SOLUTION: http://media261.acellus.com/Library/00001230_161_1401_A1.png Copy and Paste Link to see the problem!

Question 1018082: http://media261.acellus.com/Library/00001230_161_1401_A1.png
Copy and Paste Link to see the problem!
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
your equations are:

y = x^2 - 5x - 4
y = -2x

these are two equations that need to be solved simultaneously.

in the first equation, replace y with -2x from the second equation to get:

-2x = x^2 - 5x - 4

add 2x to both sides of this equation to get:

0 = x^2 - 5x - 4 + 2x

combine like terms to get 0 = x^2 - 3x - 4

this is the same as x^2 - 3x - 4 = 0

this is a quadratic equation in standard form.

factor the quadratic to get (x-4) * (x+1) = 0

solve for x to get x = 4 or x = -1.

go back to your original equations and solve for y.

you should get the same value of y for both equations.

when x = 4,y = -x b3comes y = -2*4 which becomes -8.

when x = -1, y = -2x becomes y = -2*-1 which becomes y = 2.

your coordinate points are (4,-8) and (-1,2).

since the leftmost point is x = -1, then you list your solutions as:

(-1,2) and (4,-8).

the graph of both your equations is shown below.

$$$

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