# How do you solve the following system: 5x + 8y = -2, 4x + 7y = 6x?

Aug 5, 2017

$x = - \frac{14}{39} \mathmr{and} y = - \frac{4}{39}$

#### Explanation:

We have: $5 x + 8 y = - 2 \mathmr{and} 4 x + 7 y = 6 x$

Let's solve both equations for $y$:

$R i g h t a r r o w 8 y = - 5 x - 2 \mathmr{and} 7 y = 6 x - 4 x$

$R i g h t a r r o w y = \frac{- 5 x - 2}{8} \mathmr{and} y = \frac{2 x}{7}$

Then, let's set the two expressions for $y$ equal to each other:

$R i g h t a r r o w \frac{- 5 x - 2}{8} = \frac{2 x}{7}$

$R i g h t a r r o w 7 \times \left(- 5 x - 2\right) = 8 \times 2 x$

$R i g h t a r r o w - 35 x - 14 = 4 x$

$R i g h t a r r o w - 35 x - 4 x - 14 = 0$

$R i g h t a r r o w - 39 x - 14 = 0$

$R i g h t a r r o w - 39 x = 14$

$\therefore x = - \frac{14}{39}$

Now that we have a value of $x$, let's find $y$ using one of the expressions:

$R i g h t a r r o w y = \frac{2 x}{7}$

$R i g h t a r r o w y = \frac{2 \times - \frac{14}{39}}{7}$

$R i g h t a r r o w y = \frac{- \frac{28}{39}}{7}$

$R i g h t a r r o w y = \frac{- \frac{28}{39}}{\frac{7}{1}}$

$R i g h t a r r o w y = - \frac{28}{39} \times \frac{1}{7}$

$R i g h t a r r o w y = - \frac{28 \times 1}{39 \times 7}$

$R i g h t a r r o w y = - \frac{28}{273}$

$\therefore y = - \frac{4}{39}$

Therefore, the solutions to the equation are $x = - \frac{14}{39}$ and $y = - \frac{4}{39}$.