# Thread: Solving for the inverse of a function.

1. ## Solving for the inverse of a function.

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.

2. ## Re: Solving for the inverse of a function.

Originally Posted by sepoto

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 $x=\frac{y-1}{2y+3}$ for $y~?$

3. ## 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.

4. ## Re: Solving for the inverse of a function.

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

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

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

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

Divide both sides by 2x- 1: $y= -\frac{3x+ 1}{2x- 1}$.

5. ## Re: Solving for the inverse of a function.

Originally Posted by sepoto
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.
$x=\frac{y-1}{2y+3}$
$\\x(2y+3)={y-1}\\2xy-y=-3x-1\\y(2x-1)=-(3x+1)\\y=-\frac{3x+1}{2x-1}$