Question

Solve. 1/6r+1/4=1/3+3/4r

Answer 1

\[\frac{1}{6}r+\frac{1}{4}=\frac{1}{3}+\frac{3}{4}r\] like that?

Answer 2

multiply both sides by the least common multiple of 6 3 and 4, which is 12
that way, all the denominators will cancel and you will have whole numbers to work with

Answer 3

yes

Answer 4

\[12\left(\frac{1}{6}r+\frac{1}{4}\right)=12\left(\frac{1}{3}+\frac{3}{4}r\right)\]

Answer 5

6 goes in to 12 twice, 4 goes in to 12 3 times and 3 goes in to 12 4 times, so the denominators all go and you get
\[2r+3=4+3\times 3r\] or
\[2r+3=4+9r\] is that step ok or confusing?

Answer 6

i'm trying to get it lol processing

Answer 7

why did you multiply 12?

Answer 8

ok i can write out in more detail if you like, but sometimes more detail is more confusing
the idea is that you really don't want to do all the work with fractions, so if you multiply both sides by the least common multiple of the denominators they will go away

Answer 9

in other words, find a number that 6, 3 and 4 go in to evenly

Answer 10

it is pretty clear that they all divide 12 evenly, so if you multiply both sides of the equation by 12 there will be no more fractions

Answer 11

oh! ok i got it

Answer 12

you have to multiply everything by 12, which really means cancel as you go

Answer 13

\[12\left(\frac{1}{6}r+\frac{1}{4}\right)=12\left(\frac{1}{3}+\frac{3}{4}r\right)\]
\[12\times \frac{1}{6}r+12\times \frac{1}{4}=12\times \frac{1}{3}+12\times \frac{3}{4}r\]
\[2r+3=4+9r\] you do the middle step in your head

Answer 14

then it should be easier
\[3=4+7r\]
\[-1=7r\]
\[-\frac{1}{7}=r\]

Answer 15

hope all steps are clear
you could work with the fractions but that would require adding and subtracting fractions, and dividing by fractions, which is more work

Answer 16

It was a bit confusing before but you made it a lot easier thank you!

Answer 17

yw