## Algebra: A Combined Approach (4th Edition)

Use the substitution method and solve the second equation for x $x+y=4$ $x=4-y$ So substitute x to the first equation $(4-y)^2+y^2=1$ $16-8y+y^2=1$ $y^2-8y+15=0$ By the quadratic formula, where a=0, b=-8, and c=15, we have $y=\frac{8+/-\sqrt (-8)^2-2\times0\times15}{2\times0}$ Since denominator is 0, there is no solution