Solve the system: `y = -x^2 + 2x + 9 and y = -5x^2 + 10x + 12.`

`-x^2 + 2x + 9 = -5x^2 + 10x + 12` Bring everything to right side.

`4x^2 - 8x - 3 = 0`

Use quadratic formula. `(8+-sqrt(8^2 - 4(4)(-3)))/(2*4)`

`(8+-sqrt(112))/8` = `(8+-4sqrt(7))/8` = `(2+-sqrt(7))/2` = 2.323 and -0.323

x...

## See

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

Solve the system: `y = -x^2 + 2x + 9 and y = -5x^2 + 10x + 12.`

`-x^2 + 2x + 9 = -5x^2 + 10x + 12` Bring everything to right side.

`4x^2 - 8x - 3 = 0`

Use quadratic formula. `(8+-sqrt(8^2 - 4(4)(-3)))/(2*4)`

`(8+-sqrt(112))/8` = `(8+-4sqrt(7))/8` = `(2+-sqrt(7))/2` = 2.323 and -0.323

x = 2.323, -0.323. Substitute into either equation to solve for y.

`y = -x^2 + 2x + 9`

`y = -(2.323)^2 + 2(2.323) + 9`

`y = 8.25`

`y = -x^2 + 2x + 9`

`y = -(-0.323)^2 + 2(-0.323) + 9`

`y = 8.25`

**#1) The solutions are (2.323, 8.25) and (-0.323, 8.25).**

#2) Let the first number be "x", and the second number be "y".

The **system of equations** would be:

`x + y^2 = 18` and `y^2 - 2x = 12.`

Solve for y^2. The 2 equations would be:

`y^2 = 18 - x` and `y^2 = 2x + 12` .

Therefore, `18 - x = 2x + 12.`

Solve for x.

`3x = 6 `

`x = 2`

Substitute to find y.

`x + y^2 = 18`

`2 + y^2 = 18`

`y^2 = 16`

`y = +-4`

**There are 2 possible solutions that work. The 2 numbers could be 2 and 4, or 2 and -4.**