Solve algebraically the simultaneous equations $x^2 - 4y = 9$ $3x + 4y = 7$ - Edexcel - GCSE Maths - Question 21 - 2019 - Paper 3

Substitute this expression for $y$ into the second equation:

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

Simplifying gives:

$3x + (x^2 - 9) = 7$

Combine like terms to form a quadratic equation:

$x^2 + 3x - 16 = 0$

Now, apply the quadratic formula to solve for $x$:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1$, $b = 3$, and $c = -16$:

$x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-16)}}{2 \cdot 1} = \frac{-3 \pm \sqrt{81}}{2}$

This results in:

$x = \frac{-3 + 9}{2} = 3 \quad \text{or} \quad x = \frac{-3 - 9}{2} = -6$

Substituting both values of $x$ back into the equation for $y$ to find corresponding $y$ values:

For $x = 3$:

$y = \frac{3^2 - 9}{4} = \frac{0}{4} = 0$

For $x = -6$:

$y = \frac{(-6)^2 - 9}{4} = \frac{27}{4}$

Thus, the solutions for the simultaneous equations are:

1. $(3, 0)$
2. $\left(-6, \frac{27}{4}\right)$