Given f(x)=3x+2/4x-7. Write the equation for f^-1(x). (It's inverse)
@tom982
\[y=\frac{3x+2}{4x-7}\] solve this equation for x
First step: multiply both sides by 4x-7 Second step: distribute anything that needs to be distribute Third step: Put your terms that have a y in it on one side and anything without a y on the other side
Wouldn't you multiply it by 4x+7? To remove fractions?
4x-7
Multiplying by 4x+7 would do nothing - try it :)
(4x-7)/(4x-7)=1
@tom982 Remember the question we did a few days ago where you crossed out fractions by using the conjugate to multiply? x+5/x-2=5/x+2+28/x^3-4 I'm a little confused, why is this different? @freckles I see that now.
\[\text{ i think you mean } x^2-4 \\ x^2-4=(x-2)(x+2)\] you need to multiply both sides of that equation you are presenting by (x-2)(x+2)
I don't remember your question specifically, but I assume we were doing the difference of two squares there, where we know (a+b)(a-b)=a^2-b^2. Link me to it and I'll have a look :)
@freckles Figured out how you did the last one. Can you finish this one for me?
have you tried following those steps gave you
All I know is that I switch the x and y. Then I multiply the numerator and denomenator by 4y-7?
So I'm solving the following equation for x: (liked to switch x and y at the end but it doesn't matter) \[y=\frac{3x+2}{4x-7} \\ \text{ apply step 1} \\ \text{ multiply both sides by } (4x-7) \\ (4x-7)y=3x+2 \\ \text{ apply step 2 } \\ \text{ distribute anything that needs to be distribute } \\ 4xy-7y=3x+2 \\ \text{ apply step 3} \\ \text{ put all your terms with x on one side and all your terms without x on the other side }\] see what that gives you:
we are doing that to try and isolate the x part
if we can get all terms with x on one side then we can factor the x out and divide by what is being multiplied by x on both sides
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