A solution to the given system of equations is [tex](\frac{100}{3}, \frac{8}{3} )[/tex]
Given the following system of equations:
[tex]y=-2x-64[/tex] ...equation 1.
[tex]x-y=-36[/tex] ...equation 2.
To find a solution to the given system of equations, we would use the substitution method:
First of all, we would make x the subject of formula in equation 2.
[tex]x=-36+y[/tex] ...equation 3.
Substituting eqn. 3 into eqn. 1, we have:
[tex]y = -2(-36+y)-64\\\\y=72-2y-64\\\\y+2y=72-64\\\\3y=8\\\\y=\frac{8}{3}[/tex]
For the value of x:
[tex]x=-36+y\\\\x=-36+\frac{8}{3} \\\\x=\frac{100}{3}[/tex]
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