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

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1. anonymous

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

2. anonymous

cause its lol

3. anonymous

you still have x's on both sides

4. anonymous

you didnt completly isolate the x

5. anonymous

oh. wtf. i'm retarded

6. anonymous

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

7. anonymous

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

8. anonymous

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

9. anonymous

i missed the first step...?

10. 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

11. anonymous

subtract ax from both side

12. 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

13. anonymous

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

14. anonymous

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

15. 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)

16. anonymous

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

17. anonymous

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

18. anonymous

got it basketmath? or need an explanation?

19. anonymous

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

20. 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...?

21. 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.

22. anonymous

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

23. 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?

24. 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

25. 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 :)

26. 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

27. anonymous

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

28. 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?

29. 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?

30. 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!

31. 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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