Please help me!
Problem:
x+2 x-3
_________ - _________
2x^2+5x+2 2x^2-5x-3
*Simplify the expression
2x^2+5x+2 =(x+2)(2x+1)
2x^2-5x-3 =(2x+1)(x-3)
So x+2/(x+2)(2x+1) = 1/2x+1
x-3/(2x+1)(x-3) = 1/2x+1
Finally 1/2x+1 - 1/2x+1 = 0
Let's take a look at the problem:
(x+2)/(2x^2+5x+2) - (x-3)/(2x^2-5x-3)
Let's try to factor the denominators first:
2x^2+5x+2 factors are (2x+1)(x+2)
2x^2-5x-3 factors are (2x+1)(x-3)
Now, let's rewrite the problem:
(x+2)/(2x+1)(x+2) - (x-3)/(2x+1)x-3)
Reduce each fraction, then you will have the same denominator for both fractions. Once you do those steps, you should easily see what the answer will be.
I hope this helps.
To simplify the expression,
(x+2)/(2x^2+5x+2) - (x-3)/(2x^2-5x-3)
First, factor the denominators:
2x^2+5x+2 = (2x+1)(x+2)
2x^2-5x-3 = (2x+1)(x-3)
The expression now becomes:
(x+2)/[(2x+1)(x+2)] - (x-3)/[(2x+1)(x-3)]
Next, reduce each fraction:
(x+2) cancels out with one (x+2) from the denominator, leaving:
1/(2x+1) - (x-3)/[(2x+1)(x-3)]
Now, find a common denominator for the two fractions, which is (2x+1)(x-3):
[(x-3)]/(2x+1)(x-3)
Subtract the two fractions:
1/(2x+1) - [(x-3)]/(2x+1)(x-3)
To subtract fractions, you need a common denominator. So, multiply the first fraction by (x-3)/(x-3) and the second fraction by 1/(2x+1):
[x-3]/[(2x+1)(x-3)] - [(x-3)]/(2x+1)(x-3)
Combine the terms:
[x-3 - (x-3)]/[(2x+1)(x-3)]
The numerator becomes zero:
0/[(2x+1)(x-3)]
Finally, simplify the zero numerator. The expression simplifies to:
0
So, the simplified expression is 0.