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

$$4x^{2} - 3x^{2} - 36 + 36 = -12 + 36$$ $$x^{2} = 24$$

$$x = \pm \sqrt{24}$$ $$x = \pm \sqrt{4 \cdot 6} = \pm 2\sqrt{6}$$

$2\sqrt{6} \approx 4.90 \neq \pm 2$ and $-2\sqrt{6} \approx -4.90 \neq \pm 2$. Both valid.

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

Question 1(b)

First term: $2x(2x - 1)$
Second term: $-3(x + 3)$
Right side: $x(x + 3)$

$$2x(2x - 1) - 3(x + 3) = x(x + 3)$$

Left: $4x^{2} - 2x - 3x - 9 = 4x^{2} - 5x - 9$
Right: $x^{2} + 3x$

$$4x^{2} - 5x - 9 = x^{2} + 3x$$

$$4x^{2} - 5x - 9 - x^{2} - 3x = 0$$ $$3x^{2} - 8x - 9 = 0$$

Discriminant: $b^{2} - 4ac = 64 - 4(3)(-9) = 64 + 108 = 172$

$$x = \frac{8 \pm \sqrt{172}}{6} = \frac{8 \pm 2\sqrt{43}}{6} = \frac{4 \pm \sqrt{43}}{3}$$

$\sqrt{43} \approx 6.56$, 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}