Question

Solve the linear system. $$ \begin{aligned} &\frac{1}{2} x-y=-5\\\ &x-\frac{1}{3} y=0 \end{aligned} $$

Short answer

The solution to the system of equations is \(x=10, y=30\).

Step-by-step solution

Rearrange the second equation

Rearrange the second equation \(x-\frac{1}{3} y=0\) to solve for \(x\). This gives \(x=\frac{1}{3} y\).

Substitute \(x\) into the first equation

Now that we know \(x=\frac{1}{3} y\), substitute this value into the first equation. This gives \(\frac{1}{2} (\frac{1}{3} y)-y=-5\) which simplifies to \(-\frac{1}{6} y=-5\).

Solve for \(y\)

Now solve for \(y\) in the equation \(-\frac{1}{6} y=-5\) to get \(y=30\).

Substitute \(y\) into the first equation to solve for \(x\)

Substitute \(y=30\) into the equation \(x=\frac{1}{3}y\) to evaluate \(x\). This gives \(x=\frac{1}{3}*30=10\).