Question 1

We will solve each pair of linear equations using the substitution method.

Step 1 — Solve (i)

Let's write down the given equations. Equation (1) is $x + y = 14$. Equation (2) is $x - y = 4$. We express $x$ from Equation (2).

$x = 4 + y$

Now, we substitute this expression for $x$ into Equation (1). This helps us find the value of $y$.

$(4 + y) + y = 14$

$4 + 2y = 14$

$2y = 14 - 4$

$2y = 10$

$\boxed{y = 5}$

We substitute $y = \textbf{5}$ back into the expression for $x$. This gives us the value of $x$.

$x = 4 + 5$

$\boxed{x = 9}$

Step 2 — Solve (ii)

Let's write down the given equations. Equation (1) is $s - t = 3$. Equation (2) is $\frac{s}{3} + \frac{t}{2} = 6$. First, we simplify Equation (2). We multiply Equation (2) by 6 to clear fractions.

$6 \left(\frac{s}{3} + \frac{t}{2}\right) = 6 \times 6$

$2s + 3t = 36$

Let's call this new equation Equation (3). Now, we express $s$ from Equation (1).

$s = 3 + t$

We substitute this expression for $s$ into Equation (3). This helps us find the value of $t$.

$2(3 + t) + 3t = 36$

$6 + 2t + 3t = 36$

$6 + 5t = 36$

$5t = 36 - 6$

$5t = 30$

$\boxed{t = 6}$

We substitute $t = \textbf{6}$ back into the expression for $s$. This gives us the value of $s$.

$s = 3 + 6$

$\boxed{s = 9}$

Step 3 — Solve (iii)

Let's write down the given equations. Equation (1) is $3x - y = 3$. Equation (2) is $9x - 3y = 9$. We express $y$ from Equation (1).

$y = 3x - 3$

Now, we substitute this expression for $y$ into Equation (2). This helps us find the solution.

$9x - 3(3x - 3) = 9$

$9x - 9x + 9 = 9$

$9 = 9$

This statement is always true. This means the two equations are dependent. They represent the same line. Therefore, there are infinitely many solutions.

$\boxed{\text{Infinitely many solutions}}$

Step 4 — Solve (iv)

Let's write down the given equations. Equation (1) is $0.2x + 0.3y = 1.3$. Equation (2) is $0.4x + 0.5y = 2.3$. First, we simplify both equations. We multiply Equation (1) by 10 to clear decimals.

$10(0.2x + 0.3y) = 10(1.3)$

$2x + 3y = 13$

Let's call this Equation (3). We multiply Equation (2) by 10 to clear decimals.

$10(0.4x + 0.5y) = 10(2.3)$

$4x + 5y = 23$

Let's call this Equation (4). Now, we express $x$ from Equation (3).

$2x = 13 - 3y$

$x = \frac{13 - 3y}{2}$

We substitute this expression for $x$ into Equation (4). This helps us find the value of $y$.

$4\left(\frac{13 - 3y}{2}\right) + 5y = 23$

$2(13 - 3y) + 5y = 23$

$26 - 6y + 5y = 23$

$26 - y = 23$

$-y = 23 - 26$

$-y = -3$

$\boxed{y = 3}$

We substitute $y = \textbf{3}$ back into the expression for $x$. This gives us the value of $x$.

$x = \frac{13 - 3(3)}{2}$

$x = \frac{13 - 9}{2}$

$x = \frac{4}{2}$

$\boxed{x = 2}$

Step 5 — Solve (v)

Let's write down the given equations. Equation (1) is $\sqrt{2}x + \sqrt{3}y = 0$. Equation (2) is $\sqrt{3}x - \sqrt{8}y = 0$. First, we simplify Equation (2). We know that $\sqrt{8}$ is $2\sqrt{2}$. So, Equation (2) becomes $\sqrt{3}x - 2\sqrt{2}y = 0$. Now, we express $x$ from Equation (1).

$\sqrt{2}x = -\sqrt{3}y$

$x = -\frac{\sqrt{3}}{\sqrt{2}}y$

We substitute this expression for $x$ into the simplified Equation (2). This helps us find the value of $y$.

$\sqrt{3}\left(-\frac{\sqrt{3}}{\sqrt{2}}y\right) - 2\sqrt{2}y = 0$

$-\frac{3}{\sqrt{2}}y - 2\sqrt{2}y = 0$

We multiply the entire equation by $\sqrt{2}$.

$-3y - 2\sqrt{2}(\sqrt{2})y = 0$

$-3y - 4y = 0$

$-7y = 0$

$\boxed{y = 0}$

We substitute $y = \textbf{0}$ back into the expression for $x$. This gives us the value of $x$.

$x = -\frac{\sqrt{3}}{\sqrt{2}}(0)$

$\boxed{x = 0}$

Step 6 — Solve (vi)

Let's write down the given equations. Equation (1) is $\frac{3x}{2} - \frac{5y}{3} = -2$. Equation (2) is $\frac{x}{3} + \frac{y}{2} = \frac{13}{6}$. First, we simplify both equations. We multiply Equation (1) by 6 to clear fractions.

$6\left(\frac{3x}{2} - \frac{5y}{3}\right) = 6(-2)$

$9x - 10y = -12$

Let's call this Equation (3). We multiply Equation (2) by 6 to clear fractions.

$6\left(\frac{x}{3} + \frac{y}{2}\right) = 6\left(\frac{13}{6}\right)$

$2x + 3y = 13$

Let's call this Equation (4). Now, we express $x$ from Equation (4).

$2x = 13 - 3y$

$x = \frac{13 - 3y}{2}$

We substitute this expression for $x$ into Equation (3). This helps us find the value of $y$.

$9\left(\frac{13 - 3y}{2}\right) - 10y = -12$

We multiply the entire equation by 2.

$9(13 - 3y) - 20y = -24$

$117 - 27y - 20y = -24$

$117 - 47y = -24$

$-47y = -24 - 117$

$-47y = -141$

$y = \frac{-141}{-47}$

$\boxed{y = 3}$

We substitute $y = \textbf{3}$ back into the expression for $x$. This gives us the value of $x$.

$x = \frac{13 - 3(3)}{2}$

$x = \frac{13 - 9}{2}$

$x = \frac{4}{2}$

$\boxed{x = 2}$

Answer

(i) $x = 9, y = 5$ (ii) $s = 9, t = 6$ (iii) Infinitely many solutions (iv) $x = 2, y = 3$ (v) $x = 0, y = 0$ (vi) $x = 2, y = 3$