Given the system of equations :
[tex]\begin{gathered} 2x-3y=4 \\ 2x+8y=11 \end{gathered}[/tex]
By subtract the first equation from the second equation :
[tex]\begin{gathered} (2x+8y)-(2x-3y)=11-4 \\ (2x-2x)+(8y+3y)=7 \end{gathered}[/tex][tex]\begin{gathered} 11y=7 \\ y=\frac{7}{11} \end{gathered}[/tex]
substitute with y at the first equation to find x
[tex]\begin{gathered} 2x-3\cdot\frac{7}{11}=4 \\ 2x-\frac{21}{11}=4 \\ 2x=4+\frac{21}{11}=\frac{65}{11} \\ \\ x=\frac{65}{22} \end{gathered}[/tex]
So, the solution of the system is :
[tex]\begin{gathered} x=\frac{65}{22} \\ \\ y=\frac{7}{11} \end{gathered}[/tex]
