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pastaBest ResponseYou've already chosen the best response.0
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
 one year ago

MathPhysicsBest ResponseYou've already chosen the best response.1
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 \[3x+3x=13\rightarrow 6x=2\rightarrow x=\frac{ 1 }{ 3 }\]
 one year ago

pastaBest ResponseYou've already chosen the best response.0
MathPhysics i think you just changed the question
 one year ago

MathPhysicsBest ResponseYou've already chosen the best response.1
pasta! Why are you saying this?!! Quite the contrary, I think this is you who did so!!
 one year ago

pastaBest ResponseYou've already chosen the best response.0
I THINK THE "X" IS PART OF THE DENOMINATOR NOT THE NUMERATOR
 one year ago

MathPhysicsBest ResponseYou've already chosen the best response.1
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.
 one year ago
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