Question

help me solve this equation...

Answer 1

ok im ready lady

Answer 2

\[\frac{ 12r }{ r+2 } = \frac{ 4 }{ r+2 }-6\]

Answer 3

get the 6 in terms of r+2
then you can cancel the denominator

Answer 4

wait what?

Answer 5

\(\huge 6*\frac{r+2}{r+2}\)

Answer 6

wouldnt it be -6?

Answer 7

@bibby

Answer 8

same difference

Answer 9

soo is it like 6r+12?

Answer 10

the way i see it it's -(6)
so if we subtract 6(r+12) it'd be
\(12r=4-(6(r+12))\)

Answer 11

so what do i do after that?

Answer 12

\(\large \frac{ 12r }{ r+2 } = \frac{ 4 }{ r+2 }-\frac{6(r+2)}{r+2}\) distribute the 6
\(\large \frac{ 12r }{ r+2 } = \frac{ 4 }{ r+2 }-\frac{6r+12}{r+2}\) combine fractions
\(\large \frac{ 12r }{ r+2 } = \frac{ 4-(6r+12) }{ r+2 }= \frac{ 12r }{ r+2 } = \frac{ (6r-8) }{ r+2 }\)

multiply both sides by r+2

Answer 13

both fractions?

Answer 14

what gets done to one side must be done to the other side. this is the essence of algebra
\(\large \cancel{r+2}*\frac{ 12r }{ \cancel{r+2} } = \frac{ (6r-8) }{ \cancel{r+2} }*\cancel{r+2}\)

Answer 15

so then its 12r=6r-8

Answer 16

you subtract 6r on both sides right?

Answer 17

yeah, 6r=-8
r=\(\frac{-8}{6}\)

Answer 18

i want to plug it in and check

Answer 19

so wait thats the answer?

Answer 20

\(-16=4-6*\frac{-8}{6}\)
\(-16=4+8=12\) I think I messed up somewhere but I can't see where