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factor

Solved. How? Well ...

With h. One has to first factor the 8p + 2pq (denominator). Then one can eliminate common terms. That's how one is left with 4 + q (after factoring). If I forget to factor first, then I can get an answer that might feel right but isn't.

Woth C, we need to factor the numerator first. The denominator is already factored.

Last is the one in black and white. Both the numerator and the denominator can be factored. Then, we can solve by eliminated common terms. That's why we get a 1 + 2b as the correct answer.

My lesson from this is that it's good to keep up with learning. These questions were from a textbook I'm using to supplement my education.

Dinesh was right. "factor".

H.

1. for this equation we have 12p/(8p+2pq)

2. factor denominator

8p+2pq -> 2p(4+q)

we now have 12p/(2p(4+q))

3. simplify

12p/2p(4+q)

4. we now have 6/(4+q)

bottom right problem.

1. factor out "a" from both expressions

(3ac+5a)/(a+2ab) -> (a(3c+5))/(a(1+2b))

2. notice that because "a"is factoring out itself, we place a "1" because a(1)=a, hence "1+2b"

3. simplify

(a(3c+5))/(a(1+2b))

we now have (3c+5)/(1+2b)

C.

1. factor numerator

((7x^2y)-(5x^2y))/12xy

7x^2y -> 7x(xy)

5x^y->5x(xy)

2. we can now factor out (xy) from both the numerator and denominator

(7x(xy)-5x(xy))/12xy

we are left with

(7x-5x)/12

3. simplify

2x/12 -> x/6

4. we now have x/6

hope this helps