Question

PLEASE HELP ADDITION SUBTRACTION DIVISION

Answer 1

If you could PLEASE help me w/ these i would really appreciate please show work so i can actually learn it :)

Answer 2

Sure. I'll help. Hold on.

Answer 3

\[\frac{8}{x + 2} + \frac{4}{x + 3} = \frac{8}{x + 2} \times \frac{x + 3}{x + 3}+ \frac{4}{x + 3} \times \frac{x + 2}{x + 2} = \frac{8(x + 3) + 4(x + 2)}{(x + 2)(x + 3)} = \]\[\frac{8x + 24 + 4x + 8}{x^{2} + 5x + 6} = \frac{12x + 32}{x^{2} + 5x + 6}\]

Answer 4

The restrictions are the same as last time. Just set it equal to 0 and solve.
x + 2 ≠ 0
x + 3 ≠ 0
x ≠ -2, -3

Answer 5

Next Problem:
\[\frac{2}{x - 4} - \frac{7x - 28}{(x - 4)^{2}} = \frac{2}{x - 4} - \frac{7(x - 4)}{(x - 4)^{2}} = \frac{2}{x - 4} - \frac{7\cancel{(x - 4)}}{(x - 4)\cancel{^{2}}} = \frac{2 - 7}{x - 4} = \frac{-5}{x - 4}\]

Answer 6

Restrictions, same.
x - 4 ≠ 0
x ≠ 4

Answer 7

Try doing the last one?

Answer 8

ok for the first one
x ≠ 2, 3
or
x ≠ −6, −3, −2

Answer 9

@Calcmathlete

Answer 10

The restrictions for the last one is strange to me...
@zepp Can you help with the last one for me?

Answer 11

no for the first one the restrictions the 12x+32 one

Answer 12

Which?

Answer 13

the first 1

Answer 14

\[\large \frac{t+3}{2t}\div\frac{2t+6}{t-4}\]?

Answer 15

oh

Answer 16

@zepp can you help me w/ the division one please!!

Answer 17

You'd have the flip it, and multiply

\[\large \frac{t+3}{2t}*\frac{t-4}{2t+6}=\frac{(t+3)}{2t}*\frac{(t-4)}{2(t+3)}=\frac{1}{2t}*\frac{(t-4)}{2}=\frac{(t-4)}{4t}\]

Answer 18

Then restrictions would be \(t \ne4,0\)

Answer 19

oh i forgot abot the flipping thing thnx you are a life saver!

Answer 20

You are welcome :)