Question

I need help... I don't m is how to do this

Answer 1

U have to find what is x to find the y

Answer 2

4)

$\begin{cases} x-2y=10\\ \frac{1}{2}x-2y=4 \end{cases}\\\\ -----------------------------\\\\ \textit{now, we solve for the 1st equation for either, in this case, "y"}\\\\ x-2y=10\implies -2y=10-x\implies y=\cfrac{10-x}{-2}\implies \boxed{y}=\cfrac{x}{2}-5\\\\ -----------------------------\\\\ \textit{and we use that "y" from the 1st one, IN the 2nd one}\\\\ \cfrac{1}{2}x-2\boxed{y}=4\implies \cfrac{1}{2}x-2\left( \boxed{\cfrac{x}{2}-5}\right)=4$

$\bf \cfrac{x}{2}-2\cdot \cfrac{x}{2}-2\cdot -5=4\implies \cfrac{x}{2}-x+10=4 \\\\\\ \textit{now, we multiply both sides by 2, to do away with the denominator}\\\\ 2\left( \cfrac{x}{2}-x+10 \right)=2\cdot 4\implies x-2x+20=8 \\\\\\ -x=-12\implies 12=x$

solving for either variable on either equation, and using that equivalent on the other, is "substituting" the variable in the other equation, thus is called "substitution"