SOLUTION: 2-(a-7/a+4)=(a^2-5/a+4) can you help me solve that problem please? thanks

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Question 427110: 2-(a-7/a+4)=(a^2-5/a+4) can you help me solve that problem please? thanks
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
I'm guessing your equation is:
2-%28a-7%29%2F%28a%2B4%29=%28a%5E2-5%29%2F%28a%2B4%29
even though what you posted means:
2-%28a-7%2Fa%2B4%29=%28a%5E2-5%2Fa%2B4%29
(You're still not putting numerators and denominators in parentheses. Click on the "Show source" link just above this solution to see what I have typed.)

The first thing I like to so with these problems is to eliminate the fractions. After all, most people think working without fractions is easier that with fractions. The fractions in your equation can be eliminated by multiplying both sides of the equation by the Lowest Common Denominator (LCD) of all the denominators on both sides of the equation. Since the two denominators in your equation are the same, a+4, the LCD is simply x+4. Multiplying both sides by a+4:
%282-%28a-7%29%2F%28a%2B4%29%29%28a%2B4%29=%28%28a%5E2-5%29%2F%28a%2B4%29%29%28a%2B4%29
On the left side we need to use the Distributive Property to multiply:

All the denominators cancel:

leaving
%282%29%28a%2B4%29-%28a-7%29=a%5E2-5
Now we simplify. Multiply first:
2a%2B8-%28a-7%29=a%5E2-5
Subtract. (Be careful. Subtract both the a and the -7!):
a+%2B+15+=+a%5E2-5
Now we solve for a. This is a quadratic equation so we want one side to be zero. Subtracting a nd 15 from each side we get:
0+=+a%5E2-a-20
Now we factor (or use the Quadratic Formula). This factors fairly easily:
0 = (a-5)(a+4)
From the Zero Product Property we know that one of these factors must be zero. So:
a-5 = 0 or a+4 = 0
Solving these we get:
a = 5 or a = -4

Now we check our answers. When we multiplied both sides of the equation by the lCD, a+4, we multiplied by something that might be zero (depending on the value of a). Whenever you multiply both sides of an equation by an expression that might be zero, you must check your answer(s).

Use the original equation to check:
2-%28a-7%29%2F%28a%2B4%29=%28a%5E2-5%29%2F%28a%2B4%29
Checking a = 5:
2-%28%285%29-7%29%2F%28%285%29%2B4%29=%28%285%29%5E2-5%29%2F%28%285%29%2B4%29
We can already see that the denominators will not be zero when a = 5. This is the required part of the check. The rest of the check will tell us is we made a mistake. You are welcome to finish the check.

Checking a = -4:
2-%28%28-4%29-7%29%2F%28%28-4%29%2B4%29=%28%28-4%29%5E2-5%29%2F%28%28-4%29%2B4%29
We can already see that the denominators will be zero if a = -4. So we must reject this solution. (This kind of thing can happen whenever you multiply both sides of an equation by an expression that might be zero. It is why you must check your answer(s).)

So there is just one solution to your equation: a = 5.