How do you solve the following system: x+3y=-2, 5x + 2y = 5 ?

Jun 13, 2018

By arranging the equation

Explanation:

Expand the first equation by $- 5$:

$- 5 x - 15 y = 10$

Now add this to the second equation:

$5 x + 2 y = 5$

Yielding:

$- 13 y = 15$

$y = - \frac{15}{13}$

Put this value in the first original equation:

$x + 3 \left(- \frac{15}{13}\right) = - 2$

$x - \frac{45}{13} = - 2$

$x = - 2 + \frac{45}{13}$

$x = \frac{- 26 + 45}{13}$

$x = \frac{19}{13}$

Jun 13, 2018

$\text{the solution is shown below.}$

Explanation:

• You can use the "elimination" method to solve the equation system. To do this, you must destroy one of two variables (x, y). You have to equalize the coefficients in both equations by choosing one of the variables 'x' or 'y'.

• x+3y=-2$\text{ }$ (1) ; 5x+2y=5 $\text{ }$ (2)

• If both sides of equation (1) are multiplied by 5, the coefficients of 'x' are equal in both equations.

$\textcolor{red}{5} \left(x + 3 y\right) = - \textcolor{red}{5} \left(2\right) x +$
$5 x + 15 y = - 10 \text{ } \left(3\right)$

• Now let's subtract equation (3) from equation (2).

• 5x+2y-(5x+15y)=5-(-10)
$\cancel{5 x} + 2 y \cancel{- 5 x} - 15 y = 5 + 10$
$- 13 y = 15$

$y = - \frac{15}{13}$

• Finally, in (1) or (2), write -15/13 instead of 'y'.

$x + 3 \left(- \frac{15}{13}\right) = - 2$

$x - \frac{45}{13} = - 2$

$x = - 2 + \frac{45}{13}$

$x = \frac{- 26 + 45}{13}$

$x = - \frac{19}{13}$