# How do you solve the system 2x+5y=4 and x+y=-1 by substitution?

May 26, 2015

$x = - 3$ and $y = 2$ .

Solve $x + y = - 1$ for $x$.

$x = - y - 1$

Substitute $- y - 1$ for $x$ into the other equation and solve for $y$.

$2 x + 5 y = 4$ =

$2 \left(- y - 1\right) + 5 y = 4$ =

$- 2 y - 2 + 5 y = 4$ =

$3 y - 2 = 4$ =

$3 y = 6$

Divide both sides by $3$.

$y = 2$

Substitute $2$ for $y$ into the first equation, and solve for $x$.

$x + y = - 1$ =

$x + 2 = - 1$ =

$x = - 3$

Check by substituting the $x$ and $y$ values into both equations.

$x + y = - 1$ =

$- 3 + 2 = - 1$ =

$- 1 = - 1$

$2 x + 5 y = 4$ =

#2(-3)+5(2)=4

$- 6 + 10 = 4$ =

$4 = 4$