Thursday
July 24, 2014

Homework Help: Algebra, Solving equations

Posted by Amy on Saturday, April 21, 2007 at 11:45am.

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.

Thanks in advance,
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!

Answer this Question

First Name:
School Subject:
Answer:

Related Questions

Algebra - I dont quite understand how to solve matrices with three variables. If...
Math - Solve each equation write answer on simplest form 1/4x = 6 2t = 4/7 Can ...
algebra 1 - I'm stuck on this problem. Can you help? -4x-9y=1 -x + 2y =-4 WHat ...
algebra II - this is called equations utilizng inverses how would i solve this ...
ALGEBRA 1!PLZ HELP! - Use the Substitution method to solve the system of ...
Algebra - solving linear equations w/ substitution - The question in the book ...
Algebra - Systems of equation 1. +2y=5 8 -6y=3x-15 2).p-2q=4 2q=p+2 Solve the ...
algebra - how would i find an equation of a line that goes through points(1,6) ...
Engineering - I need help solving the following simultaneous equations, for the ...
Math Linear Algebra - Find the inverses of the matrices. Let A = 1 2 5 12 b1 = -...

Search
Members