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eyust707 Group TitleBest ResponseYou've already chosen the best response.0
You bring all the terms with x's to one side and the numerical term to the other side. Then you find the least common multiple of the denominators. You multiply each term by the appropriate factor to make the denominators all have the same LCM. Then you add the terms. Finally you multiply each side by the denominator so that x is isolated.
 2 years ago

dddan Group TitleBest ResponseYou've already chosen the best response.0
can you explain further?
 2 years ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
Well, you want x to be on only one side of the equation when you're finished, so that's actually a good place to start.
 2 years ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
Given\[\frac{x}{2}+\frac{x}{5}=\frac{x}{6}\frac{1}{3}\],
 2 years ago

dddan Group TitleBest ResponseYou've already chosen the best response.0
i put all the x values on one side and i made the LCD 60 is that good so far?
 2 years ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
Correct, GeekChic_! 60 is a good denominator! 30 also works if you'd like, but 60 will be fine too!
 2 years ago

dddan Group TitleBest ResponseYou've already chosen the best response.0
yea thats what i got but shouldn't it be 5/8?
 2 years ago

dddan Group TitleBest ResponseYou've already chosen the best response.0
yea lol i meant 30 as LCD
 2 years ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
So you have\[\frac{30x}{60}+\frac{12x}{60}\frac{10x}{60}=\frac{1}{3}\]
 2 years ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
Oh yeah, definately\[\frac{5}{8}\]
 2 years ago

GeekChic_ Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{ 5x+2x }{ 10 }=\frac{ 3x6 }{ 18 }\] then crossmultiply \[18(5x+2x)=10(5x+2x)\]
 2 years ago

dddan Group TitleBest ResponseYou've already chosen the best response.0
alright thanks guys!
 2 years ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
No problem! I'm glad you could work through the problem! Take care!
 2 years ago
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