Question

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

Answer 1

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))`