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you posted this question earlier
Im pretty sure it was involving different equations
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?
Yes I am with ya
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?
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?
Now move 80 to the other side and you get 4y^2+4y-80=0 Then factor
thats where I get confused because what adds to get 4 but multiplies to get -80???
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
thats a lot cleaner and easier to deal with
Well, how does that factor?
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
no, I bet its really easy but I just cant wrap my head around it.
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|
5 and -4
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|
And yeah, you got it, haha
y=-5 and y=4
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
-5 doesnt work when plugged in so its got to be 4
After plugging in 4 to the equation I get x=6
How does it not work?
II get x^2=0
Which just means x = 0
x^2+4(-5)^2=100 x^2+100=100 -100 -100 x^2=0 but you cannot take the square root of 0
Like how does that work?
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?
Wait no its just (0, -5)
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/
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
Currently plugging that into the first equatiom
So final answers are (4,6) (0,-5) (6,4) (-6,4)
So we actually have 3 solutions. |dw:1433534800787:dw| Sorry for the bad drawing, but yeah.
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.
Okay so we only use the solutions (0,-5), (6,4), and (-6,4) Which gives a sum of 0 for the x coordinates