} 2(x + y) &= 2(5) \\ 2x + 2y &= 10 \quad \dots (3) \end{aligned}

Subtract equation $(3)$ from equation $(2)$ :

\begin{aligned} (2x - 3y) - (2x + 2y) &= 4 - 10 \\ 2x - 3y - 2x - 2y &= -6 \\ -5y &= -6 \\ y &= \dfrac{6}{5} \end{aligned}

Substitute $y = \dfrac{6}{5}$ into equation $(1)$ :

\begin{aligned} x + \dfrac{6}{5} &= 5 \\[0.6em] x &= 5 - \dfrac{6}{5} \\[0.6em] x &= \dfrac{25 - 6}{5} = \dfrac{19}{5} \end{aligned}

Step 2 · Substitution Method

From equation $(1)$ , express $x$ in terms of $y$ : $x = 5 - y \quad \dots (4)$

Substitute $x = 5 - y$ into equation $(2)$ :

\begin{aligned} 2(5 - y) - 3y &= 4 \\ 10 - 2y - 3y &= 4 \\ 10 - 5y &= 4 \\ -5y &= 4 - 10 \\ -5y &= -6 \\ y &= \dfrac{6}{5} \end{aligned}

Substitute $y = \dfrac{6}{5}$ into equation $(4)$ :

\begin{aligned} x &= 5 - \dfrac{6}{5} \\[0.6em] x &= \dfrac{25 - 6}{5} = \dfrac{19}{5} \end{aligned}

Answer

(i) $x = \dfrac{19}{5}, \quad y = \dfrac{6}{5}$

(ii) $3x + 4y = 10$ and $2x - 2y = 2$

Step 1 · Elimination Method

Given equations: $3x + 4y = 10 \quad \dots (1)$ $2x - 2y = 2 \quad \dots (2)$

Multiply equation $(2)$ by $2$ :

\begin{aligned} 2(2x - 2y) &= 2(2) \\ 4x - 4y &= 4 \quad \dots (3) \end{aligned}

Add equation $(1)$ and equation $(3)$ :

\begin{aligned} (3x + 4y) + (4x - 4y) &= 10 + 4 \\ 3x + 4y + 4x - 4y &= 14 \\ 7x &= 14 \\ x &= 2 \end{aligned}

Substitute $x = 2$ into equation $(1)$ :

\begin{aligned} 3(2) + 4y &= 10 \\ 6 + 4y &= 10 \\ 4y &= 10 - 6 \\ 4y &= 4 \\ y &= 1 \end{aligned}

Step 2 · Substitution Method

Divide equation $(2)$ by $2$ : $x - y = 1 \implies x = 1 + y \quad \dots (4)$

Substitute $x = 1 + y$ into equation $(1)$ :

\begin{aligned} 3(1 + y) + 4y &= 10 \\ 3 + 3y + 4y &= 10 \\ 3 + 7y &= 10 \\ 7y &= 10 - 3 \\ 7y &= 7 \\ y &= 1 \end{aligned}

Substitute $y = 1$ into equation $(4)$ : $x = 1 + 1 = 2$

Answer

(ii) $x = 2, \quad y = 1$

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

Step 1 · Elimination Method

Rearrange the given equations: $3x - 5y = 4 \quad \dots (1)$ $9x - 2y = 7 \quad \dots (2)$

Multiply equation $(1)$ by $3$ :

\begin{aligned} 3(3x - 5y) &= 3(4) \\ 9x - 15y &= 12 \quad \dots (3) \end{aligned}

Subtract equation $(3)$ from equation $(2)$ :

\begin{aligned} (9x - 2y) - (9x - 15y) &= 7 - 12 \\ 9x - 2y - 9x + 15y &= -5 \\ 13y &= -5 \\ y &= -\dfrac{5}{13} \end{aligned}

Substitute $y = -\dfrac{5}{13}$ into equation $(1)$ :

\begin{aligned} 3x - 5\left(-\dfrac{5}{13}\right) &= 4 \\[0.6em] 3x + \dfrac{25}{13} &= 4 \\[0.6em] 3x &= 4 - \dfrac{25}{13} \\[0.6em] 3x &= \dfrac{52 - 25}{13} \\[0.6em] 3x &= \dfrac{27}{13} \\[0.6em] x &= \dfrac{27}{13 \times 3} = \dfrac{9}{13} \end{aligned}

