Find the exact solutions of each system of equations 4x+y^2=20 4x^2+y^2=100
We need to find the exact solutions of each system of equations written as ordered pairs.
1) 4x+y^2=20
2) 4x^2+y^2=100
Subtract equation one from equation two, eliminating y^2.
4x^2 + y^2 = 100
-( 4x + y^2 = 20)
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4x^2 - 4x = 80
4x^2 - 4x = 80
3) 4x^2 - 4x - 80 = 0
Solve equation three for x. Factor the left side.
4x^2 - 4x - 80 = 0
(4x - 20)(x + 4) = 0
One or both factors must equal zero.
Factor one:
(4x - 20) = 0
4x = 20
x = 5
Factor two:
(x + 4) = 0
x = -4
Plug both of those values into equation one to solve for y.
4x + y^2 = 20
4(5) + y^2 = 20
20 + y^2 = 20
y^2 = 0
y = 0
(5, 0)
4x + y^2 = 20
4(-4) + y^2 = 20
-16 + y^2 = 20
y^2 = 20 + 16
y^2 = 36
y = ±6 (you can get 36 by squaring 6 and by squaring negative 6)
(-4, -6)
(-4, 6)
Use equation two to verify the various values.
(5, 0)
4x^2 + y^2 = 100
4(5^2) + 0^2 = 100
4(25) + 0 = 100
100 = 100
(-4, -6)
4x^2 + y^2 = 100
4(-4^2) + (-6^2) = 100
4(16) + (36) = 100
64 + 36 = 100
100 = 100
(-4, 6)
4x^2 + y^2 = 100
4(-4^2) + 6^2 = 100
4(16) + 36 = 100
64 + 36 = 100
100 = 100
All three pairs are valid:
(5, 0), (-4, -6) and (-4, 6)