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

Substituting into the second equation gives:

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

This simplifies to:

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

Which further simplifies to:

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

We can factor this quadratic:

$$(x - 4)(x + 4) = 0$$

Thus, we find:

$$x = 4 ext{ or } x = -4$$

Now substituting each value of x back into the expression for y:

1. For $x = 4$:

   $$y = \frac{4^2 - 9}{4} = \frac{16 - 9}{4} = \frac{7}{4}$$

2. For $x = -4$:

   $$y = \frac{(-4)^2 - 9}{4} = \frac{16 - 9}{4} = \frac{7}{4}$$

The two pairs of solutions are:

• For $x = 4$, $y = \frac{7}{4}$
• For $x = -4$, $y = \frac{7}{4}$

Thus, the solutions are (4, \frac{7}{4}) and (-4, \frac{7}{4}).