Posted by **Sara** on Monday, June 13, 2011 at 9:31pm.

I was wondering if I solved these problems right?

Simplify the following:

3xy^2 4xz

______ + ______ = 7xyz

2y 3x _____

5

2(x-1) 3(x^2-4)

______ * ________ = 6(x-2)

4(x+2) 5x-5 ______

20

x^2-x-6 3(x^2-4)

________ / __________ = (x-3) (x+5)

7x+7 x^2+6x+5 ______________

21(x-2)

- Math -
**Reiny**, Monday, June 13, 2011 at 10:10pm
Wow, did you look how your typing showed up ??

I will take a "guess" what the last one is ....

(x^2 - x - 6)/(7x+7) ÷ 3(x^2-4)/(x^2+6x+5) = (x-3)(x+5)/(21(x-2))

(x-3)(x+2)/(7(x+1)) (x+1)(x+5)/(3(x+2)(x-2)) = (x-3)(x+5)/(21(x-2))

(x-3)(x+5)/(21(x-2) = (x-3)(x+5)/(21(x-2))

everything cancels for

1 = 1

so the equation is an identity and true for all values of x, except x ≠ -1, ± 2,

retype the others using brackets.

- Math -
**Sara**, Monday, June 13, 2011 at 11:06pm
1)

(3xy^2)/(2y) + (4xz)/(3x)

2)

(2(x-1))/(4(x+2)) *(3(x^2-4))/(5x-5)

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