Rearrange 5y-3x-4=0 to get y=mx+c? Help please

Original post by Pinkberry_y

That's what you think

lmao

Reply 22

clarkey500

Original post by vector12

Thank you. I wasn't sure what happened to the 0, but it seems that once you move the -4 to the other side and it becomes +4 to cancel it out, that then makes the 0 disappear from the equation and gets rid of it?

So basically, the 0 is gotten rid of once you move one of the subjects to the right side of the equation, and it is just a placeholder after the = sign?

You're welcome!
Yeah you can think of it in a way of 'zeroing' a unit on a side so when you add 4

5y - 3x - 4 = 0
(+4) 5y - 3x -4 + 4 = 0 + 4 (+4)
obviously 4-4 = 0. So it essentially 'zeroes' it on the y side - you do this till you get the y on it's own.

Reply 23

Original post by Pinkberry_y

That's what you think

It is.

Reply 24

Paranoid_Glitch

Original post by vector12

How do you do it? I can do it normally, just I cannot rearrange it with the 0.

Can someone also explain to me what the 0 does please, and if it means anything?

Thanks!

Add 3x to the right hand side and left hand side. Add 4 to the right hand side and left hand side.

You should end up with 5y=3x+4. because the 3x's and 4's subtract and give zero on the left hand side and the 3x and 4 add to the zero on the right hand side.

Divide everything by 5.

You should end up with y=3/5x + 4/5.

Makes sense?

(edited 7 years ago)

Reply 25

zainyyyyy

where does it say that?

Original post by RDKGames

Whether you agree with me or not is irrelevant. Posting full solutions on this forum is against its rules. Just hint the OP in the right direction next time.

Reply 26

Original post by zainyyyyy

where does it say that?

Sticky at the top of the forum. The one you're supposed to read before posting.

http://www.thestudentroom.co.uk/showthread.php?t=4066671&p=64637319#post64637319

Reply 27

Zacken

Original post by zainyyyyy

where does it say that?

http://www.thestudentroom.co.uk/showthread.php?t=4066671&p=64637319#post64637319

I'm guessing you didn't read the big red "Please Read - Posting Guide" thing at the top of the page...?

Reply 28

anosmianAcrimony

Original post by Pinkberry_y

Y = 3/5x +20

BTW it's against the rules to just give the answer. Also, I would guess that it's super against the rules to give the wrong answer.

Reply 29

Original post by anosmianAcrimony

BTW it's against the rules to just give the answer. Also, I would guess that it's super against the rules to give the wrong answer.

hahahhahahahhahaha

Reply 30

Pinkberry_y

Original post by anosmianAcrimony

BTW it's against the rules to just give the answer. Also, I would guess that it's super against the rules to give the wrong answer.

Original post by Naruke

hahahhahahahhahaha

Can someone tell me how my answer is wrong seeing as this is a learning process for all.
FYI I meant 3x/5

Reply 31

Original post by Pinkberry_y

Can someone tell me how my answer is wrong seeing as this is a learning process for all.
FYI I meant 3x/5

So you want me to teach you?

Reply 32

Original post by Pinkberry_y

Can someone tell me how my answer is wrong seeing as this is a learning process for all.
FYI I meant 3x/5

3x/5 is correct, but 4/5 is not 20.

Reply 33

EricPiphany

Well

y = mx+c = mx + c+ 0

, but

0=5y-3x-4

, so, substituting,

y = mx+c+5y-3x-4

.
The rest is easy.

Reply 34

Original post by EricPiphany

Well

y = mx+c = mx + c+ 0

, but

0=5y-3x-4

, so, substituting,

y = mx+c+5y-3x-4

.
The rest is easy.

Er, no.

That would get him

y=\frac{1}{4}(3-m)x+\frac{1}{4}(4-c)

which is not what we want.

Reply 35

Pinkberry_y

Original post by RDKGames

3x/5 is correct, but 4/5 is not 20.

Ohhhhh OMG I multiplied by accident.

Original post by Naruke

So you want me to teach you?

See I did know what I was doing

Reply 36

EricPiphany

Original post by RDKGames

Er, no.

That would get him

y=\frac{1}{4}(3-m)x+\frac{1}{4}(4-c)

which is not what we want.

But you forget that

5y-3x-4 = 0

Reply 37

Original post by Pinkberry_y

Ohhhhh OMG I multiplied by accident.

See I did know what I was doing

if you say so bae

Reply 38

Zacken

Original post by RDKGames

Er, no.

That would get him

y=\frac{1}{4}(3-m)x+\frac{1}{4}(4-c)

which is not what we want.

It was a joke. :tongue:

(edited 7 years ago)

Reply 39