[tex]Here's\ the\ solution; \\ \\ Elimination\ method; \\ \\ 3x-\not{y}=5\to equation\ 1 \\ 2x+\not{y}=10\to equation\ 2 \\ ^{--------} \\ \frac{\not5x}{\not5}=\frac{15}{5} \\ \\ \boxed{x=3} \\ \\ Solving\ for\ y: \\ by\ using\ equation\ 1 \\ \\ 3x-y=5 \\ \\ 3(3)-y=5\ \ \ \ \ \ \ \ |\ Substitute\ x\to3 \\ \\ 9-y=5 \\ \\ -y=5-9 \\ \\ \frac{-y}{-1}=\frac{-4}{-1} \\ \\ \boxed{y=4} [/tex]
[tex]You\ can\ use\ equation\ 2\ if\ you\ want\ too\ in\ finding \\ the\ value\ of\ y. \\ Well,\ here's\ the\ solution\ by\ using\ equation\ 2. \\ \\ 2x+y=10 \\ \\ 2(3)+y=10\ \ \ \ \ \ \ \ \ |\ Substitute\ x\to3 \\ \\ 6+y=10 \\ \\ y=10-6 \\ \\ \boxed{y=4} \\ \\ \\ Conclusion: \\ Therefore,\ the\ value\ of\ x\ and\ y\ is\ 3\ and\ 4\ respectively. \\ \\ Hope\ it\ Helps\ :) \\ Domini[/tex]
