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Mathematics 21 Online
OpenStudy (anonymous):

Solve for y Y^2/4=9 y^2+x=49 y^2-x^2=0

OpenStudy (anonymous):

\[\frac{ y^2 }{ 4 }=9\]

OpenStudy (anonymous):

we want to isolate the y so lets multiply both sides of the equation by 4 to get rid of the fraction

OpenStudy (anonymous):

so now we have \[y^2=36\] , now just take the square root of both sides so we have \[y=\sqrt{36}\]

OpenStudy (anonymous):

when you have a square root its + or - so for this answer it would be \[y=\pm6\]

OpenStudy (anonymous):

hey taht works thanks can you explain other problems too that would be great

OpenStudy (anonymous):

or atleast tell me what to do when theres X and have to solve for y?

OpenStudy (anonymous):

i mean do i just cut it off the equation?

OpenStudy (anonymous):

\[y^2+x=49\]

OpenStudy (anonymous):

if you need to solve for y on this one then set the equation to \[y^2=-x+49\]

OpenStudy (anonymous):

then we would need to take the square root of both sides

OpenStudy (anonymous):

what did u get

OpenStudy (anonymous):

y=? +7

OpenStudy (anonymous):

dunno what to do with x

OpenStudy (anonymous):

the answer will be left in square root form on this one so it will be \[y=\pm \sqrt{-x+49}\]

OpenStudy (anonymous):

there was no value given for x so we just have to leave it as x

OpenStudy (anonymous):

hmm that makes sense let me do the last problem and tell me if i was right

OpenStudy (anonymous):

wait there can be - in square root?

OpenStudy (anonymous):

ya but we dont know the value of x, what if it was 3

OpenStudy (anonymous):

so the answer for last wuestion will be y= square root of x^2

OpenStudy (anonymous):

or y=x?

OpenStudy (anonymous):

it would be \[y=\pm x\]

OpenStudy (anonymous):

when you take the square root of something you have to include the +or-

OpenStudy (anonymous):

alright thanks bro :) now my omework is done i really appreciate it

OpenStudy (anonymous):

np

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