help please; keep getting wrong answer..

[bobisaka]

Can anybody tell me what I am doing wrong?



LCD = (v+1)(v-1)

v (v-1)/(v+1)(v-1) + 3(v+1)/(v-1)(v+1) - 6/(v-1)(v+1)

v-1/(v+1)(v-1) + 3v+3/(v+1)(v-1) -6/(v+1)(v-1)

Numerator = v+3v-1+3-6 = 4v-4

4(v-1)/(v+1)(v-1)

4/v+1

 

[Deleted member 4993]

Can anybody tell me what I am doing wrong?

View attachment 18574

LCD = (v+1)(v-1)

v (v-1)/(v+1)(v-1) + 3(v+1)/(v-1)(v+1) - 6/(v-1)(v+1)

v-1/(v+1)(v-1) + 3v+3/(v+1)(v-1) -6/(v+1)(v-1) <<<< I believe I am doing something wrong with the subtraction and distributing it to the 6 to create a -6?

Numerator = v+3v-1+3-6 = 4v-4

4(v-1)/(v+1)(v-1)

4/v+1

In the 3rd. line of your solution:

v(v-1) ...... after multiplication should turn out to be v² - v

 

[bobisaka]

AH YES, i will come back after i redo it.

Edit:

v^2 + 3v - v + 3 - 6

v^2 + 2v -3

(v+3)(v-1)/(v+1)(v-1)

v+3/v+1


Thank you!

 

[HallsofIvy]

One comment: what you have written, \(\displaystyle v+ 3/v+ 1= v+ \frac{3}{v}+ 1\) is wrong! You should have \(\displaystyle \frac{v+ 3}{v+ 1}= (v+3)/(v+ 1)\). Parentheses are important!

 

[Deleted member 4993]

One comment: what you have written, \(\displaystyle v+ 3/v+ 1= v+ \frac{3}{v}+ 1\) is wrong! You should have \(\displaystyle \frac{v+ 3}{v+ 1}= (v+3)/(v+ 1)\). Parentheses are important!

The original post also has the same mistake. You wrote:

v (v-1)/(v+1)(v-1) + 3(v+1)/(v-1)(v+1) - 6/(v-1)(v+1)

It should have been:

v (v-1) / [(v+1)(v-1)] + 3(v+1) / [(v-1)(v+1)] - 6 / [(v-1)(v+1)]

Always remember to follow PEMDAS (Hierarchy of arithmetic operations) - these mistakes can "knock" your grade down during an exam.