Solve algebraically the simultaneous equations $$2x^2 - y^2 = 17$$ $$x + 2y = 1$$ - Edexcel - GCSE Maths - Question 1 - 2018 - Paper 1

Now substitute $x = 1 - 2y$ into the first equation:

$2(1 - 2y)^2 - y^2 = 17$.

Expanding the equation gives:

$2(1 - 4y + 4y^2) - y^2 = 17$.

This simplifies to:

$2 - 8y + 8y^2 - y^2 = 17$, which becomes:

$7y^2 - 8y - 15 = 0$.

Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 7$, $b = -8$, and $c = -15$:

$y = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \times 7 \times (-15)}}{2 \times 7}$.

Calculating the discriminant:

$(-8)^2 - 4 \times 7 \times (-15) = 64 + 420 = 484$, thus,

$y = \frac{8 \pm \sqrt{484}}{14} = \frac{8 \pm 22}{14}$.

So,

$y = \frac{30}{14} = \frac{15}{7}$ or $y = \frac{-14}{14} = -1$.

Using the values of $y$ to find corresponding $x$:

1. For $y = \frac{15}{7}$:

$x = 1 - 2 \times \frac{15}{7} = 1 - \frac{30}{7} = \frac{7 - 30}{7} = -\frac{23}{7}$.

2. For $y = -1$:

$x = 1 - 2(-1) = 1 + 2 = 3$.

The solutions are:

1. $x = -\frac{23}{7}$ and $y = \frac{15}{7}$.
2. $x = 3$ and $y = -1$.