1. Solve the following pair of linear equations by the substitution method. (i) (ii) (iii) (iv) (v) (vi)

Pair of Linear Equations in Two Variables | Exercise 3.2

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) $\sqrt{2}x + \sqrt{3}y = 0$
$\sqrt{3}x - \sqrt{8}y = 0$

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

Solution

Understand the Question

The substitution method solves a pair of linear equations by expressing one variable in terms of the other from one equation and substituting it into the second equation.
This reduces two equations in two variables to a single linear equation in one variable.
After solving for the first variable, substitute its value back into either equation to find the second variable.
If substitution leads to an identity that is always true (such as $9 = 9$ ) with no variables left, the equations represent coincident lines and have infinitely many solutions.

(i) Solve the pair of linear equations by substitution method: $x + y = 14$ $x - y = 4$

Step 1 · Express $x$ and Substitute to Find $x$ and $y$

Given $x + y = 14 \quad \dots (1)$ $x - y = 4 \quad \dots (2)$ Diagram 1

From equation $(2)$ $x = 4 + y$

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

\begin{aligned} (4 + y) + y &= 14 \\ 4 + 2y &= 14 \\ 2y &= 14 - 4 \\ 2y &= 10 \\ y &= 5 \end{aligned}

Substitute $y = 5$ into $x = 4 + y$ $x = 4 + 5 = 9$

Answer

(i) $x = 9, \quad y = 5$

(ii) Solve the pair of linear equations by substitution method: $s - t = 3$ $\dfrac{s}{3} + \dfrac{t}{2} = 6$

Step 1 · Clear Fractions and Solve for $s$ and $t$

Given $s - t = 3 \quad \dots (1)$ $\dfrac{s}{3} + \dfrac{t}{2} = 6 \quad \dots (2)$

Multiply equation $(2)$ by $6$ to clear fractions

\begin{aligned} 6 \left(\dfrac{s}{3} + \dfrac{t}{2}\right) &= 6 \times 6 \\[0.6em] 2s + 3t &= 36 \quad \dots (3) \end{aligned}

From equation $(1)$ $s = 3 + t$

Substitute $s = 3 + t$ into equation $(3)$

\begin{aligned} 2(3 + t) + 3t &= 36 \\ 6 + 2t + 3t &= 36 \\ 6 + 5t &= 36 \\ 5t &= 36 - 6 \\ 5t &= 30 \\ t &= 6 \end{aligned}

Substitute $t = 6$ into $s = 3 + t$ $s = 3 + 6 = 9$

Answer

(ii) $s = 9, \quad t = 6$

(iii) Solve the pair of linear equations by substitution method: $3x - y = 3$ $9x - 3y = 9$

Step 1 · Substitute and Determine the Nature of Solutions

Given $3x - y = 3 \quad \dots (1)$ $9x - 3y = 9 \quad \dots (2)$

From equation $(1)$ $y = 3x - 3$

Substitute $y = 3x - 3$ into equation $(2)$

\begin{aligned} 9x - 3(3x - 3) &= 9 \\ 9x - 9x + 9 &= 9 \\ 9 &= 9 \end{aligned}

This statement is always true for all values of $x$ . Therefore, the equations represent coincident lines and have infinitely many solutions.

Answer

(iii) Infinitely many solutions

(iv) Solve the pair of linear equations by substitution method: $0.2x + 0.3y = 1.3$ $0.4x + 0.5y = 2.3$

Step 1 · Clear Decimals and Solve for $x$ and $y$

Given $0.2x + 0.3y = 1.3 \quad \dots (1)$ $0.4x + 0.5y = 2.3 \quad \dots (2)$

Multiply equations $(1)$ and $(2)$ by $10$ to remove decimals $2x + 3y = 13 \quad \dots (3)$ $4x + 5y = 23 \quad \dots (4)$

From equation $(3)$

\begin{aligned} 2x &= 13 - 3y \\[0.6em] x &= \dfrac{13 - 3y}{2} \end{aligned}

Substitute $x = \dfrac{13 - 3y}{2}$ into equation $(4)$

\begin{aligned} 4\left(\dfrac{13 - 3y}{2}\right) + 5y &= 23 \\[0.6em] 2(13 - 3y) + 5y &= 23 \\ 26 - 6y + 5y &= 23 \\ 26 - y &= 23 \\ -y &= 23 - 26 \\ -y &= -3 \\ y &= 3 \end{aligned}

