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Thread: Solving for the inverse of a function.

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    Solving for the inverse of a function.

    Solving for the inverse of a function.-1a-5question.png

    I am having trouble solving a) for x so I can graph the inverse function. I am told to cross multiply but I'm not sure how. I tried to solve for x a few times but I'm not seeing the solution yet.

    Thanks in advance...
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    Re: Solving for the inverse of a function.

    Quote Originally Posted by sepoto View Post
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    I am having trouble solving a) for x so I can graph the inverse function. I am told to cross multiply but I'm not sure how. I tried to solve for x a few times but I'm not seeing the solution yet.
    Are you saying that you cannot solve $\displaystyle x=\frac{y-1}{2y+3}$ for $\displaystyle y~?$
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    Re: Solving for the inverse of a function.

    I'm trying to cross multiply both sides by 2y + 3 / 1 so then I am left with an equation x(2y+3) = y -1. I'm sorry to ask such a basic question but I am stuck.
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    Re: Solving for the inverse of a function.

    You have $\displaystyle x= \frac{y- 1}{2y+ 3}$.

    Multiply both sides by2y+3: $\displaystyle x(2y+ 3)= 2yx+ 3x= y- 1$.

    Subtract 3x and y from both sides: $\displaystyle 2yx- y= -3x- 1$

    Factor y out on the left: $\displaystyle y(2x- 1)= -3x- 1$

    Divide both sides by 2x- 1: $\displaystyle y= -\frac{3x+ 1}{2x- 1}$.
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    Re: Solving for the inverse of a function.

    Quote Originally Posted by sepoto View Post
    I'm trying to cross multiply both sides by 2y + 3 / 1 so then I am left with an equation x(2y+3) = y -1. I'm sorry to ask such a basic question but I am stuck.
    $\displaystyle x=\frac{y-1}{2y+3}$
    $\displaystyle \\x(2y+3)={y-1}\\2xy-y=-3x-1\\y(2x-1)=-(3x+1)\\y=-\frac{3x+1}{2x-1}$
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