well, you know that x=2y-5
so, substitute that into the other equation, solve for y, and then you can get x.
Solve the system by the substitution method. Show your work.
2y - x = 5
x2 + y2 - 25 = 0
3 answers
2y - x = 5
2y - 5 = x
so x = 2y - 5
(2y - 5)² + y² - 25 = 0
(2y - 5)(2y - 5)
4y² - 10y - 10y + 25
4y² - 20y + 25
4y² - 20y + 25 + y² - 25 = 0
4y² + y² - 20y = 0
4y² + y² - 20y = 0
y(4y + y - 20) = 0
y=0
x= -5
4y + y - 20 = 0
5y - 20 = 0
5y=20
y = 4
x = 3
2y - 5 = x
so x = 2y - 5
(2y - 5)² + y² - 25 = 0
(2y - 5)(2y - 5)
4y² - 10y - 10y + 25
4y² - 20y + 25
4y² - 20y + 25 + y² - 25 = 0
4y² + y² - 20y = 0
4y² + y² - 20y = 0
y(4y + y - 20) = 0
y=0
x= -5
4y + y - 20 = 0
5y - 20 = 0
5y=20
y = 4
x = 3
or should I just do this instead of all my factors and simplifies
2y - x = 5
2y - 5 = x
so x = 2y - 5
(2y - 5)² + y² - 25 = 0
(2y - 5)(2y - 5)
2y - x = 5
2y - 5 = x
so x = 2y - 5
(2y - 5)² + y² - 25 = 0
(2y - 5)(2y - 5)
2 answers