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solve the given linear-quadratic system graphically. how many points of intersection are on the graph? show all work.

5 – y = x^2 + x
y + 1 = 3/4x

x = sqrt(433/64) - 7/8 and x = -sqrt(433/64) - 7/8 So there are two.  Let me know if there's something you don't understand.

Step-by-step explanation:

First you want to get them into the form where just y is on one side.

5-y=x^2+x

-y=x^2+x-5

y = -x^2 - x + 5

y+1 = 3/4x

y = 3/4x - 1

Let me know if you don't get how I did that.

To find where they intersect you just set them equal to each other and solve for x.

-x^2 - x + 5 = 3/4x - 1

Before you go solving you are going to want to avoid multiplying and dividing.  Especially dividing here.  if you divide something by (x-a) you make it so a cannot be used.  take 1 for instance.  graphing 1 is just a horizontal line at y=1.  If you divide it by (x-a) in particular lets say (x-1) if you try and solve for x=1 you get 1/0 and you can't divide by 0.  If you graph it on a calculator or computer you'll see the graph doesn't exist at x=1 either.

So yeah, no multiplying or dividing.  So we turn this into a quadratic.

-x^2 - x + 5 = 3/4x - 1

-x + 5 = x^2 + 3/4x - 1

5 = x^2 + 7/4x - 1

0 = x^2 + 7/4x - 6

There are a few ways to solve but I am going to complete the square.  If you don't understand how to solve it let me know and I will more specifically walk through it.

(x + 7/8)^2 - 433/64 = 0

(x + 7/8)^2 = 433/64

x + 7/8 = +/- sqrt(433/64)

x = +/- sqrt(433/64) - 7/8

So there are two answers here.  Positive sqrt(433/64) - 7/8 and negative sqrt(433/64) - 7/8 Or approximately 1.726 and -3.476 respectively.  To check try plugging these into the original two functions y = -x^2 - x + 5 and y = 3/4x-1 and make sure they give the same answer.

Let me know again if there's something you don't get.  When oing intersections though you set the two equations equal to each other and solve for x however you can.

answer: 19 is   2

step-by-step explanation: i just looked at your answer and wrote it down

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step-by-step explanation: