Question

Part 1: Explain how you calculate the restricted value(s) of x for the equation (See attachment) and what those value(s) represent.
Part 2: Explain which method (cross-multiplication or LCD) you would use to solve the equation and why.
Part 3: Using complete sentences, explain the steps to solve the equation using the method you chose in step 2. Provide the solution to the equation.

Answer 1

Part one :
A restriction is any value of x that would make the denominator 0 . In this case : -2 would make the denominator zero .

I need help with the rest please !!!!!!

Answer 2

also 0 would make a denominator 0, so both \(-2\) and 0 are out

Answer 3

Okay , thank you !! Can you help me with the rest please ?

Answer 4

ok

Answer 5

you have a choice, but i would multiply everything on both sides by \(2x(x+2)\) cancelling as you go

Answer 6

\[\frac{3}{x+2}+\frac{1}{2x}=\frac{4}{x+2}\]
\[2x(x+2)\left(\frac{3}{x+2}+\frac{1}{2x}\right)=2x(x+2)\frac{4}{x+2}\]

Answer 7

What method would that be man ?

Answer 8

that would be the "LCD" method because i multiplied both sides by the LCD of \(2x\) and \(x+2\) namely \(2x(x+2)\)

Answer 9

we get after cancelling
\[6x+x+2=8x\]

Answer 10

solve to get
\[7x+2=8x\] or
\[x=2\]

Answer 11

I see . . . What would be my reason for using that method though ?

Answer 12

because no one likes fractions, so it is best to clear them if you can

Answer 13

Lol thats true , let me look over this . .

Answer 14

Thanks man ! @satellite73