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Nomad123
Sep 7, 2007, 11:55 AM
I no I shudnt be but I am kind of stuck... havent done this in a while n I'm meant to be taking gcse a year early so it's a hardcore thing... I just haven't been properly taught how to answer this one...

the line with the equation 6y+5x=15...
rearrange 6y+5x=15
to make y the subject...
I've tried the ways I no to work it out but I'm not getting the right answer...
help please:confused: ;)

Capuchin
Sep 7, 2007, 11:59 AM
minus 5x from both sides.
divide both sides by 6.

MimiGirl
Sep 7, 2007, 12:18 PM
it'll probably be like something below..

6y+5x=15
-5 -5

6y 0 = 10
6 = 6
y=1.66

the answer I think might be 1.66 is the subject of y...
y=1.66

let me know if I am wrong... thanks and I hope this was helpful

Capuchin
Sep 7, 2007, 12:43 PM
Received in e-mail:


thanks for your help
however
i tried that, and i got that answer first and i dont think it works that way
ive come on this as a last resort
i no the answer is y=-5/6x+2 1/2
as we have the answer but i need to no how to get to the answer
as i have only jus startd schoool again and havent done this in a while
id really appreciate any more help you could offer
thanks :)

If you follow my steps you should get the answer you quote here. I just tried it and I got the same answer.

Nomad123
Sep 7, 2007, 12:54 PM
Recieved in e-mail:



If you follow my steps you should get the answer you quote here. I just tried it and I got the same answer.

hrrrm could you actually write how you did it
I keep getting to the dsame point yet it is not working...
thanks for the help though
OK
so I have 6y+5x=15
then from there I minus 5x from both sides
which gets to
6y=15-5x
divide both sides by 6
goes to
y=15/6-5/6
then I don't no what to do from there..?

Capuchin
Sep 7, 2007, 12:59 PM
well, the answer you got is

y = 15/6-5/6x

and the answer you'r supposed to get is

y = -5/6x+2 1/2

... those are exactly the same.

s_cianci
Sep 8, 2007, 10:33 AM
Do you mean solve it for y in terms of x? If so, then two steps will do it ; a subtraction and a division.