# Algebra, Solving equations

I'm stuck with this equation:

y = (4x-1)/(2x+3)

We're working on inverses, which I understand, but my mind is blanking on how to solve this for x.

Amy :)

after you interchanged the x and y variables, and expanded you should have had
x = 8y^2 + 10y - 3

arrange to look like a quadratic equation
8y^2 + 10y - 3 - x = 0

now a=8, b=10 and c=-3-x

use the formula to solve for y

notice you get 2 different equations.
Are you also learning about functions?

Oh, this is embarassing...I'm studying inverse functions in calculus and I'm getting hung up on algebra I did in high school. :)

This particular problem's directions say to find the inverse of the given function, but I thought I read that if it's not a one-to-one function, it doesn't have an inverse. So if there's two equations once I use the quadratic formula, then this equation doesn't have an inverse, right?

Hold on a sec...There's division between those two terms, so I wouldn't end up getting two different answers, right?

I got y = (-10 + sqrt(100-32(-3-x)))/16 for one and
y = (-10 - sqrt(100-32(-3-x)))/16 for the other, thus 2 distinct equations

recall that taking the inverse of a relation results in a reflection in the line y=x
Your original equation was a parabola opening upwards, so its inverse is a parabola with axis parallel to the x-axis.

so for a given value of x (in the new domain) you now have 2 different values of y, making it NOT a function.

Do you recall something called the "vertical line test" ?

But it's division. I'm dividing (4x-1) by (2x+3). I guess my main question is how do I simplify the original equation so that I can solve it for x. I know I can't just factor it because there's no common factors. When I graph the equation on my calculator, it passes both the VLT and the HLT, so it's definitely one-to-one.

oops, oops, oops, my mistake, lets back up
I read that as a multiplication, sorry

now it's actually easier:

step 1 of finding inverse: interchange the x and y variables...

y = (4x-1)/(2x+3) turns into x = (4y-1)(/2y+3)

cross multiply
2xy+3x=4y-1
2xy-4y=-1-3x or
4y-2xy = 1+3x
y(4-2x) = 1+3x
y = (1+3x)/(4-2x)

now clearly the original was a function and so is the new one.

test (-1,-5) in original and (-5,-1) in new one

Thanks. I was getting so confused!

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