How do you solve 3x-y=-4 and x+3y=-28 using substitution?

Mar 31, 2016

The solution is $x = - 4$ and $y = - 8$.
Or, $\left(- 4 , - 8\right)$

Explanation:

One of the variables, either $x$ or $y$ , must be isolated . In this case, I will rearrange the second equation, $x + 3 y = - 28$ and isolate the $x$ variable.

$x + 3 y = - 28$ Now subtract $3 y$ from each side of the equation

$x + 3 y - 3 y = - 28 - 3 y$

$x = - 3 y - 28$

Now that I have isolated the $x$ variable in your second equation, I will substitute it in your first equation.

$3 x - y = - 4$

$3 \left(- 3 y - 28\right) - y = - 4$

$- 9 y - 84 - y = - 4$ collect like terms

$- 10 y - 84 = - 4$ Add 84 to each side

$- 10 y - 84 + 84 = - 4 + 84$

$- 10 y = 80$ Divide by $- 10$

$y = - 8$
Now simply substitute $- 8$ in for $y$ in either equation.

$x + 3 y = - 28$

$x + 3 \left(- 8\right) = - 28$

$x - 24 = - 28$ Now add $24$ to each side of the equation

$x - 24 + 24 = - 28 + 24$

$x = - 4$

$\left(- 4 , - 8\right)$

Now there is one small problem, and that is, how do I know this is the correct solution? Since I used your second equation, $x + 3 y = - 28$ to calculate the value of $y$, I can't use this same equation to check my answer. It would always appear to be correct. I must go to the other equation, in your case the first one, to verify that I have the correct answer. So:

$3 x - y = - 4$

$3 \left(- 4\right) - \left(- 8\right) = - 4$

$- 12 + 8 = - 4$

$- 4 = - 4$
This confirms that I have the solution correct.