SOLUTION: Solving rational equations. There are two fractions on one side of the equal sign, one on the other side. (5/x-3) + 1/6 = 7/x-2 The numerators are 5, 1, and 7 The denominators

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Solving rational equations. There are two fractions on one side of the equal sign, one on the other side. (5/x-3) + 1/6 = 7/x-2 The numerators are 5, 1, and 7 The denominators      Log On


   

Question 260664: Solving rational equations.
There are two fractions on one side of the equal sign, one on the other side.
(5/x-3) + 1/6 = 7/x-2
The numerators are 5, 1, and 7
The denominators are x-3, 6, and x-2
The question is NOT 5 divided by x -3 plus 1 divided by 6 = 7divided by x minus 2
I have multiplied by LCD and tried to find solutions by solving for 0. I got:
x^ -5x -27=0
But I cannot factor this to solve! Please help, I have spent a lot of time on this!
Found 2 solutions by jsmallt9, mananth:
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
5%2F%28x-3%29+%2B+1%2F6+=+7%2F%28x-2%29
You have the right idea in multiplying by the LCD. But somehow your multiplication has one or more errors because, as you will soon see, the equation we get after multiplying, simplifying and getting one side equal to zero is:
x%5E2+-+17x+%2B+72+=+0
not
x%5E2+-5x+-27
(NOTE: If your equation had been correct, then use the Quadratic Formula when factoring doesn't work.)

Let's see how we get the correct equation. The LCD is (x-3)(6)(x-2) so we will multiply both sides of the equation bby this:

On the left side we need to use the Distributive Property:

Now we can cancel:

leaving:
%286%29%28x-2%29%285%29+%2B+%28x-3%29%28x-2%29%281%29+=+%28x-3%29%286%29%287%29
which simplifies as follows:
%2830x+%2B+60%29+%2B+%28x%5E2+-5x+%2B+6%29+=+%2842x+-+126%29
x%5E2+%2B25x+-54+=+42x+-+126
Get one side equal to zero by subtracting 42x from and adding 126 to each side:
x%5E2+-17x+%2B+72+=+0
This does factor:
%28x-8%29%28x-9%29+=+0
Using the Zero Product Property which tells us that a product can be zero only if one of the factors is zero:
x-8+=+0 or x-9+=+0
Solving these we get:
x+=+8 or x+=+9

We should check these answers because we need to make sure that none of the denominators turn out to be zero for these x values. A quick check shows that when x=8 or x=9 none of the denominators are zero. So there are two solutions to your equation. (If any denominator, even just one, turns out to be zero for a certain x value, we must reject that solution.)



Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
5/x-3) + 1/6 = 7/x-2
5/(x-3) -7/(x-2)= - 1/6
(5x-10-7x+14)/(x-2)(x-3) =-1/6 LCD
(-2x+4)/(x-2)(x-3) = -1/6
6*(-2x+4)/(x-2)(x-3) =-1
6*(-2x+4)= -1 *(x-2)(x-3)
-12x+24= -1( x^2-5x+6)
-12x+24= - x^2+5x-6
x^2-17x+30=0
x^2-15x-2x+30=0
x(x-15)-2(x-15)=0
(x-15)(x-2)=0
x=15 or 2
M Ananth-------- mananth@hotmail.com