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

Next, substitute y with the expression obtained from the first equation:

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

Expanding the squared term:

$(16x^2 + 8x + 1) + 5x^2 + 2x = 0$.

Combine like terms:

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

Now we will use the quadratic formula to solve for x:

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

where (a = 21), (b = 10), and (c = 1). Substituting these values gives:

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

Simplifying further:

$x = \frac{-10 \pm \sqrt{100 - 84}}{42}$

$x = \frac{-10 \pm \sqrt{16}}{42}$

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

This results in two possible solutions for x:

1. (x = \frac{-6}{42} = -\frac{1}{7})
2. (x = \frac{-14}{42} = -\frac{1}{3})

Now substitute both x values back into the equation for y:

For (x = -\frac{1}{7}):

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

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 solutions for the simultaneous equations are:

1. (\left(-\frac{1}{7}, -\frac{3}{7}\right))
2. (\left(-\frac{1}{3}, \frac{1}{3}\right))