algebra help please....

Hi,

...4..........5
--------.-.--------- =
4 - a²....6 + 3a

...4................5
--------------.-.------------ =
(2-a)(2+a)...3(2 + a)

Multiply the top and bottom of the first fraction by 3 and multiply the top and bottom of the second fraction by (2 - a) to make a common denominator.

4(3)................5(2-a)
----------------.-.---------------- =
3(2-a)(2+a)...3(2+a)(2-a)


12.-.5(2-a)
----------------- =
3(2-a)(2+a)


12.-.10 + 5a
------------------ =
3(2-a)(2+a)

2 + 5a
---------------- <==ANSWER
3(2-a)(2+a)

I hope that helps!! :-)

= (4/[4 - a²]) - (5/[6 + 3a])
= (- 4/[{a + 2}{a - 2}]) - (5/[3{a + 2}])
= (- 12/[3{a + 2}{a - 2}]) - ([5a - 10]/[3{a + 2}{a - 2}])
= (- 12 - 5a + 10)/(3[a + 2][a - 2])
= - (5a + 2)/(3[a + 2][a - 2])

Answer: - (5a + 2)/(3[a + 2][a - 2])

There is a sign missing in question, unfortunately.

4/(4 - a^2) - 5/(6 + 3a)
= 4/(2 + a)(2 - a) - 5/3(2 + a)
= 4(3)/3(2 + a)(2 - a) - 5(2 - a)/3(2 + a)(2 - a)
= 12/3(2 + a)(2 - a) - (10 - 5a)/3(2 + a)(2 - a)
= (12 - 10 + 5a)/3(2 + a)(2 - a)
= (2 + 5a)/3(2 + a)(2 - a)

4 - 5
------------------
4-a^2 6+3a

4 - 5
--------------
2-a 2+a 3(2+a)

3(2-a)(2+a) is the lcd

12- 5(2-a)
12-10-5a

2+5a/ 3(2-a)(2+a)

Factor the denominators:
4 - a^2 --> (2 - a)(2 + a) and 6 + 3a --> 3(2 + a)

LCD would be 3(2 - a)(2 + a) so the left side needs a 3 and the right side needs a (2 - a).

4 * 3 - 5(2 - a) --> 12 - 10 + 5a --> 2 + 5a (just the numerator)

(2 + 5a) / {3(2 - a)(2 + a)}

It's important that you make sure all the signs are present or we cannot solve your equation.

You have (4)/(4-a^2)-(5)/(6 3a)

But is that last part (6+3a) or is it (6-3a) or just plain (63a)??

Can't simplify this unless we know so please edit and resubmit.