Question

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

Answer 1

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

what am i doing wrong??!??

Answer 2

cause its lol

Answer 3

you still have x's on both sides

Answer 4

you didnt completly isolate the x

Answer 5

oh. wtf. i'm retarded

Answer 6

2(cx +d)= ax+b
2cx +2d = ax +b

get Xs into the same side and solve

Answer 7

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

Answer 8

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

Answer 9

i missed the first step...?

Answer 10

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

Answer 11

subtract ax from both side

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

(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)

Answer 16

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

Answer 17

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

Answer 18

got it basketmath? or need an explanation?

Answer 19

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

Answer 20

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...?

Answer 21

(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.

Answer 22

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

Answer 23

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?

Answer 24

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

Answer 25

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 :)

Answer 26

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

Answer 27

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

Answer 28

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?

Answer 29

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?

Answer 30

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!

Answer 31

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'