Can someone Subtract this problem for me...I'm soooo sleepy and tired.?

Question

(3v over v^2 - 7v+10) subtracted from (2v over v^2 -8v+15)

( ) = fractions

The difference is?

Answer 1

Factor the polynomials in the denominator:

2*v / (v-3)*(v-5) - 3*v / (v-2)*(v-5)

multiply the expression to make the denominators the same

2*v*(v-2) / (v-5)*(v-3)*(v-2) - 3*v*(v-3) / (v-5)*(v-3)*(v-2)

2*v*(v-2) - 3*v*(v-3)) ] / (v-5)*(v-3)*(v-2)

[2*v^2 - 4*v - 3*v^2 + 9*v] / (v-5)*(v-3)*(v-2)

[-v^2 + 5*v] / (v-5)*(v-3)*(v-2)

-v*(v-5) / (v-5)*(v-3)*(v-2)

-v / (v-3)*(v-2)

Answer 2

2v/(v^2 - 8v + 15) - 3v/(v^2 - 7v + 10) =

=2v/((v - 3)(v - 5)) - 3v/((v - 2)(v - 5)) =

(2v(v - 2) - 3v(v - 3))/((v-2)(v-3)(v-5)) =

= (2v^2 - 4v - 3v^2 + 9v)/((v-2)(v-3)(v-5)) =

= (v(5 - v))/((v-2)(v-3)(v-5)) = -v/((v-2)(v-3)(v-5))

Answer 3

Does the ^2 mean an exponent in computer terms?

I'm taking a crack at this:
Could the answer be (v over v^2 - 15v - 5) ?

Just a guess.

Answer 4

2v / (v- 5).(v - 3) - 3v / ((v - 5).(v - 2)
2v(v - 2) - 3v.(v - 3) / (v - 5).(v - 3).(v - 2)
(5v - v²) / (v - 5).(v - 3).(v - 2)
-v.(v - 5) / (v - 5).(v - 3).(v - 2)
- v / (v - 3).(v - 2)

Answer 5

-v / {(v-2)(v-3)}

Answer 6

you first factor the denominators... then find the common denominator, which is (v-3)(v-5)(v-2)...

anyway... just to give you the answer...

-v over (v-3)(v-2)

Answer 7

2v/(v-3)(v-5)-3v / (v-2)(v-5)

2v(v-2)-3v(v-3)/v-2;v-3;v-5

2v^2-4v-3v^2+9v/v-2;v-3;v-5

-v^2+5v/v-2;v-3;v-5
-v(v-5).v-2;v-3;v-5
-v/(v-2)(v-3)

Answer 8

whts the meaning of over