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)
5495 + 200.75x = 7500 + 100.50x
100.25x = 2005
x=20 <- the amount of sugar at the equilibrium point.
7500 + 100.50(20)
7500 + 2005
$9505 <- the cost at that point. ■
