In the system shown below, what is the sum of all of the x-coordinates of all solutions? x^2+4y^2=100 4y-x^2=-20 please please explain I am very stuck
Try to use elimination method: \[ ~~~~x^2+4y^2=100 \\ \underline{+~-x^2+4y=-20}\\~~~~~~~~~~~~~~?\]
add equations together, you should eliminate \(x^2\), leaves you \(4y^2+4y=80\)
They want the sum of all the x-coordinates of the solution. Eliminate y. Find y from the second equation, substitute it in y and you will have an equation in x.
Explain how to eliminate y please @aum
4y-x^2=-20 add x^2 4y = x^2-20 divide by 4 y = (x^2-20)/4 Put this in the first equation and simplify.
it's more of a substitution than elimination, but that's good move though
I put (x^2-20)/4 back into the equation and my answer was bizarre
I know, right lol... So you have \(x^2+4\left(\dfrac{x^2-20}{4}\right)^2=100\) You can simplify \(\left(\dfrac{x^2-20}{4}\right)^2\) to \(\dfrac{(x^2-20)^2}{4^2}\) What is \((x^2-20)^2\) ?
x^4-40x^2-400
Almost. it's \(\Large x^4-40x^2+400\)
sorry typo haha
so I multiplied it out and got 4x^4-160x^2+1600
so it would be 4x^4-158x^2+1600=100?
lol okay, so you now have \(x^2+4\cdot\dfrac{x^4-40x^2+400}{4^2}=100 \\ x^2+\cancel{4}\cdot\dfrac{x^4-40x^2+400}{4^{\cancel2}}=100\\x^2+\dfrac{x^4-40x^2+400}{4}=100\) Hey, you hate denominator, right? me too, so let multiply both sides by 4 to get \(4x^2+x^4-40x^2+400=400\)
wow im way off, so then we solve for x okay
do you understand what I just did, right?
Yes, I see it now
ok great, so we now have \(4x^2+x^4-40x^2+400=400 \\\rightarrow x^4-36x^2+400=400\\\rightarrow x^4-36x^2=0\) Do you know what to do next?
take out a greatest common factor (x^2)
so the answers are x=6 x=-6?
and x=0 too. Question wants sum of all of the x-coordinates of all solutions So you have -6+0+6=?
So zero
right. your answer is 0
thank you so much!
love how lengthy and harsh problem ended up get us 0 haha
welcome!
I know right
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