Question

5/7y- 15=5y+30

Answer 1

Get all the ys on once side, all the non-ys on the others. Know how to do that?

Answer 2

no

Answer 3

5/7y- 15=5y+30

5/7y-5y = 30 + 15

(5/7-5)y = 45

simplify the left hand side (5/7-5) and then divide both sides by that result to get y by itself on the left hand side.

Answer 4

OK, you have this: \(\frac{5}{7}y- 15=5y+30\).... I was going to start by making sure the y was not on the bottom of that fraction....

Answer 5

i am still confused

Answer 6

OK. first, it has a -15 on the left, correct? If you add 15, it will balance that. But if you do it to the left you must do it to the right.

\(\frac{5}{7}y- 15+15=5y+30+15\)
Then simplify
\(\frac{5}{7}y=5y+45\)

That part make sense?

Answer 7

yes it makes sense

Answer 8

OK. Now, the same sort of thing can be done to the 5y, but it needs to be subtracted.

Answer 9

i still can not get the answer that is -7/2

Answer 10

That is OK, we will get there. If you subtract off the 5y, what does it become on the right and left?

Answer 11

30/7

Answer 12

\(\frac{5}{7}y-5y=45\)
\(\frac{5}{7}y-\frac{35}{7}y=45\)
\(-\frac{30}{7}y=45\)

Answer 13

That is what i have so what is next?

Answer 14

Hmmm... I see what you mean. Make sure you have the problem copied right. I see a -21/2 coming.

Answer 15

i have it copied right

Answer 16

\((-\frac{7}{30})(-\frac{30}{7})y=45(-\frac{7}{30})\)
\(y=3(-\frac{7}{2})\)
\(y=-\frac{21}{2}\)

I also do not get the book answer.

Answer 17

how did you get the -7/2 to part ?

Answer 18

I simplified the 45 and 30. 45/15=3, 30/15=2.

Answer 19

what does that mean

Answer 20

\(\frac{4}{5}\times \frac{1}{4}\implies \frac{\not 4}{5}\times \frac{1}{\not 4}\implies \frac{1}{5}\times \frac{1}{1}\implies \frac{1}{5}\)

Same sort of thing.

Answer 21

\[y=45(-\frac{7}{30})\\
\implies y=\frac{45}{1}(-\frac{7}{30})\\
\implies y=\frac{3\times 15}{1}(-\frac{7}{2\times 15})\\
\implies y=\frac{3}{1}(-\frac{7}{2})\]

Answer 22

Thanks very muchi got it now

Answer 23

No problem. It is simplifying out the comon factor, which can really help at times.