How do you solve the system 2x+y=-9 and 3x+y=11?

May 24, 2015

Noticing the same term in $y$ in both equations, we can eliminate $y$ quickly as follows:

Subtract the first equation from the second to get:

$x = \left(3 x + y\right) - \left(2 x + y\right) = 11 - \left(- 9\right) = 20$

Substitute this value of $x$ into the first equation to get:

$- 9 = \left(2 \times 20\right) + y = 40 + y$

Subtract $40$ from both sides to get:

$y = - 9 - 40 = - 49$

As a quick check, substitute these values for $x$ and $y$ into the left hand side of the second equation:

$3 x + y = \left(3 \times 20\right) + \left(- 49\right) = 60 - 49 = 11$