Step 2 · Substitution Method

From equation $(1)$ , express $x$ in terms of $y$ :

\begin{aligned} 3x &= 4 + 5y \\[0.6em] x &= \dfrac{4 + 5y}{3} \quad \dots (4) \end{aligned}

Substitute $x$ into equation $(2)$ :

\begin{aligned} 9\left(\dfrac{4 + 5y}{3}\right) - 2y &= 7 \\[0.6em] 3(4 + 5y) - 2y &= 7 \\ 12 + 15y - 2y &= 7 \\ 12 + 13y &= 7 \\ 13y &= 7 - 12 \\ 13y &= -5 \\ y &= -\dfrac{5}{13} \end{aligned}

Substitute $y = -\dfrac{5}{13}$ into equation $(4)$ :

\begin{aligned} x &= \dfrac{4 + 5\left(-\dfrac{5}{13}\right)}{3} \\[1.1em] &= \dfrac{4 - \dfrac{25}{13}}{3} \\[1.1em] &= \dfrac{\dfrac{52 - 25}{13}}{3} \\[1.1em] &= \dfrac{\dfrac{27}{13}}{3} \\[1.1em] &= \dfrac{27}{13 \times 3} = \dfrac{9}{13} \end{aligned}

Answer

(iii) $x = \dfrac{9}{13}, \quad y = -\dfrac{5}{13}$

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

Step 1 · Simplify and Solve using Elimination Method

Clear denominators by multiplying $\dfrac{x}{2} + \dfrac{2y}{3} = -1$ by $6$ :

\begin{aligned} 6\left(\dfrac{x}{2}\right) + 6\left(\dfrac{2y}{3}\right) &= 6(-1) \\[0.6em] 3x + 4y &= -6 \quad \dots (1) \end{aligned}

Multiply $x - \dfrac{y}{3} = 3$ by $3$ :

\begin{aligned} 3(x) - 3\left(\dfrac{y}{3}\right) &= 3(3) \\[0.6em] 3x - y &= 9 \quad \dots (2) \end{aligned}

Subtract equation $(2)$ from equation $(1)$ :

\begin{aligned} (3x + 4y) - (3x - y) &= -6 - 9 \\ 3x + 4y - 3x + y &= -15 \\ 5y &= -15 \\ y &= -3 \end{aligned}

Substitute $y = -3$ into equation $(1)$ :

\begin{aligned} 3x + 4(-3) &= -6 \\ 3x - 12 &= -6 \\ 3x &= -6 + 12 \\ 3x &= 6 \\ x &= 2 \end{aligned}

Step 2 · Substitution Method

From equation $(2)$ , express $y$ in terms of $x$ :

\begin{aligned} -y &= 9 - 3x \\ y &= 3x - 9 \quad \dots (3) \end{aligned}

Substitute $y = 3x - 9$ into equation $(1)$ :

\begin{aligned} 3x + 4(3x - 9) &= -6 \\ 3x + 12x - 36 &= -6 \\ 15x - 36 &= -6 \\ 15x &= -6 + 36 \\ 15x &= 30 \\ x &= 2 \end{aligned}

Substitute $x = 2$ into equation $(3)$ :

\begin{aligned} y &= 3(2) - 9 \\ y &= 6 - 9 = -3 \end{aligned}

Answer

(iv) $x = 2, \quad y = -3$

Common Mistakes

Sign Errors on Subtraction: Forgetting to distribute the negative sign to every term when subtracting one equation from another (e.g. $-(3x - y) = -3x + y$ , not $-3x - y$ ).
Self-Substitution Loop: In the substitution method, substituting an expression back into the same equation it was derived from results in a trivial identity like $0 = 0$ .
Incomplete Multiplication: Forgetting to multiply the constant term on the RHS when scaling an equation to match coefficients or clear fractions.

Question 1

More questions in Exercise 3.3