Linear Algebra and Its Applications (5th Edition)

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