Solve algebraically the simultaneous equations $$2x^2 - y^2 = 17$$ $$x + 2y = 1$$ - Edexcel - GCSE Maths - Question 20 - 2018 - Paper 3

Next, we substitute this expression for $x$ into the first equation:

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

Now we expand and simplify:

\Rightarrow 2 - 8y + 8y^2 - y^2 = 17 \\ \Rightarrow 7y^2 - 8y - 15 = 0$$

We can use the quadratic formula to solve for $y$:

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

For our equation $7y^2 - 8y - 15 = 0$, we have:

• $a = 7$
• $b = -8$
• $c = -15$

Calculating the discriminant:

$D = (-8)^2 - 4 \cdot 7 \cdot (-15) = 64 + 420 = 484$

Now applying the quadratic formula:

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

This gives us two possible values for $y$:

\text{and} \\ y_2 = \frac{-14}{14} = -1$$

Now we will find the corresponding values of $x$ for both values of $y$:

1. For $y_1 = \frac{15}{7}$:

$x = 1 - 2\left(\frac{15}{7}\right) = 1 - \frac{30}{7} = -\frac{23}{7}$

2. For $y_2 = -1$:

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

Thus we have two pairs of solutions:

• $\left(-\frac{23}{7}, \frac{15}{7}\right)$
• $(3, -1)$