SOLUTION: hello, I am having trouble finiding the lcd for this equation. Id appreciate the help on this problem but also tips on how to find it in problems like this. thank you, carla

Question 643834: hello,
I am having trouble finiding the lcd for this equation. Id appreciate the help on this problem but also tips on how to find it in problems like this.
thank you, carla
x/(2x+2)=-2x/(4x+4)+(2x-3)/(x-4)
so far i know i have to get the lcd for 2x+2,4x+4 and x-4 but since there was that -4 im at a loss, thank you
Found 3 solutions by Alan3354, MathLover1, stanbon:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
x/(2x+2)=-2x/(4x+4)+(2x-3)/(x-4)
---------------
The LCD is not relevant.
You need to solve for x
------------
If you mean
Reduce -2x/(4x+4)

Add x/(2x+2)

Reduce again

Cross multiply
x(x-4) = (x+1)*(2x+3)

Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=69 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: -0.346688068540963, -8.65331193145904. Here's your graph:

========================

Answer by MathLover1(20850)   (Show Source): You can put this solution on YOUR website!
...first factor out in and in

...right side has denominators and ...their LCD is their product (they have no common factors)

Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
x/(2x+2)=-2x/(4x+4)+(2x-3)/(x-4)
------
Factor:
x/(2(x+1)) = -2x/(4(x+1) + (2x-3)/(x-4)
-------
Multiply thru by lcm = 4(x+1)(x-4)
--------------------
2x(x-4) = (-2x)(x-4) + (2x-3)(4(x+1))
------
2x^2-8x = -2x^2+8x + 4(2x^2-x-3)
----
4x^2-16x = 8x^2-4x - 12
4x^2 + 12x -12 = 0
x^2 + 3x - 3 = 0
---
x = [-3 +- sqrt(9+12)]/2
----
x = [-3+sqrt(21)]/2 or x = [-3-sqrt(21)]/2
========================================
Cheers,
Stan H.

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