Solve the system by the substitution method. show your work. 2y - x = 5, x2 + y2 - 25 = 0

Subsitution

2y-x=5
minus 2y both sides
-x=5-2y
times -2 both sides
x=2y-5

subsitute 2y-5 for x in other equation

(2y-5)²+y²-25=0
expand
4y²-20y+25+y²-25=0
5y²-20y=0
factor
5y(y-4)=0
set to zero

5y=0
y=0

y-4=0
y=4

subsitute back

for y=0
x=2y-5
x=2(0)-5
x=-5
one solution is (-5,0)

for x=4
x=2y-5
x=2(4)-5
x=8-5
x=3
one solution is (3,4)

the 2 soltuions are (-5,0) and (3,4)