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

Substituting $x = 2y + 1$ into the second equation:

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

Expanding the equation:

$(4y^2 + 4y + 1) + y^2 = 29$

Combine like terms:

5y^2 + 4y - 28 = 0.$$

Applying the quadratic formula, where $a = 5$, $b = 4$, and $c = -28$:

$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 5 \cdot (-28)}}{2 \cdot 5}$

Calculating the discriminant:

$b^2 - 4ac = 16 + 560 = 576,$

thus,

$y = \frac{-4 \pm 24}{10}.$

This gives us:

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

Using the values of y to find x:

1. For $y = 2: x = 2(2) + 1 = 5$
2. For $y = -2.8: x = 2(-2.8) + 1 = -4.6.$

Thus, the pairs $(x, y)$ are:

1. (5, 2)
2. (-4.6, -2.8).