## Linear Algebra and Its Applications (5th Edition)

$x_{1}=x=\frac{4}{7}$ $x_{2}=y=\frac{9}{7}=1\frac{2}{7}$ Ordered pair $(\frac{4}{7}, \frac{9}{7})$.
This problem is a different wording of "What are the $x_{1}$ and $x_{2}$ values that work for both equations?" This is because the intersection of two lines means that the x values are the same ($x_{1}$) and the y values are the same ($x_{2}$). Now that you know this, let's get started. The principle is the same from the first two problems, eliminate one variable and solve for the other, and then substitute. In this case, you can eliminate $x_{1}$ immediately. $x_{1}+5x_{2}=7$ $x_{1}-2x_{2}=-2$ Subtract and you will get $7x_{2}=9$. Now you can solve for $x_{2}$. $7x_{2}=9$ $x_{2}=\frac{9}{7}$ Don't get bogged down by the fact that this number is tricky. Continue to solve for $x_{1}$. $x_{1}+5x_{2}=7$ $x_{1}+5(\frac{9}{7})=7$ (Substitution) $x_{1}+\frac{45}{7}=\frac{49}{7}$ (Simplifying and changing 7 into $\frac{49}{7}$ for the next step of subtraction). $x_{1}=\frac{49}{7}-\frac{45}{7}=\frac{4}{7}$ Check your results with the other equation. $x_{1}-2x_{2}=-2$ $(\frac{4}{7})-2(\frac{9}{7})=-2$ (Substitution) $\frac{4}{7}-\frac{18}{7}=-2$ $\frac{-14}{7}=-2$ (Simplify) $-2=-2$ This solution means that the lines will intersect at point $(\frac{4}{7}, \frac{9}{7})$.