Add or subtract

Question

1) 4b-2/3b + b/b+2

 

2)9/n - 8/n+1

Answer 1

For adding or subtracting fractions, you need to have a common denominator. The easiest way to to that is to multiply the first term by the denominator of the second over itself, and multiply the second term by the denominator of the first over itself. I'll explain what that means below. Then you can add, subtract and simplify the fractions.

 

1) 4b-2/3b + b/b+2

 

Don't worry about adding or subtracting yet, first you need a common denominator.

 

So first, take (4b-2/3b) and times by (b+2/b+2), since b+2 is the denominator of the other fraction. Since (b+2/b+2) is really just 1 (anything over itself is 1, like 4/4, or 4356/4356), you're technically not changing the value of the fraction. That gets you (4b^2)-2b+8b-4 over (3b^2)+6b.

 

On the other side, where you have b/b+2, multiply the top and bottom by 3b. That gets you 3b^2 over (3b^2)+6b.

 

Now, you see you have common denominators for both fractions, which allows you to add the numerators. That gets you (7b^2)+6b-4 over (3b^2)+6b.

 

2) Same principles apply.

 

Multiply the top and bottom of 9/n by the other fraction's denominator, in this case n+1. That gets you 9n+9 over (n^2)+n.

 

 

Then multiple both 8 and n+1 by n, the denominator of the first fraction. That gets you 8n over (n^2)+n. Now you have another common denominator, so you can start adding and subtracting.

 

9n+9 over (n^2)+n minus 8n over (n^2)+n equals n+9 over (n^2)+n.

Answer 2

Hi Sierra,

 

1) 4b-2/3b  + b/b+2 = (4b-2)(b+2) + b (3b) /3b(b+2)

 

                               =4b² +8b -2b -4 + 3b² / 3b(b+2)

 

                                ⁼7b² +6b-4 / 3b² +6b

²⁾

   9/n - 8/n+1 = 9(n+1)-8n /n(n+1)

 

                     =9n+9 -8n /n(n+1)

 

                     =n+9 /n² + n