Help me
Solve this system of equations to get the sum of all of the x coordinates from the solutions.
x^2+4y^2=100
4y-x^2=-20

- cutiecomittee123

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- anonymous

you posted this question earlier

- cutiecomittee123

Im pretty sure it was involving different equations

- anonymous

The one from before were different equations. But the same idea holds. You want to make some sort of substitution that will let you reduce one of the equations to only x or only y.
So you're looking for something in common in both equations that will make for an easy substitution. I see that both equations have an x^2 term. So if I were to solve the first equation for x^2, I would get
x^2 = 100 - 4y^2
With me so far?

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- cutiecomittee123

Yes I am with ya

- cutiecomittee123

wait wouldnt you get 4y^2+4y=80
because you can cancel out the two x^2's since one is negative and if you add the equations it works right?

- anonymous

Well, if we do the substitution for x^2 into the 2nd equation. we get
4y - (100-4y^2) = -20
4y - 100 + 4y^2 = -20
4y^2 + 4y = 80
SO yes, you're right, good job :)
Would you know what to do from here?

- cutiecomittee123

Now move 80 to the other side
and you get 4y^2+4y-80=0
Then factor

- cutiecomittee123

thats where I get confused because what adds to get 4 but multiplies to get -80???

- anonymous

Well, conveniently all of those terms are divisible by 4, so let's just divide out the 4 from all terms. Doing this would reduce the quadratic into
y^2 + y - 20 = 0

- cutiecomittee123

thats a lot cleaner and easier to deal with

- cutiecomittee123

now what

- anonymous

Well, how does that factor?

- cutiecomittee123

IDK???

- anonymous

Well, -20 would need to factor into 2 numbers that would also add up to a positive 1. This means:
One of the numbers has to be positive and the other has to be negative (otherwise they would never multiply to a negative 20)
The largest of those 2 numbers must be positive.
So can you come up with the two numbers? Two factors that multiply to -20 yet add to positive 1

- cutiecomittee123

no, I bet its really easy but I just cant wrap my head around it.

- anonymous

It is easy. And youll think its easy also once you get used to the process. Okay, so let's look at how 20 could possibly break apart:
|dw:1433533847258:dw|

- cutiecomittee123

5 and -4

- anonymous

So those are the possible combinations we need to consider. Now, in order for any of those 2 numbers to add to a positive value, the largest one must be positive. Thus we can more specifically consider these 3 choices:
|dw:1433533919775:dw|

- anonymous

And yeah, you got it, haha

- cutiecomittee123

y=-5 and y=4

- anonymous

Correct :) Now since y must be either of those values, you can substitute each one into one of your given equations and solve for x

- cutiecomittee123

-5 doesnt work when plugged in so its got to be 4

- cutiecomittee123

After plugging in 4 to the equation I get x=6

- anonymous

How does it not work?

- cutiecomittee123

II get x^2=0

- anonymous

Which just means x = 0

- cutiecomittee123

x^2+4(-5)^2=100
x^2+100=100
-100 -100
x^2=0
but you cannot take the square root of 0

- cutiecomittee123

Like how does that work?

- cutiecomittee123

nevermind. so now we know that (4,6) is a solution, Next we have (0, and we have to solve for the y coordinate right?

- cutiecomittee123

Wait no its just (0, -5)

- cutiecomittee123

So since there are only 2 intersections right? its safe to go ahead and add up the x coordinates. which are; 4 and 0. So the answer is 4 as the sum of all x coordinates for the solutions/

- anonymous

SOrry for the late response. But yeah, the square root of 0 is just 0. So you get the coordinate point (0,-5) as mentioned. Now when you plug in y = 4 however, we have to be a bit more careful. Let me plug y = 4 into the 1st equation:
x^2 + 4(4)^2 = 100
x^2 + 64 = 100
x^2 = 36
Now at this point, need to remember that when we square root both sides of an equation, we need to use a plus or minus. So we actually have x = 6 and x = -6

- cutiecomittee123

Okay

- cutiecomittee123

Currently plugging that into the first equatiom

- cutiecomittee123

So final answers are
(4,6)
(0,-5)
(6,4)
(-6,4)

- anonymous

So we actually have 3 solutions.
|dw:1433534800787:dw|
Sorry for the bad drawing, but yeah.

- cutiecomittee123

Makes sense

- anonymous

Remember, we found y = -5 and y = 4. Using y = -5, we got x = 0 and only x = 0. So that means the coordinate is (0, -5).
Now we have y= 4. We plugged that in and got 2 x-values, 6 and -6. Now be careful, these are x-values, gotta put the coordinates in the correct order. This gives us the two coordinates
(6,4), (-6,4)
So 3 solutions total and the graph above shows the 3 intersections.

- cutiecomittee123

Okay so we only use the solutions (0,-5), (6,4), and (-6,4)
Which gives a sum of 0 for the x coordinates

- cutiecomittee123

0+6-6

- anonymous

Correct :)

- cutiecomittee123

Okay cool

- anonymous

Mhm :3

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