alg 2 help .. please ?

Question

Add the following:

[(4-x)/(9-x^2)] + [(6)/(5x+15)]

a. (38-11x)/5(x+3)(3-x)
b. (30-10x)/5(x+3)(3-x)
c. (x+3)(x-3)/5(x+3)
d. 11(3+x)/5(x+3)(3-x)




Multiply the following:

[(2-x)/(8x-12)]*[(4x)/(x-2)]

a. x/(2x-3)
b. -x/(2x-3)
c. 2x/(2x-3)
d. x/(x-3)

Answer 1

(1)



Here, the denominator terms are (9 - x^2 ) and (5x+15)

We hav e to find the LCD.

( 9 - x ^2 ) : 1* ( 3 + x ) *( 3 - x )

(5x + 15 ) : 5 * ( x + 3 )
-------------------------------------------------------
5 * ( 3+x ) * ( 3 - x )
--------------------------------------------------------
So, [ 5 ( 4 - x ) / 5 * (3 +x)*(3-x) ] + [ 6 (3 - x ) / 5 * (3+x)*(3-x) ]

= [ 20 - 5x + 18 - 6x ] / [ 5 *(3+x)*(3-x) ]

= [ 38 - 11x] / [ 5 * (3+x)*(3-x) ]

Hence, Choice (a) is the right answer.

******************************************************************

(2)



Here, (8x-12) = 4 ( 2x - 3 )

So,

[ ( 2-x ) / (8x - 12 ) ] * [ (4x) / (x-2) ]

= [ (2 -x) * 4 x ] / [ 4 ( 2x - 3 ) ( x - 2 ) ]

Cancel out 4 in both numerator and denominator. and we can

write ( 2 - x ) as - ( x - 2 ) .

So, the expression becomes,

[ - ( x - 2) x ] / [ ( 2x - 3 ) ( x-2) ]

Now, cancel out ( x-2).

Therfore we have, - ( x - 2 ) / ( 2x - 3 )

Thus the choice is (b).

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Answer 2

Do some factoring to help get a common denominator:

[(4-x)/(9-x^2)] + [(6)/(5x+15)]
[(4-x) / (3-x)(3+x) ] + [ 6 / 5(x+3)]
[ 5(4-x) / 5(3-x)(3+x) ] + [ 6(3-x) / 5(x+3)(3-x)]
[ 5(4-x) + 6(3-x) ] / 5(3+x)(3-x)
[ 20 - 5x + 18 - 6x ] / 5(3+x)(3-x)
[ 38 - 11x ] / 5(3+x)(3-x)

Now do the second one using the same kind of reasoning.

Answer 3

(4-x)/[(3+x)(3-x)] + 6/[5(x+3)] =
5(4-x)/[5(3+x)(3-x)] + 6(3-x)/[5(x+3)(3-x)] =
(20 - 5x + 18 - 6x)/[5(3+x)(3-x)] =
(38 - 11x)/[5(3+x)(3-x)]
that's a.

(2-x)/[4(2x-3)] • 4x/(x-2) =
-x/(2x-3)
that's b.

Answer 4

c

c