# How do you determine the solution in terms of a system of linear equations for x^2 + y^2 = 13, 2x-3y = -6?

Sep 23, 2015

Solutions are
$x = \frac{- 12 + 3 \sqrt{133}}{13}$, $y = \frac{18 + 2 \sqrt{133}}{13}$
and
$x = \frac{- 12 - 3 \sqrt{133}}{13}$, $y = \frac{18 - 2 \sqrt{133}}{13}$

#### Explanation:

Solutions may be found by substitution of the suitably rearranged first order equation (the second one in your particular system of equations) into the second order equation (the first one in your system). The quadratic will typically have two solutions for each substitution.

Taking the first equation

$2 x - 3 y = - 6$

And rearranging to find the value of one of the variables in terms of the other, for example (it would be possible to do this in either of two ways)

$2 x = 3 y - 6$

$\implies x = \frac{3}{2} y - 3$

This value of x may be substituted into the other equation (which is a quadratic), that is

${\left(\frac{3}{2} y - 3\right)}^{2} + {y}^{2} = 13$

$\implies \left(\frac{9}{4} {y}^{2} - \frac{18}{2} y + 9\right) + {y}^{2} - 13 = 0$

$\implies \frac{13}{4} {y}^{2} - \frac{18}{2} y - 4 = 0$

Multiplying throughout by 4 for convenience

$\implies 13 {y}^{2} - 36 y - 16 = 0$

This is a messy equation; it would be quickest in an exam to apply the quadratic formula rather than searching for factors. So, values of y satisfying this equation are given by:

$\frac{- \left(- 36\right) \pm \sqrt{{\left(- 36\right)}^{2} - 4 \left(13\right) \left(- 16\right)}}{2 \left(13\right)}$

$= \frac{36 \pm \sqrt{{\left(- 36\right)}^{2} - 4 \left(13\right) \left(- 16\right)}}{26}$

$= \frac{36 \pm \sqrt{2128}}{26}$

$= \frac{36 \pm \sqrt{2 \cdot 2 \cdot 2 \cdot 2 \cdot 133}}{26}$

$= \frac{36 \pm 4 \sqrt{133}}{26}$

$= \frac{18 \pm 2 \sqrt{133}}{13}$

The corresponding values of $x$ may be found by substituting these solutions into the first order equation (after it has been rearranged to give $x$ in terms of $y$; this has already been done).

When $y = \frac{18 + 2 \sqrt{133}}{13}$,

$x = \frac{3}{2} \left(\frac{18 + 2 \sqrt{133}}{13}\right) - 3$

$= 3 \left(\frac{9 + \sqrt{133}}{13}\right) - 3$

$= \left(\frac{27 + 3 \sqrt{133}}{13}\right) - \frac{39}{13}$

$= \frac{- 12 + 3 \sqrt{133}}{13}$.

When $y = \frac{18 - 2 \sqrt{133}}{13}$,

$x = \frac{3}{2} \left(\frac{18 - 2 \sqrt{133}}{13}\right) - 3$

$= 3 \left(\frac{9 - \sqrt{133}}{13}\right) - 3$

$= \left(\frac{27 - 3 \sqrt{133}}{13}\right) - \frac{39}{13}$

$= \frac{- 12 - 3 \sqrt{133}}{13}$.

Remember to check workings, and to check consistency of solutions by substitution into the given equations.