Solve the simultaneous equations $$ x+y=2$$ $$ 4y^2 - x^2 = 11$$ - Edexcel - A-Level Maths Pure - Question 5 - 2011 - Paper 1

Next, we substitute $y$ from the first equation into the second equation:

$4(2 - x)^2 - x^2 = 11$.

Expanding this gives: $4(4 - 4x + x^2) - x^2 = 11$ $16 - 16x + 4x^2 - x^2 = 11$ $3x^2 - 16x + 16 - 11 = 0$.

Thus, we simplify to: $3x^2 - 16x + 5 = 0$.

We can solve this quadratic using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 3$, $b = -16$, and $c = 5$:

1. Calculate the discriminant: $b^2 - 4ac = (-16)^2 - 4(3)(5) = 256 - 60 = 196$
2. Substitute into the formula: $x = \frac{16 \pm \sqrt{196}}{6} = \frac{16 \pm 14}{6}$ This gives us:
   • For $x = \frac{30}{6} = 5$
   • For $x = \frac{2}{6} = \frac{1}{3}$.

Now that we have $x$ values, we can find corresponding $y$ values:

1. If $x = 5$: $y = 2 - 5 = -3$.
2. If $x = \frac{1}{3}$: $y = 2 - \frac{1}{3} = \frac{6}{3} - \frac{1}{3} = \frac{5}{3}$.

The solutions to the simultaneous equations are:

1. $\mathbf{(5, -3)}$
2. $\mathbf{\left(\frac{1}{3}, \frac{5}{3}\right)}$.