# How do you solve the simultaneous equations x – 2y = 1 and x^2 + y^2 = 29?

Jul 28, 2015

I found:
$x = 5$; $y = 2$
$x = - \frac{23}{5}$;$y = - \frac{15}{5}$

#### Explanation:

We can try isolate $x$ from the first equation:
$x = 1 + 2 y$
substitute into the second (for $x$):
${\left(\textcolor{red}{1 + 2 y}\right)}^{2} + {y}^{2} = 29$
$1 + 4 y + 4 {y}^{2} + {y}^{2} = 29$
$5 {y}^{2} + 4 y - 28 = 0$
solve for $y$ (using the Quadratic Formula) gives you:
${y}_{1 , 2} = \frac{- 4 \pm \sqrt{16 + 560}}{10} = \frac{- 4 \pm 24}{10}$
you get two solutions:
${y}_{1} = - \frac{28}{10} = - \frac{14}{5}$
${y}_{2} = \frac{20}{10} = 2$
Substituted back to find $x$ you get:
if $y = 2$ then $x = 1 + 4 = 5$
if $y = - \frac{15}{5}$ then $x = 1 - 2 \cdot \frac{14}{5} = - \frac{23}{5}$