Question

Algebra 2 Help please! @DisplayError Mind helping with another, please?? Will attach it.

Answer 1

Generally when we're looking for the least common denominator, we're looking to make the denominator the same so we can carry out addition or subtraction. You're essentially multiplying the numerator and denominator by the same value (which equals 1), which doesn't change the value of the number you're multiplying. In your problem:
\[\frac{x+1}{3} - \frac{x+2}{2x}\]
We're told that the LCD to use is 6x. Generally, if you multiply all of the fractions by the product of all the denominators, you should get the LCD. Thus:
\[\frac{x+1}{3} \times \frac{3 \times 2x}{3 \times {2x}}\]
The 3 in the numerator cancels out the 3 in the denominator, and we're left with
\[(x+1) \times \frac{2x}{3 \times {2x}} = \frac{(x+1) \times 2x}{6x}\]
Doing the same with the second part of the fraction:
\[\frac{x+2}{2x} \times \frac{3 \times 2x}{3 \times {2x}} = (x+2) \times \frac{3}{3 \times 2x} = \frac{(x+2) \times 3}{6x}\]
Our expression then becomes
\[\frac{(x+1) \times 2x}{6x} - \frac{(x+2) \times 3}{6x}\]
Distribute the 2x and 3 into the expression in parentheses, subtract, and done!

Answer 2

My grandfather says "What's the final answer?"

Answer 3

I think it should be
\[\frac{2x^2 + 2x}{6x} - \frac{3x + 6}{6x}\]
So combine the expression (remember to distribute the negative sign first).

Answer 4

Okay, we think we got it. Thank you, Error! :)