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\[y^2-4x=4 and 4x-y=16\]

Hint:
y = 4x - 16

i did but i dont know whre i was wrong

i know dude to solve for x and set it eual and then solve for x

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Don't make it complicated bro

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solve for y am i right this is to find area between curve sooo i must to with respect y

Actually, that's a good approach after all.

but something is wrong in solving hmm i am not getting the answer

multiply both sides by 4

What did you get for the answer?

.08 and -12.08

That's what you got for y values?

nope

i gote it 5 and -4

Okay, so what's the problem?

arithmetic mistake lol

now i have to integrate thats it

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but not sure which one goes top and which goes bottom with respect to y

Here's another way to do it to avoid confusion:
\[\frac{y^2 - 4}{4} - \frac{16 + y}{4} = 0\]

ya i tried that worked well

Bro, a is always the lowest value. b is always the highest value.

i mean the function not the points (curves)

\[x=\frac{ y^2-4 }{ 4 }\]

Are you sure you're supposed to be using y values for a and b?

or\[x=\frac{ 16+y }{ 4 }\]

ya i am damn sure

if \[\sqrt{y}\]

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alright thx

ya i noticed it thx