How do you solve y=2x-3/5x+4

Question

We have to learn how to slove for x when we are given the equation in terms of y. I dont know how to do this with an equation with a variable fraction.  any help?

Answer 1

This is kind of a tricky one b/c there's gonna be a few steps, but I can help you break this down.  To do this were going to need to use A) distributive property, B) Some cross multiplication and C) factoring a GCF from a binomial.

 

K so you have y = (2x-3) / (5x+4)?

 

1) cross multiply --> make it y/1 = (2x-3) / (5x+4) and cross multiply, this will get rid of that fraction.

                              you should have y(5x+4) = 2x-3

 

2) distribute the y --> 5xy + 4y = 2x-3

 

3) no since we are solving for X, I want them all on the same side of the equation. I vote we subtract 2x from both sides.  --> 5xy - 2x + 4y = -3

 

 

4) Since I want X's on the left, lets subtract 4y on both sides ---> 5xy -2x = -3 - 4y

 

 

5) OK so here's the tricky part, in  5xy -2x, both of the terms have an X, that means its our GCF (Greatest Common Factor) in that binomial.  So if you factor out the X you get something like --> X(5y - 2) =-7

 

6) Almost done (Promise!) What you have on the left is X times (5y-2). Now you only want the X, not what you are multiplying it by. So you have to thing of what operation is opposite multiplication........

So, you can actually divide (5y-2) on both sides. It will completely cancel out on the left so you will have

x = (-3 - 4y)/(5y-2)

 

7) Circle x = (-3 - 4y)/(5y-2)) as your answer b/c you're done! 

Answer 2

Is it 3/(5x) or (3/5)*x?

 

In the second case you just collect like terms.

 

2x-(3/5)*x+4=(2-3/5)*x+4=7/5*x+4

Then y=7/5*x+4, from which you get x=(y-4)*5/7

 

For the first case, y=2x-3/(5x)+4; Multiply both sides by x, provided it is not zero (if you get x=0 as a solution, it must be thrown away).

 

yx=2x²-3/5+4x; Rewrite it as follows:

 

2x²-(y-4)x-3/5=0

 

Solve using quadratic formula:

 

x_1,2=[(y-4)±√{(y-4)²+24/5}]/4