Solve the equation: $$\frac{9x - 6}{2} = \frac{3x - 14}{3} + \frac{9x}{4}$$ Solve the simultaneous equations: $$3x - y = 4$$ $$4x^2 - 3xy = 4.$$ - Leaving Cert Mathematics - Question 2 - 2020

From the first equation, we can isolate y:

$y = 3x - 4$

Now, substitute this expression for y into the second equation:

$4x^2 - 3x(3x - 4) = 4$

Expanding this gives:

$4x^2 - 9x^2 + 12x = 4$

Combining like terms yields:

$-5x^2 + 12x - 4 = 0$

Multiplying through by -1 for simplicity, we have:

$5x^2 - 12x + 4 = 0$

Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting in a = 5, b = -12, c = 4:

$x = \frac{12 \pm \sqrt{(-12)^2 - 4(5)(4)}}{2(5)}$

Calculating the discriminant gives:

$x = \frac{12 \pm \sqrt{144 - 80}}{10} = \frac{12 \pm \sqrt{64}}{10}$

This results in:

$x = \frac{12 \pm 8}{10}$

Thus, the two possible values for x are:

1. $x = \frac{20}{10} = 2$
2. $x = \frac{4}{10} = \frac{2}{5}$

Substituting these values back to find y: For $x = 2$: $y = 3(2) - 4 = 2$ For $x = \frac{2}{5}$: $y = 3(\frac{2}{5}) - 4 = \frac{6}{5} - 4 = \frac{6}{5} - \frac{20}{5} = -\frac{14}{5}$

The solutions to the simultaneous equations are therefore:

1. (2, 2)
2. $\left(\frac{2}{5}, -\frac{14}{5}\right)$.