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July 25th, 2014, 08:05 PM  #1 
Newbie Joined: Jul 2014 From: Mexico Posts: 2 Thanks: 0  System of equations (practical problem)
It looks like a simple problem but I'm not sure how to going on. In a pastry class each student prepare a dough of water and flour weighing 8 ounces. The amounts of flour and water for each student were varied but never 0. One student spent one quarter of the available water and 1/6 of the available flour. How many students were in the class? My attempt: $$ \frac{1}{4}mass_{water} + \frac{1}{6}mass_{flour} = 8 \\ \frac{3}{4}mass_{water} + \frac{5}{6}mass_{flour} = 8 \times(n1), $$ being $n$ the number of students. It seems that I need one equation more. What do you think? Last edited by dapias09; July 25th, 2014 at 08:07 PM. 
July 25th, 2014, 08:53 PM  #2  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 407  Hello, dapias09! You're off to a good start. Quote:
I used your equations. Let $n$ = number of students. $\quad \frac{1}{4}W + \frac{1}{6}F \:=\:8$ $\quad \frac{3}{4}W + \frac{5}{6}F \:=\:8(n1)$ I cleared out the fractions: $\quad 3W + 2F \:=\:96$ $\quad 9W + 10F \:=\:96n96$ I solved the system: $\begin{Bmatrix}W &=& 96  16n \\ F &=& 24n  96\end{Bmatrix}$ Since $W > 0$, we have: $\:9616n \:>\:0 \quad\Rightarrow\quad n \:<\:6$ Since $F > 0$, we have: $\: 24n  96 \:>\:0 \quad\Rightarrow\quad n \:>\:4$ Therefore: $\:n \:=\:5.$  
July 26th, 2014, 08:11 AM  #3 
Newbie Joined: Jul 2014 From: Mexico Posts: 2 Thanks: 0 
Hello soroban, thank you very much, your explanation makes sense for me


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equations, fractions, inequalities, practical, problem, system, system of equatioms 
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