MHB Solving Linear Equations with Fractions

AI Thread Summary
To solve linear equations with fractions, it's effective to eliminate the denominators by multiplying through by the least common multiple, which in this case is 20. The equation $$\frac{y}{5}+\frac{7}{20}=\frac{5-y}{4}$$ can be simplified by multiplying each term by 20, resulting in a more manageable equation. After expanding and rearranging the terms, the solution leads to $$y=2$$. Participants express appreciation for the guidance received in understanding the process. The discussion highlights the importance of clear steps in solving equations involving fractions.
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If I have a similar question $$\frac{y}{5}+\frac{7}{20}=\frac{5-y}{4}$$ should I go about the same process as the http://mathhelpboards.com/pre-algebra-algebra-2/solve-following-equation-12605.html? Try and cancel out the denominators of 5, 20 and 4? Which would be 20? So times the equation by 20?
 
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yep and that would get rid of the denominator 4y+7=5(5-y)
 
mathsheadache said:
If I have a similar question $$\frac{y}{5}+\frac{7}{20}=\frac{5-y}{4}$$ should I go about the same process as the http://mathhelpboards.com/pre-algebra-algebra-2/solve-following-equation-12605.html? Try and cancel out the denominators of 5, 20 and 4? Which would be 20? So times the equation by 20?

I have moved this question into its own thread. We ask that new questions not be tagged onto existing threads, rather we ask that a new thread be started for a new question. This keeps threads from becoming convoluted and hard to follow. :D
 
ineedhelpnow said:
yep and that would get rid of the denominator 4y+7=5(5-y)

Understood that part, so from here expand to get $$4y+7=25-5y?$$ Move like terms together so, $$4y+5y=25-7?$$ $$\therefore9Y=18. y=2?! $$ :D
 
yep
 
mathsheadache said:
Understood that part, so from here expand to get $$4y+7=25-5y?$$ Move like terms together so, $$4y+5y=25-7?$$ $$\therefore9y=18\implies y=2?! $$ :D

Yes, that's correct. :D
 
Thank you guys, I have learned so much from your help. I would still be stuck on those questions if it wasn't for your help and this forum! :D
 
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