Solution
Consider The following system of equations
2x+y=8
x-2y=-1
[tex]\begin{gathered} 2x+y=8...........(1) \\ x-2y=-1.............(2) \end{gathered}[/tex]
using substitution method:
[tex]\begin{gathered} 2x+y=8 \\ y=8-2x.........(3) \end{gathered}[/tex]
sub the value of y in equ(2)
[tex]\begin{gathered} x-2y=-1 \\ x-2(8-2x)=-1 \\ x-16+4x=-1 \\ 5x=-1+16 \\ 5x=15 \\ x=\frac{15}{5} \\ x=3 \end{gathered}[/tex]
Sub the value of x in equ(1)
[tex]\begin{gathered} 2x+y=8 \\ 2(3)+y=8 \\ 6+y=8 \\ y=8-6 \\ y=2 \end{gathered}[/tex]
Therefore the solution of the equations are
[tex]\begin{gathered} x=3 \\ y=2 \end{gathered}[/tex]
