SOLUTION: "What is the solution of (2/r+2)+(3/r+3)=(5/r+2)"

Algebra ->  Rational-functions -> SOLUTION: "What is the solution of (2/r+2)+(3/r+3)=(5/r+2)"      Log On


   

Question 271240: "What is the solution of (2/r+2)+(3/r+3)=(5/r+2)"
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
2%2F%28r%2B2%29%2B3%2F%28r%2B3%29=5%2F%28r%2B2%29
Equations without fractions are generally easier to solve than those with fractions. So we will start by getting rid of the fractions. To get rid of the fractions in an equation, multiply both sides of the equation by the Lowest Common Denominator (LCD). (The LCD is normally used when you try to add or subtract fractions. But it is also useful in eliminating fractions in an equation.

The LCD of all the fractions is (r+2)(r+3). So this is what we will multiply by:

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

Now we can start canceling:

leaving:
%28r%2B3%29%282%29%2B%28r%2B2%29%283%29+=+%28r%2B3%29%285%29
This is a fairly simple equation to solve. First we simplify:
2r%2B6%2B3r%2B6+=+5r%2B15
5r%2B12+=+5r%2B15
Subtracting 5r from both sides we get:
12+=+15
The variable has disappeared!? This means that either there is no solution or the solution is all numbers. The fact that the equation we now have is a false statement (12 is NOT equal to 15!) tells us that there is no solution to this equation. IOW, 2%2F%28r%2B2%29%2B3%2F%28r%2B3%29=5%2F%28r%2B2%29 can never be true, no matter what number you use for r.