How do you solve the system 2x^2-4y=22 and y=-2x+3?

1 Answer
Douglas K.
May 7, 2017

Substitute the right side of equation [2] for y in equation [1].
Solve the resulting quadratic for the two x values.
Use equation [2] to find the corresponding y values.

Explanation:

Given:
2x^2-4y=22" [1]"
y=-2x+3" [2]"

Substitute the right side of equation [2] for y in equation [1].

2x^2-4(-2x+3)=22

Solve the resulting quadratic for the two x values.

Use the distributive property:

2x^2+8x-12=22

Combine like terms:

2x^2+8x-34=0

Divide both sides by 2

x^2+4x-17=0

Use the quadratic formula:

x = (-4+ sqrt(4^2-4(1)(-17)))/2 and x = (-4- sqrt(4^2-4(1)(-17)))/2

x = -2+ sqrt(4+17) and x = -2- sqrt(4+17)

x = -2+ sqrt(21) and x = -2- sqrt(21)

Use equation [2] to find the corresponding y values.

y = -2(-2+ sqrt(21))+3 and y = -2(-2- sqrt(21))+3

y = -1- 2sqrt(21) and y = -1+ 2sqrt(21)

The two points are:

(-2+ sqrt(21), -1- 2sqrt(21)) and (-2- sqrt(21),-1+ 2sqrt(21))

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