find intersection in terms of y of y^2-4x=4and 4x-y=16

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find intersection in terms of y of y^2-4x=4and 4x-y=16

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

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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
|dw:1357491125167:dw|
|dw:1357491191134:dw|
Don't make it complicated bro
|dw:1357491240591:dw|
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
|dw:1357491558149:dw|
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
I remember these. They can be tricky. What you do is just turn your graph sideways. And it will be the same as like when you're working with x-axis.
if \[\sqrt{y}\]
|dw:1357491948129:dw|
?
alright thx
Bro, I graphed it as \[\frac{16 + x}{4}\] and \[\frac{x^2 - 4}{4}\] You get the same area when integrating from a = -4 to b = -5 The only difference is you're using variable y instead of variable x
ya i noticed it thx

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