SOLUTION: I need to figure out how to solve for x in this problem: (x+1)/(x+1) + (5)/(x+2) = (3)/(x+1)

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Question 263727: I need to figure out how to solve for x in this problem:
(x+1)/(x+1) + (5)/(x+2) = (3)/(x+1)

Found 2 solutions by JBarnum, unlockmath:
Answer by JBarnum(2044) About Me  (Show Source):
You can put this solution on YOUR website!
(x+1)/(x+1) + (5)/(x+2) = (3)/(x+1) times every thing by ((x+1)(X+2))
((x+1)(x+2))+((5(x+1))= ((3(X+2)) multiply together
(x^2+3x+2)+(5x+5)=(3x+6) add like terms
x^2 + 8x + 7= 3x + 6 Subtract 3x and 6 from both sides
. -3x - 6 -3x - 6
x^2 +5x+1=0 now u have a quadratic equation
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B5x%2B1+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%285%29%5E2-4%2A1%2A1=21.

Discriminant d=21 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-5%2B-sqrt%28+21+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%285%29%2Bsqrt%28+21+%29%29%2F2%5C1+=+-0.20871215252208
x%5B2%5D+=+%28-%285%29-sqrt%28+21+%29%29%2F2%5C1+=+-4.79128784747792

Quadratic expression 1x%5E2%2B5x%2B1 can be factored:
1x%5E2%2B5x%2B1+=+1%28x--0.20871215252208%29%2A%28x--4.79128784747792%29
Again, the answer is: -0.20871215252208, -4.79128784747792. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B5%2Ax%2B1+%29






Answer by unlockmath(1614) About Me  (Show Source):
You can put this solution on YOUR website!
Hello,
First off let's get rid of the fractions by multiplying each (x+1)(x+2) to get:
(x+1)(x+2) + (5)(x+1) = (3)(x+2) This can be expanded out tobe:
x^2+3x+2+5x+5=3x+6 Now combine terms and subtract 3x and 6 from both sides to get 5:
x^2+5x+1=0
Now apply the quadratic equation: {{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }
x = (-5 +- sqrt(21))/2
There we go!
RJ
www.math-unlock.com