Solving Linear Equations with Fractions

[ai93]

If I have a similar question \(\displaystyle \frac{y}{5}+\frac{7}{20}=\frac{5-y}{4}\) 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?

 

[ineedhelpnow]

yep and that would get rid of the denominator 4y+7=5(5-y)

 

[MarkFL]

If I have a similar question \(\displaystyle \frac{y}{5}+\frac{7}{20}=\frac{5-y}{4}\) 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

 

[ai93]

yep and that would get rid of the denominator 4y+7=5(5-y)

Understood that part, so from here expand to get \(\displaystyle 4y+7=25-5y?\) Move like terms together so, \(\displaystyle 4y+5y=25-7?\) \(\displaystyle \therefore9Y=18. y=2?! \) :D

 

[ineedhelpnow]

yep

 

[MarkFL]

Understood that part, so from here expand to get \(\displaystyle 4y+7=25-5y?\) Move like terms together so, \(\displaystyle 4y+5y=25-7?\) \(\displaystyle \therefore9y=18\implies y=2?! \) :D

Yes, that's correct. :D

 

[ai93]

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