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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this is what i get and it's wrong: \[x= {2cx+2d \over a+b}\] what am i doing wrong??!??

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you still have x's on both sides

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you didnt completly isolate the x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh. wtf. i'm retarded

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02(cx +d)= ax+b 2cx +2d = ax +b get Xs into the same side and solve

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i missed the first step...?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait. 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
 5 years ago
Best ResponseYou've already chosen the best response.0subtract ax from both side

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02cx +2d = ax +b subtract ax from both side 2cxax+2d=b subtract 2d from both side 2cxax=b2d Factor out x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i'm getting the same thing as you now. the book shows the answer as: \[x= {2d  b \over a  2c}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(ax+b)/ (cx+d)=2 ax + b = 2(cx+d) ax + b = 2cx + 2d ax  2cx = 2d b x(a2c) = (2db) x = (2db) / (a2c)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you factor out x in 5th line in my above post.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I made a mistake x(2ca)=b2d

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0got it basketmath? or need an explanation?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i'm just following along. i am trying to take all this in. thanks for your guys help!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so.... 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
 5 years ago
Best ResponseYou've already chosen the best response.0(ax+b)/ (cx+d)=2 [ORIGINAL EQ] ax + b = 2(cx+d) ax + b = 2cx + 2d b  2d = 2cx  ax (b2d) = x (2c  a) (b2d)/ (2c a) = x Both the answers are correct, you can do either way.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02db is the same as b2d? That's where i am confused...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and 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
 5 years ago
Best ResponseYou've already chosen the best response.0dont see 2db in entirety, actually in linear eq, result can be represented in different ways, for example, in this case[ (b2d)/ (2ca) ]= [ (2db) / (a2c) ] = x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay. 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
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, that's good.. always good to go in detail & understand every single bit :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this: b  2d = 2cx  ax (b2d) = 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 \[b2d= {2cxax \over 2c  a} \] 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
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, 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
 5 years ago
Best ResponseYou've already chosen the best response.0right. 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
 5 years ago
Best ResponseYou've already chosen the best response.0yes, 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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