Solve the simultaneous equations y + 4x + 1 = 0 y^2 + 5x^2 + 2x = 0 - Edexcel - A-Level Maths Pure - Question 7 - 2016 - Paper 1

Substituting for y gives us:

$(-4x - 1)^2 + 5x^2 + 2x = 0$

Expanding the left side:

$16x^2 + 8x + 1 + 5x^2 + 2x = 0$

Combine like terms:

$21x^2 + 10x + 1 = 0$

This is now a quadratic equation in the form of $ax^2 + bx + c = 0$. We can use the quadratic formula:

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

Substituting a = 21, b = 10, and c = 1:

$x = \frac{-10 \pm \sqrt{10^2 - 4 \cdot 21 \cdot 1}}{2 \cdot 21} = \frac{-10 \pm \sqrt{100 - 84}}{42} = \frac{-10 \pm \sqrt{16}}{42}$

This simplifies to:

$x = \frac{-10 \pm 4}{42}$

Thus, we find the two values for x:

1. When we add:

$x = \frac{-6}{42} = -\frac{1}{7}$

2. When we subtract:

$x = \frac{-14}{42} = -\frac{1}{3}$

Now substituting these x values back into the equation for y:

1. For $x = -\frac{1}{7}$:

$y = -4(-\frac{1}{7}) - 1 = \frac{4}{7} - 1 = \frac{4}{7} - \frac{7}{7} = -\frac{3}{7}$

2. For $x = -\frac{1}{3}$:

$y = -4(-\frac{1}{3}) - 1 = \frac{4}{3} - 1 = \frac{4}{3} - \frac{3}{3} = \frac{1}{3}$

Thus, the final pairs of solutions are:

• For $x = -\frac{1}{7}$, $y = -\frac{3}{7}$
• For $x = -\frac{1}{3}$, $y = \frac{1}{3}$.