Question

Hey, i'm a little stuck on rearranging equations. I'll post the question below..

Answer 1

\[y=\frac{ 3(x-5) }{ 6-2x }\]
make x the subject

Answer 2

as a first step, multiply both sides by (6-2x)
what do you get ?

Answer 3

y(6-2x)=3(x-5)

Answer 4

The idea is you want the x's to all be "up top" (even if on the wrong side)

Answer 5

next step, distribute the y on the left side and the 3 on the right side
what do you get ?

Answer 6

by distribute, do you mean expand the brackets??

Answer 7

The idea is to get the x's out of the parens.

yes, distribute means expand the brackets.

Answer 8

if so, 6y-2yx=3x-15 ?

Answer 9

next add -2yx to both sides (this gets the x's on the same side)
add +15 to both sides

what do you get ?

Answer 10

oops, +2xy not -2xy

Answer 11

6y+15=3x+2yx

Answer 12

any idea what to do next ?

Answer 13

ermmmmm... just gotta get that 'y' away so either divide by y or put in brackets or something??

Answer 14

how about factor out the x so you get (stuff) * x
then divide both sides by (stuff) to get x by itself?

Answer 15

so.. 6y+15=x(3+2y)
and finally
\[\frac{ 6y+15 }{ 3+2y}=x\]

Answer 16

what you have is good, but it might be written differently
often people would write the bottom as 2y+3 (putting the variable first)

but good work!

Answer 17

ahh i see, yeh that's that order thing where theres x" then ?x then ? etc... yeh thanks alot :)

Answer 18

would \[\sqrt{\frac{ 5 }{ x+a }} \]
be
\[\frac{ 5 }{ Y ^{2} }-A\]

Answer 19

yes

on a multiple choice you might see
\[ \frac{ 5- a y^2}{y^2} \] or
\[ -\frac{ a y^2-5 }{y^2} \]
if they changed a to a y^2/y^2 (to get a common denominator)
and then added the two fractions.

in the last form, you can "factor out" a -1 to that (5-ay^2) is written as
-(ay^2-5)
(this makes the ay^2 positive)
of course
\[ \frac{ -(a y^2-5) }{y^2} = -\frac{ a y^2-5 }{y^2} \]

Answer 20

next time make it a new post, so you can get a medal.

Answer 21

haha okayy thanks, well i have a couple more questions, just writing them now :)