## Algebra 1: Common Core (15th Edition)

The error is in step 2. See below for correction. Solution: $(\displaystyle \frac{7}{51},\frac{133}{51})$
Step 1 is good because the second equation has been chosen to be solved for x. Step 2 is faulty, because it does not represent the second equation with x substituted with $21-8y.$ $x$ has been substituted into the first equation instead of into the second. It should be: $7( 21-8y)+5y=14$ $147-56y+5y=14\quad$... add $-147$, simplify $-51y=-133$ $y= \displaystyle \frac{133}{51}$ Substitute $y$ in $\quad x=21-8y.$ $x=21-8\displaystyle \cdot\frac{133}{51}$ $x=\displaystyle \frac{1071- 1064}{51}$ $x=\displaystyle \frac{7}{51}$ Solution: $(\displaystyle \frac{7}{51},\frac{133}{51})$