Use an inverse matrix approach to solve the system of linear equation: 3x + 4y = -2 and 5x + 3y = 4

Question 1(a)

Step 1: Cross-multiply to eliminate fractions.
$4(x^{2} - 9) = 3(x^{2} - 4)$

Step 2: Expand both sides.
$4x^{2} - 36 = 3x^{2} - 12$

Step 3: Subtract $3x^{2}$ from both sides and add 36 to both sides.
$4x^{2} - 3x^{2} - 36 + 36 = -12 + 36$
$x^{2} = 24$

Step 4: Take square root of both sides.
$x = \pm \sqrt{24}$
$x = \pm \sqrt{4 \cdot 6} = \pm 2\sqrt{6}$

Step 5: Verify solutions satisfy $x \neq \pm 2$.
$2\sqrt{6} \approx 4.90 \neq \pm 2$ and $-2\sqrt{6} \approx -4.90 \neq \pm 2$. Both valid.

x = 2\sqrt{6,\ -2\sqrt{6}}

Question 1(b)

Step 1: Multiply both sides by $2x(x + 3)$ to clear denominators ($x \neq 0$, $x \neq -3$).
$2x(x + 3) \left( \frac{2x - 1}{x + 3} - \frac{3}{2x} \right) = 2x(x + 3) \cdot \frac{1}{2}$

Step 2: Distribute on left side.
First term: $2x(2x - 1)$
Second term: $-3(x + 3)$
Right side: $x(x + 3)$
$2x(2x - 1) - 3(x + 3) = x(x + 3)$

Step 3: Expand all terms.
Left: $4x^{2} - 2x - 3x - 9 = 4x^{2} - 5x - 9$
Right: $x^{2} + 3x$
$4x^{2} - 5x - 9 = x^{2} + 3x$

Step 4: Bring all terms to left side.
$4x^{2} - 5x - 9 - x^{2} - 3x = 0$
$3x^{2} - 8x - 9 = 0$

Step 5: Solve using quadratic formula $x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$, where $a=3$, $b=-8$, $c=-9$.
Discriminant: $b^{2} - 4ac = 64 - 4(3)(-9) = 64 + 108 = 172$
$x = \frac{8 \pm \sqrt{172}}{6} = \frac{8 \pm 2\sqrt{43}}{6} = \frac{4 \pm \sqrt{43}}{3}$

Step 6: Verify solutions satisfy $x \neq 0$, $x \neq -3$.
$\sqrt{43} \approx 6.56$, so $x \approx \frac{4 + 6.56}{3} \approx 3.52 \neq 0, -3$
$x \approx \frac{4 - 6.56}{3} \approx -0.85 \neq 0, -3$. Both valid.

x = \dfrac{4 + \sqrt{43}{3},\ \dfrac{4 - \sqrt{43}}{3}}