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OpenStudy (freckles):
can you solve y=mx+b for x ?
OpenStudy (freckles):
are you there?
OpenStudy (anonymous):
I think I have the answer.
OpenStudy (freckles):
oh okay
do you want to post it
OpenStudy (anonymous):
y= -bm(1+m) / (m^2-1) + b
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OpenStudy (freckles):
Hmm... solving y=mx+b should only be a two step process
Shouldn't result in all of that
OpenStudy (freckles):
Solving y=mx+b for x:
1st step: Subtract b on both sides
2nd Step: Divide both sides by m (of course we are assuming m isn't 0 you know since we can't divide by 0)
OpenStudy (anonymous):
y = (x-b) / m
OpenStudy (freckles):
right y=mx+b
step 1: we have y-b=mx
step 2: we have (y-b)/m=x
interchanging x and y gives
y=(x-b)/m
OpenStudy (anonymous):
so I would get that for my final answer
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OpenStudy (freckles):
the inverse is what we have found above
OpenStudy (freckles):
If you aren't sure you test it
OpenStudy (anonymous):
thank you.
OpenStudy (freckles):
f(x)=mx+b and g(x)=(x-b)/m
check by making sure you get x for both of g(f(x)) and f(g(x))
\[f(g(x))=x \text{ and } g(f(x))=x\]