Answer:
The answer is B) [tex]\frac{2x-3}{2x+1}[/tex]
Step-by-step explanation:
Step 1: First we have to factorize the each rational expressions.
[tex]\frac{4x^2 - 9}{6x^2 + 13x + 6}[/tex] ÷[tex]\frac{4x^2 - 1}{6x^2 + x -2}[/tex]
= [tex]\frac{(2x - 3)(2x + 3)}{(3x + 2)(2x + 3)}[/tex]÷[tex]\frac{(2x+1)(2x -1)}{(2x -1)(3x +2)}[/tex]
Now we can cancel out (2x + 3) and (2x -1) since they are common factor both in the numerator and the denominator, so we get
= [tex]\frac{2x - 3}{3x + 2}[/tex] ÷[tex]\frac{2x + 1}{3x + 2}[/tex]
Step 2: If we have fraction over fraction, we can flip the second fraction and multiply.
= [tex]\frac{2x - 3}{3x + 2}[/tex] x [tex]\frac{3x + 2}{2x + 1}[/tex]
Now we can cancel out (3x + 2), so we get
= [tex]\frac{2x-3}{2x+1}[/tex]
Thefore, the answer is B.
Thank you.
Answer:
Second choice
Step-by-step explanation:
If we factor all the numerators and denominators we get
(2x - 3)(2x + 3) (2x - 1)(2x + 1)
-------------------- ÷ -----------------
(2x + 3)(3x + 2) (3x + 2)(2x - 1)
2x - 3 3x + 2
= -------- * --------- = 2x - 3 / 2x + 1 (answer) (second choice)
3x + 2 2x + 1
