Question 1

We will solve each pair of linear equations using both elimination and substitution methods.

Part (i)

Step 1 — Elimination Method for (i) Let's write down the given equations.

$x + y = 5 \quad \text{(1)}$

$2x - 3y = 4 \quad \text{(2)}$

We will multiply equation (1) by 2.

$2(x + y) = 2(5)$

$2x + 2y = 10 \quad \text{(3)}$

Now, we subtract equation (3) from equation (2).

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

$2x - 3y - 2x - 2y = -6$

$-5y = -6$

$y = \frac{6}{5}$

Step 2 — Find x using Elimination Method for (i) We substitute the value of y into equation (1).

$x + \frac{6}{5} = 5$

$x = 5 - \frac{6}{5}$

$x = \frac{25 - 6}{5}$

$x = \frac{19}{5}$

Step 3 — Substitution Method for (i) Let's use the given equations again.

$x + y = 5 \quad \text{(1)}$

$2x - 3y = 4 \quad \text{(2)}$

From equation (1), we express x in terms of y.

$x = 5 - y \quad \text{(4)}$

Now, we substitute this expression for x into equation (2).

$2(5 - y) - 3y = 4$

$10 - 2y - 3y = 4$

$10 - 5y = 4$

$-5y = 4 - 10$

$-5y = -6$

$y = \frac{6}{5}$

Step 4 — Find x using Substitution Method for (i) We substitute the value of y into equation (4).

$x = 5 - \frac{6}{5}$

$x = \frac{25 - 6}{5}$

$x = \frac{19}{5}$

Part (ii)

Step 5 — Elimination Method for (ii) Let's write down the given equations.

$3x + 4y = 10 \quad \text{(1)}$

$2x - 2y = 2 \quad \text{(2)}$

We will multiply equation (2) by 2.

$2(2x - 2y) = 2(2)$

$4x - 4y = 4 \quad \text{(3)}$

Now, we add equation (1) and equation (3).

$(3x + 4y) + (4x - 4y) = 10 + 4$

$3x + 4y + 4x - 4y = 14$

$7x = 14$

$x = 2$

Step 6 — Find y using Elimination Method for (ii) We substitute the value of x into equation (1).

$3(2) + 4y = 10$

$6 + 4y = 10$

$4y = 10 - 6$

$4y = 4$

$y = 1$

Step 7 — Substitution Method for (ii) Let's use the given equations again.

$3x + 4y = 10 \quad \text{(1)}$

$2x - 2y = 2 \quad \text{(2)}$

From equation (2), we can simplify it by dividing by 2.

$x - y = 1$

We express x in terms of y.

$x = 1 + y \quad \text{(4)}$

Now, we substitute this expression for x into equation (1).

$3(1 + y) + 4y = 10$

$3 + 3y + 4y = 10$

$3 + 7y = 10$

$7y = 10 - 3$

$7y = 7$

$y = 1$

Step 8 — Find x using Substitution Method for (ii) We substitute the value of y into equation (4).

$x = 1 + 1$

$x = 2$

Part (iii)

Step 9 — Elimination Method for (iii) Let's write down the given equations.

$3x - 5y - 4 = 0$

$3x - 5y = 4 \quad \text{(1)}$

And,

$9x = 2y + 7$

$9x - 2y = 7 \quad \text{(2)}$

We will multiply equation (1) by 3.

$3(3x - 5y) = 3(4)$

$9x - 15y = 12 \quad \text{(3)}$

Now, we subtract equation (3) from equation (2).

$(9x - 2y) - (9x - 15y) = 7 - 12$

$9x - 2y - 9x + 15y = -5$

$13y = -5$

$y = -\frac{5}{13}$

Step 10 — Find x using Elimination Method for (iii) We substitute the value of y into equation (1).

$3x - 5\left(-\frac{5}{13}\right) = 4$

$3x + \frac{25}{13} = 4$

$3x = 4 - \frac{25}{13}$

$3x = \frac{52 - 25}{13}$

$3x = \frac{27}{13}$

$x = \frac{27}{13 \times 3}$

$x = \frac{9}{13}$

$x = \frac{9}{13}$

Step 11 — Substitution Method for (iii) Let's use the given equations again.

$3x - 5y = 4 \quad \text{(1)}$

$9x - 2y = 7 \quad \text{(2)}$

From equation (1), we express x in terms of y.

$3x = 4 + 5y$

$x = \frac{4 + 5y}{3} \quad \text{(4)}$

Now, we substitute this expression for x into equation (2).

$9\left(\frac{4 + 5y}{3}\right) - 2y = 7$

$3(4 + 5y) - 2y = 7$

$12 + 15y - 2y = 7$

$12 + 13y = 7$

$13y = 7 - 12$

$13y = -5$

$y = -\frac{5}{13}$

Step 12 — Find x using Substitution Method for (iii) We substitute the value of y into equation (4).

$x = \frac{4 + 5\left(-\frac{5}{13}\right)}{3}$

$x = \frac{4 - \frac{25}{13}}{3}$

$x = \frac{\frac{52 - 25}{13}}{3}$

$x = \frac{\frac{27}{13}}{3}$

$x = \frac{27}{13 \times 3}$

$x = \frac{9}{13}$

$x = \frac{9}{13}$

Part (iv)

Step 13 — Simplify Equations for (iv) Let's simplify the given equations first.

$\frac{x}{2} + \frac{2y}{3} = -1$

Multiply by 6 to clear fractions.

$6\left(\frac{x}{2}\right) + 6\left(\frac{2y}{3}\right) = 6(-1)$

$3x + 4y = -6 \quad \text{(1)}$

Now for the second equation.

$x - \frac{y}{3} = 3$

Multiply by 3 to clear fractions.

$3(x) - 3\left(\frac{y}{3}\right) = 3(3)$

$3x - y = 9 \quad \text{(2)}$

Step 14 — Elimination Method for (iv) We have the simplified equations.

$3x + 4y = -6 \quad \text{(1)}$

$3x - y = 9 \quad \text{(2)}$

We subtract equation (2) from equation (1).

$(3x + 4y) - (3x - y) = -6 - 9$

$3x + 4y - 3x + y = -15$

$5y = -15$

$y = -3$

Step 15 — Find x using Elimination Method for (iv) We substitute the value of y into equation (1).

$3x + 4(-3) = -6$

$3x - 12 = -6$

$3x = -6 + 12$

$3x = 6$

$x = 2$

Step 16 — Substitution Method for (iv) Let's use the simplified equations again.

$3x + 4y = -6 \quad \text{(1)}$

$3x - y = 9 \quad \text{(2)}$

From equation (2), we express y in terms of x.

$-y = 9 - 3x$

$y = 3x - 9 \quad \text{(3)}$

Now, we substitute this expression for y into equation (1).

$3x + 4(3x - 9) = -6$

$3x + 12x - 36 = -6$

$15x - 36 = -6$

$15x = -6 + 36$

$15x = 30$

$x = 2$

Step 17 — Find y using Substitution Method for (iv) We substitute the value of x into equation (3).

$y = 3(2) - 9$

$y = 6 - 9$

$y = -3$

Answer

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

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

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

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