Substituting x= 3 -7y/3 into y= 9/7 -3x/7, we have
[tex]y=\frac{9}{7}-\frac{3(3-\frac{7y}{3})}{7}[/tex]Therefore,
[tex]7y=9-3(3-\frac{7y}{3})=9-9+7y[/tex]This implies that,
[tex]\begin{gathered} 7y=7y \\ \Rightarrow0=0 \end{gathered}[/tex]So the number of solutions is infinite.
We can set x = 1, then
[tex]y=\frac{9}{7}-\frac{3}{7}=\frac{6}{7}[/tex]Hence, (1, 6/7) is a solution
We can also set y = 3, then
[tex]x=3-\frac{7\mleft(3\mright)}{3}=3-7=-4[/tex]Hence, (3, -4) is another solution
