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.
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)
2 answers
1)
(3xy^2)/(2y) + (4xz)/(3x)
2)
(2(x-1))/(4(x+2)) *(3(x^2-4))/(5x-5)
(3xy^2)/(2y) + (4xz)/(3x)
2)
(2(x-1))/(4(x+2)) *(3(x^2-4))/(5x-5)