Given the following system of equations:
[tex]\begin{gathered} x+6y=20 \\ -6x-8y=20 \end{gathered}[/tex]to solve by substitution, first notice that we can solve for x on the first equation to get the following:
[tex]x=20-6y[/tex]substituting this value on the second equation we get:
[tex]\begin{gathered} -6(20-6y)-8y=20 \\ \Rightarrow-120+36y-8y=20 \\ \Rightarrow36y-8y=20+120 \\ \Rightarrow28y=140 \\ \Rightarrow y=\frac{140}{28}=5 \\ y=5 \end{gathered}[/tex]now that we have that y=5, we can find the value of x by using our first equation:
[tex]\begin{gathered} x=20-6(5)=20-30=-10 \\ x=-10 \end{gathered}[/tex]therefore, the solution is (-10,5)
