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Solve the system by the substitution method. Show your work. 2y - x = 5 x2 + y2 - 25 = 0Question
Solve the system by the substitution method. Show your work.
2y - x = 5
x2 + y2 - 25 = 0
2y - x = 5
x2 + y2 - 25 = 0
Answers
Steve
well, you know that x=2y-5
so, substitute that into the other equation, solve for y, and then you can get x.
so, substitute that into the other equation, solve for y, and then you can get x.
Ciara
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
Ciara
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)