Substitute $y = 3$ into the expression for $x$

\begin{aligned} x &= \dfrac{13 - 3(3)}{2} \\[0.6em] &= \dfrac{13 - 9}{2} \\[0.6em] &= \dfrac{4}{2} \\[0.6em] &= 2 \end{aligned}

Answer

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

(v) Solve the pair of linear equations by substitution method: $\sqrt{2}x + \sqrt{3}y = 0$ $\sqrt{3}x - \sqrt{8}y = 0$

Step 1 · Solve for $x$ and $y$

Given $\sqrt{2}x + \sqrt{3}y = 0 \quad \dots (1)$ $\sqrt{3}x - \sqrt{8}y = 0 \implies \sqrt{3}x - 2\sqrt{2}y = 0 \quad \dots (2)$

From equation $(1)$

\begin{aligned} \sqrt{2}x &= -\sqrt{3}y \\[0.6em] x &= -\dfrac{\sqrt{3}}{\sqrt{2}}y \end{aligned}

Substitute $x = -\dfrac{\sqrt{3}}{\sqrt{2}}y$ into equation $(2)$

\begin{aligned} \sqrt{3}\left(-\dfrac{\sqrt{3}}{\sqrt{2}}y\right) - 2\sqrt{2}y &= 0 \\[0.6em] -\dfrac{3}{\sqrt{2}}y - 2\sqrt{2}y &= 0 \end{aligned}

Multiply by $\sqrt{2}$

\begin{aligned} -3y - 2\sqrt{2}(\sqrt{2})y &= 0 \\ -3y - 4y &= 0 \\ -7y &= 0 \\ y &= 0 \end{aligned}

Substitute $y = 0$ into the expression for $x$ $x = -\dfrac{\sqrt{3}}{\sqrt{2}}(0) = 0$

Answer

(v) $x = 0, \quad y = 0$

(vi) Solve the pair of linear equations by substitution method: $\dfrac{3x}{2} - \dfrac{5y}{3} = -2$ $\dfrac{x}{3} + \dfrac{y}{2} = \dfrac{13}{6}$

Step 1 · Clear Fractions and Solve for $x$ and $y$

Given $\dfrac{3x}{2} - \dfrac{5y}{3} = -2 \quad \dots (1)$ $\dfrac{x}{3} + \dfrac{y}{2} = \dfrac{13}{6} \quad \dots (2)$

Multiply both equations by $6$ to clear fractions

\begin{aligned} 6\left(\dfrac{3x}{2} - \dfrac{5y}{3}\right) &= 6(-2) \\[0.6em] 9x - 10y &= -12 \quad \dots (3) \end{aligned}

\begin{aligned} 6\left(\dfrac{x}{3} + \dfrac{y}{2}\right) &= 6\left(\dfrac{13}{6}\right) \\[0.6em] 2x + 3y &= 13 \quad \dots (4) \end{aligned}

From equation $(4)$

\begin{aligned} 2x &= 13 - 3y \\[0.6em] x &= \dfrac{13 - 3y}{2} \end{aligned}

Substitute $x = \dfrac{13 - 3y}{2}$ into equation $(3)$ $9\left(\dfrac{13 - 3y}{2}\right) - 10y = -12$

Multiply the entire equation by $2$

\begin{aligned} 9(13 - 3y) - 20y &= -24 \\ 117 - 27y - 20y &= -24 \\ 117 - 47y &= -24 \\ -47y &= -24 - 117 \\ -47y &= -141 \\[0.6em] y &= \dfrac{-141}{-47} = 3 \end{aligned}

Substitute $y = 3$ into the expression for $x$

\begin{aligned} x &= \dfrac{13 - 3(3)}{2} \\[0.6em] &= \dfrac{13 - 9}{2} \\[0.6em] &= \dfrac{4}{2} \\[0.6em] &= 2 \end{aligned}

Answer

(vi) $x = 2, \quad y = 3$

Common Mistakes

Substituting into the Same Equation: Substituting an expression back into the equation from which it was derived yields an identity (e.g. $0 = 0$ ) instead of a solution for the variable.
Fraction/Decimal Clearing Errors: Forgetting to multiply every single term (especially the RHS constant term) when clearing fractions or decimals.
Sign Errors in Parentheses Expansion: Making sign errors during distribution, such as expanding $-3(3x - 3)$ as $-9x - 9$ instead of $-9x + 9$ .

Question 1

More questions in Exercise 3.2