Solve the simultaneous equations $x - 2y = 1,$ $x^2 + y^2 = 29.$ - Edexcel - A-Level Maths Pure - Question 7 - 2005 - Paper 1

Next, substitute this expression for x into the second equation:

$x^2 + y^2 = 29$

Substituting gives:

$(2y + 1)^2 + y^2 = 29.$

Expanding the left side:

5y^2 + 4y + 1 - 29 = 0$$ This simplifies to: $$5y^2 + 4y - 28 = 0.$$

We can now apply the quadratic formula:

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

For our equation, a = 5, b = 4, and c = -28:

1. Calculate the discriminant: $b^2 - 4ac = 4^2 - 4(5)(-28) = 16 + 560 = 576.$

2. Now apply the formula:

$y = \frac{-4 \pm \sqrt{576}}{2(5)} = \frac{-4 \pm 24}{10}.$

This gives:

1. $y = \frac{20}{10} = 2,$
2. $y = \frac{-28}{10} = -2.8.$

Now finding the corresponding x values for each y:

1. For $y = 2$: $x = 2(2) + 1 = 4 + 1 = 5.$

2. For $y = -2.8$: $x = 2(-2.8) + 1 = -5.6 + 1 = -4.6.$

The solutions to the simultaneous equations are:

1. $(x, y) = (5, 2)$
2. $(x, y) = (-4.6, -2.8)$.