we have the system of equations
[tex]\begin{gathered} \frac{1}{5}x+y=\frac{6}{5}\text{ ----> equation 1} \\ \\ \frac{1}{10}x+\frac{1}{3}y=\frac{7}{10}\text{ ----> equation 2} \end{gathered}[/tex]Solve by elimination
Multiply equation 2 by -2 on both sides
[tex]-2[\frac{1}{10}x+\frac{1}{3}y=\frac{7}{10}]---->-\frac{1}{5}x-\frac{2}{3}y=-\frac{14}{10}\text{ ---> equation 3}[/tex]Adds equation 1 and equation 3
[tex]\begin{gathered} \begin{equation*} \frac{1}{5}x+y=\frac{6}{5} \end{equation*} \\ \begin{equation*} -\frac{1}{5}x-\frac{2}{3}y=-\frac{14}{10} \end{equation*} \\ --------- \\ \frac{1}{3}y=\frac{6}{5}-\frac{7}{5} \\ \\ \frac{1}{3}y=-\frac{1}{5} \\ \\ y=-\frac{3}{5} \end{gathered}[/tex]Find out the value of x
substitute the value of y in any equation
equation 1
[tex]\begin{gathered} \frac{1}{5}x+(-\frac{3}{5})=\frac{6}{5} \\ \\ \frac{1}{5}x=\frac{6}{5}+\frac{3}{5} \\ \\ \frac{1}{5}x=\frac{9}{5} \\ \\ x=9 \end{gathered}[/tex]therefore