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,
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)
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
More Related Questions
- Algebra - I don't quite understand how to solve matrices with three variables. ...
- math-rational equations - how do I solve 5/x-3+3/x+3=7x/x^2-9 please explain in ...
- Algebra - Solving Linear Equations and Inequalities Solve by eliminating x (this...
- Algebra I DONT KNOW HOW TO DO 3 VARIABLES - Solving Linear Equations and ...
- math - can someone correct these for me. 8x –4y = 16 y = 2x –4 My answer: This ...
- Algebra SOLVING LINEAR EQUATIONS AND INEQUALITIES - Please help Im am Grade 5 ...
- FOR MATHMATE - Media services charges $40 for a phone and 20/mth for its economy...
- Math - Solve each equation write answer on simplest form 1/4x = 6 2t = 4/7 Can ...
- Engineering - I need help solving the following simultaneous equations, for the ...
- Algebra - Solving equations simultaneously: .7*x+.30*y=33.1 (x*y)/(.7*y+.30*x...