Question

Algebra 2 help?

Answer 1

can you check a couple and help me with a couple?

Answer 2

@zepdrix What is the least common denominator of the fraction 2 over x plus the fraction x over x plus 1 plus the fraction x plus 1 over 3 x

Answer 3

I got 3x(x + 1)

Answer 4

Let me check your work.

Answer 5

Yes, you're correct!

Answer 6

But let @zepdrix verify it.

Answer 7

What is the least common denominator of the fraction 4 over x plus the fraction 2 x over x minus 1 plus the fraction x plus 3 over 2 x

I got 2(x-1)?

Answer 8

2x(x-1)

Answer 9

lcd of the fraction 3 x over x plus 1 plus the fraction x plus 1 over 2 x plus the fraction 5 over x

I got 2x(x+1)?

Answer 10

@IMStuck could you check my answers? :)

Answer 11

I"m not sure what's the answer for this one.

Answer 12

What is the least common denominator of the fraction 7 over 3 x plus the fraction 2 over x plus the fraction x over x minus 1


I got 3(x-1)?

Answer 13

What is the simplified form of the quantity x over 3 plus y over 4 all over the quantity x over 4 minus y over 3 I don't know how to do this :/

Answer 14

1. What is the least common denominator of the fraction 2 over x plus the fraction x over x plus 1 plus the fraction x plus 1 over 3 x

I got 3x(x+1)

Answer 15

2. What is the least common denominator of the fraction 4 over x plus the fraction 2 x over x minus 1 plus the fraction x plus 3 over 2 x

I got 2x(x-1)

Answer 16

@radar can you check ^^

Answer 17

3. What is the least common denominator of the fraction 3 x over x plus 1 plus the fraction x plus 1 over 2 x plus the fraction 5 over x

I got 2x(x+1)

Answer 18

4. What is the least common denominator of the fraction 7 over 3 x plus the fraction 2 over x plus the fraction x over x minus 1

I got 3(x-1)

Answer 19

wait puppy come back @vzfreakz

Answer 20

After the fraction x plus 1 all over 3 minus the fraction x plus 2 all over 2 x have been combined using the least common denominator of 6x, what is the numerator?

I got 2x^2 - x - 6

Answer 21

Ask, these questions on a new forum.