\[{ax + b \over cx+ d} = 2 solve for x\]

- anonymous

\[{ax + b \over cx+ d} = 2 solve for x\]

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- anonymous

this is what i get and it's wrong:
\[x= {2cx+2d \over a+b}\]
what am i doing wrong??!??

- anonymous

cause its lol

- anonymous

you still have x's on both sides

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## More answers

- anonymous

you didnt completly isolate the x

- anonymous

oh. wtf. i'm retarded

- anonymous

2(cx +d)= ax+b
2cx +2d = ax +b
get Xs into the same side and solve

- anonymous

x(2c-a) = b-2d

- anonymous

x = (b-2d)/(2c-a)

- anonymous

i missed the first step...?

- anonymous

wait. i get what imranmeah91 says in his equation... how do i get x to one side after that? that's where i'm stumped

- anonymous

subtract ax from both side

- anonymous

2cx +2d = ax +b
subtract ax from both side
2cx-ax+2d=b
subtract 2d from both side
2cx-ax=b-2d
Factor out x

- anonymous

x(2c-a)=b-2a

- anonymous

i'm getting the same thing as you now. the book shows the answer as:
\[x= {2d - b \over a - 2c}\]

- anonymous

(ax+b)/ (cx+d)=2
ax + b = 2(cx+d)
ax + b = 2cx + 2d
ax - 2cx = 2d -b
x(a-2c) = (2d-b)
x = (2d-b) / (a-2c)

- anonymous

you factor out x in 5th line in my above post.

- anonymous

I made a mistake
x(2c-a)=b-2d

- anonymous

got it basketmath? or need an explanation?

- anonymous

i'm just following along. i am trying to take all this in. thanks for your guys help!

- anonymous

so.... why do we subtract the 2cx to the left side instead of the ax to the right side? it made a difference in this equation, where it usually doesn't make much difference...?

- anonymous

(ax+b)/ (cx+d)=2 [ORIGINAL EQ]
ax + b = 2(cx+d)
ax + b = 2cx + 2d
b - 2d = 2cx - ax
(b-2d) = x (2c - a)
(b-2d)/ (2c -a) = x
Both the answers are correct, you can do either way.

- anonymous

2d-b is the same as b-2d? That's where i am confused...

- anonymous

and i know what you mean to factor out the x but will it help me later on if i factor the x instead of just dividing it out?

- anonymous

dont see 2d-b in entirety,
actually in linear eq, result can be represented in different ways,
for example, in this case[ (b-2d)/ (2c-a) ]= [ (2d-b) / (a-2c) ] = x

- anonymous

okay. i understand now. i just want to make sure that i am prepared for tests. i plan to use math for a long time :)

- anonymous

I dont understand by what you meant when you say -
"and i know what you mean to factor out the x but will it help me later on if i factor the x instead of just dividing it out?"
can you advise the step where you're finding it difficult

- anonymous

yeah, that's good.. always good to go in detail & understand every single bit :)

- anonymous

this:
b - 2d = 2cx - ax
(b-2d) = x (2c - a)
i know how to factor the x, but i would come up with this instead (how i was taught):
b - 2d = 2cx - ax
\[b-2d= {2cx-ax \over 2c - a} \]
to leave me with x.... but after writing that equation, i see why i couldn't do it that way and factoring would be the "right" way. :) otherwise i would be left with 2x right?

- anonymous

yeah, you are right, factoring is the "right" & "only" way because you are solving for 'x'.
moreover you cannot just go and divide one side by 2c -a in the step you mention,
you have to either add/divide/substract both sides of = or not at all,
for eg 1 orange = 1 fruit
1/2 orange = 1/2 fruit
1+5 orrange = 1+5 fruit
it cant be 1 orange = 1/2 fruit
got it?

- anonymous

right. i was doing it to both sides, but even if i did, it would leave me with x and x on the right side... which wouldn't equal 'x' lol. i think i got it. just have to remember to factor!

- anonymous

yes, right.. when you do both sides.. it wont solve you for x, lol.
Happy that you got it :)
remember to factor for whatever constant you are solving,
as here you are solving for 'x'

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