Algebraic Fractions

Psychguy98

Junior Member
Joined
Dec 17, 2010
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147
10y + 5/4 divided by 2y + 1/2 =

can i make the 10 a 5 leaving 5( y +1)/2 Then 2 (y +1)/2 ?
 
Hello, Psychguy98!

Which of these problems did you intend?


\(\displaystyle \displaystyle (a)\;\left(10y +\frac{5}{4}\right) \div \left( 2y + \frac{1}{2}\right)\)

\(\displaystyle \displaystyle (b)\;\left(\frac{10y+5}{4}\right) \div \left(\frac{2y+1}{2}\right)\)

\(\displaystyle \displaystyle (a)\;\text{We have: }\; \frac{10y + \frac{5}{4}}{2y + \frac{1}{2}}\)

\(\displaystyle \displaystyle \text{Multiply top and bottom by 4: }\;\frac{4\left(10y + \frac{5}{4}\right)}{4\left(2y+\frac{1}{2}\right)} \;=\;\frac{40y+5}{8y+2} \;=\;\frac{5(8y+1)}{2(4y+1)}\)


\(\displaystyle \displaystyle (b)\;\;\text{We have: }\;\frac{\dfrac{10y+5}{4}}{\dfrac{2y+1}{2}}\)

\(\displaystyle \displaystyle \text{Multiply top and bottom by 4: }\;\frac{4\left(\frac{10y+5}{4}\right)} {4\left(\frac{2y+1}{2}\right)} \;=\;\frac{10y+5}{2(2y+1)} \;=\;\frac{5(2y+1)}{2(2y+1)} \;=\;\frac{5}{2}\)

 
Thanks, but isn't there another way to do it? 10y + 5 /4 times 2 / 2y + 1

5(2Y + 1) / 2 times 1 /2(y +1) = 5/2

2y + 1 cancels out?
 
Psychguy98 said:
Thanks, but isn't there another way to do it? 10y + 5 /4 times 2 / 2y + 1

5(2Y + 1) / 2 times 1 /2(y +1) = 5/2

2y + 1 cancels out?
PG, you are still VERY NEGLIGENT with brackets:
your 10y + 5 /4 times 2 / 2y + 1 MUST BE shown this way: (10y + 5) /4 times 2 / (2y + 1)

Your 2nd line is erroneous.

Soroban showed you the 5/2 solution, and DOES have the 2y+1 cancelling out: so WHAT are you asking?
 
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