anonymous
  • anonymous
\[{ax + b \over cx+ d} = 2 solve for x\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
this is what i get and it's wrong: \[x= {2cx+2d \over a+b}\] what am i doing wrong??!??
anonymous
  • anonymous
cause its lol
anonymous
  • anonymous
you still have x's on both sides

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anonymous
  • anonymous
you didnt completly isolate the x
anonymous
  • anonymous
oh. wtf. i'm retarded
anonymous
  • anonymous
2(cx +d)= ax+b 2cx +2d = ax +b get Xs into the same side and solve
anonymous
  • anonymous
x(2c-a) = b-2d
anonymous
  • anonymous
x = (b-2d)/(2c-a)
anonymous
  • anonymous
i missed the first step...?
anonymous
  • 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
  • anonymous
subtract ax from both side
anonymous
  • 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
  • anonymous
x(2c-a)=b-2a
anonymous
  • anonymous
i'm getting the same thing as you now. the book shows the answer as: \[x= {2d - b \over a - 2c}\]
anonymous
  • 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
  • anonymous
you factor out x in 5th line in my above post.
anonymous
  • anonymous
I made a mistake x(2c-a)=b-2d
anonymous
  • anonymous
got it basketmath? or need an explanation?
anonymous
  • anonymous
i'm just following along. i am trying to take all this in. thanks for your guys help!
anonymous
  • 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
  • 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
  • anonymous
2d-b is the same as b-2d? That's where i am confused...
anonymous
  • 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
  • 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
  • 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
  • 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
  • anonymous
yeah, that's good.. always good to go in detail & understand every single bit :)
anonymous
  • 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
  • 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
  • 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
  • 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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