SOLUTION: How do I reduce the fraction (4x+3)/(20x^2+23x+6) to simplest form, and include any restrictions on x?

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Question 530528: How do I reduce the fraction (4x+3)/(20x^2+23x+6) to simplest form, and include any restrictions on x?
Answer by algebrahouse.com(1659) About Me  (Show Source):
You can put this solution on YOUR website!
4x + 3
---------------
20x² + 23x + 6

4x + 3
----------------- {factored bottom into two binomials}
(4x + 3)(5x + 2)

1
------ {cancelled 4x + 3 on top and bottom}
5x + 2

Restrictions on x, would be those that would make the denominator equal to 0.

5x + 2 = 0 {set denominator equal to 0}
5x = -2 {subtracted 2 from both sides}
x = -2/5 {divided both sides by 5}
Therefore x cannot be -2/5

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