Question

Could you show me the process? [algebra 2]

Answer 1

\[\large\rm \frac{j+4}{j+2}=2-\frac{1}{j}\]Hmm, so here is one approach.
We can multiply both sides by the `Least Common Multiple` of our denominators,
that will eliminate all fractions,
making things easier to solve.

Answer 2

One of our denominators is (j+2) and another is j,
so we should multiply both sides by \(\large\rm j(j+2)\) to get rid of all fractions.

Answer 3

\[\large\rm j\cancel{(j+2)}\frac{j+4}{\cancel{j+2}}=\left(2-\frac{1}{j}\right)j(j+2)\]On the right side we have to distribute the j(j+2) to each term,\[\large\rm j(j+4)=2j(j+2)-\frac{1}{\cancel j}\cancel j(j+2)\]Leaving us with,\[\large\rm j(j+4)=2j(j+2)-(j+2)\]What do you think?
This approach too confusing? :)

Answer 4

This is what I'm thinking XD

Answer 5

xD

Answer 6

did the approach presented by zepdrix make sense to you, Welp? : )

Answer 7

Thank you!