Math
posted by Sara .
I was wondering if I solved these problems right?
Simplify the following:
3xy^2 4xz
______ + ______ = 7xyz
2y 3x _____
5
2(x1) 3(x^24)
______ * ________ = 6(x2)
4(x+2) 5x5 ______
20
x^2x6 3(x^24)
________ / __________ = (x3) (x+5)
7x+7 x^2+6x+5 ______________
21(x2)

Math 
Reiny
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^24)/(x^2+6x+5) = (x3)(x+5)/(21(x2))
(x3)(x+2)/(7(x+1)) (x+1)(x+5)/(3(x+2)(x2)) = (x3)(x+5)/(21(x2))
(x3)(x+5)/(21(x2) = (x3)(x+5)/(21(x2))
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
1)
(3xy^2)/(2y) + (4xz)/(3x)
2)
(2(x1))/(4(x+2)) *(3(x^24))/(5x5)
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