How can you simplify these if possible?

Question

1.) (t^2 - 25) / (3t^2 - 16t + 5)
2.) (5t^2 +10t + 5) / (3t^2 + 6t + 3)
3.) (ab + a) / (a^2 -a)

Answer 1

Start of by factoring the equations:

1.) (t-5)(t+5)/ [(3t-1)(t-5)]
Then the t-5 in the denominator and the numerator would cancel out and you would have (t+5)/(3t-1)

2.) 5(t^2+2t+1) / [3(t^2+2t+1) =
the t^2+2t+1 would cancel out and you would have 5/3

If you factor the t^2+2t+1 you would have (t+1)(t+1). But it would still cancel out.

3.) a(b+1)/ [ a(a-1)]
The a would cancel out so you would have b+1/(a-1)

Answer 2

Ok, you have to remember that when doing these type of questions you have to first take out the common term.

1) (t^2-25)/(3t^2-16t+5) = (t-5)(t+5)/ (3t-1)(t-5)
= (t+5)/(3t-1)

In (1), you have to cancel out the common term in the numerator and the denominator. ie. (t-5)


2) (5t^2+10t+5)/(3t^2+16t+3) = 5(t^2+2t+1)/3(t^2+2t+1)
= 5(t+1)(t+1)/3(t+1)(t+1)
= 5/3

In (2), you have to first take out the common term in the numerator and denominator separately. That is, in the numerator, '5' is the common term whereas in the denominator, '3' is the common term. After taking out the common term, you can factorise and cancel out the common terms ie. (t+1) in the numerator and denominator.


3) (ab+a)/(a^2-a) = a(b+1)/a(a-1)

In (3), a(b+1)/a(a-1) is the final simpliest answer.

Answer 3

(1). (t^2 -- 25) / (3t^2 -- 16t + 5)
= [t^2 + 5t -- 5t + 25] / [3t^2 -- 15t -- t + 5]
= (t + 5)*(t -- 5) / (t -- 5)*(3t -- 1)
= (t + 5) / (3t -- 1)

(2) (5t^2 + 10t + 5) / (3t^2 + 6t + 3)
= 5(t + 1)*(t + 1) / 3(t + 1)*(t + 1)
=5/3

(3) (ab + a) / (a^2 -- a)
= a(b + 1) / a(a -- 1)
= (b + 1) / (a -- 1)

Answer 4

1. (t-5)(t+5)
===========
(3t-1)(t-5)

The 2 t-5's cancel out

Answer t+5/3t-1

2. (5t+5)(t+1)
============
(3t+3)(t+1)

the 2 t+1 cancel out

answer 5t+5/3t+3

3. a(b+1)
==========
a(a-1)

the 2 a's cancel out

answer b+1/a-1