#### Answer

$x_{1}=\frac{9}{4}$
$x_{2}=\frac{1}{4}$

#### Work Step by Step

This is just a different wording of "what are the $x_{1}$ and $x_{2}$ values of the equations that work for both?"
Now that you know this, we can solve the problem.
First, multiply the first equation by $3$ so we can eliminate the $x_{1}$ and solve for $x_{2}$.
$3*(x_{1}-5x_{2})=3*(1)$
$3x_{1}-15x_{2}=3$
Next, subtract one equation from the other to eliminate $x_{1}$ and then solve for $x_{2}$.
$3x_{1}-7x_{2}=5$
$3x_{1}-15x_{2}=3$
You will get $8x_{2}=2$, and simplify.
$8x_{2}=2$
$x_{2}=\frac{2}{8}=\frac{1}{4}$
Next, solve for $x_{1}$ knowing the value of $x_{2}$ to be $\frac{1}{4}$.
$x_{1}-5x_{2}=1$
$x_{1}-5(\frac{1}{4})=1$ (Substitution)
$x_{1}-\frac{5}{4}=1$
$x_{1}=(1)-(-\frac{5}{4})=1+\frac{5}{4}=\frac{9}{4}$ (Subtracting $-\frac{5}{4}$ from both sides is the same as adding $\frac{5}{4}$)
Now that you have the values, you can check to see if they are correct.
$3x_{1}-7x_{2}=5$
$3(\frac{9}{4})-7(\frac{1}{4})=5$ (Substitution)
$\frac{27}{4}-\frac{7}{4}=5$
$\frac{20}{4}=5$ (Simplification)
$5=5$
We know that these values are correct, so they are our answers.