anonymous
  • anonymous
3/8X+3/8=-3/8X+1/8
MIT 18.01 Single Variable Calculus (OCW)
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chestercat
  • chestercat
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anonymous
  • anonymous
3/8X+3/8=-3/8X+1/8
anonymous
  • anonymous
x=-3
anonymous
  • anonymous
make the LCM 8x on both sides,so that they can cancel, then the equation becomes; 3+3x=-3+x now u can get x=-3

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anonymous
  • anonymous
math2! you'd better use some parentheses to clarify! Which one do you mean? 1. \[\frac{ 3 }{ 8x }+\frac{ 3 }{ 8 }=-\frac{ 3 }{ 8x }+\frac{ 1 }{ 8 }\] 2. \[\frac{ 3 }{ 8 }x+\frac{ 3 }{ 8 }=-\frac{ 3 }{ 8 }x+\frac{ 1 }{ 8 }\] It could go on either ways. But most likely, what you mean is the second one. If so, you must do as follows: 1. Multiply all terms by 8 in order to cancel all 8's in denominators. Then it will be like this: \[3x+3=-3x+1\] Then \[3x+3x=1-3\rightarrow 6x=-2\rightarrow x=-\frac{ 1 }{ 3 }\]
anonymous
  • anonymous
MathPhysics i think you just changed the question
anonymous
  • anonymous
pasta! Why are you saying this?!! Quite the contrary, I think this is you who did so!!
anonymous
  • anonymous
I THINK THE "X" IS PART OF THE DENOMINATOR NOT THE NUMERATOR
anonymous
  • anonymous
Might be! If you read my post carefully, you'd notice that I already mentioned it. He hasn't determined what he exactly meant. Anyways, If he meant the first possibility I mentioned, you'd be right then.

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