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There is a lunch bill of $\$239.25$. $4$ students are not paying because it is their birthday. All the other students pay $1$ extra dollar to cover the bill. How many students had lunch?

I assume it means each student's meal costs the same amount, but I still don't know how to solve it. Any help?

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  • $\begingroup$ Hint: denote by $x$ number of students and by $y$ how much each of them would pay if the 4 having birthday paid as well. Turn this problem into a system of equations. $\endgroup$ – Wojowu Nov 4 '15 at 16:05
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The formula representing this problem is, if $y$ is the number of students paying, then $$\frac{239.25}{y}+1=\frac{239.25}{y-4}.$$

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Hint:

Let $x$ be the number of students and $y$ the cost of the meal. Then $xy=239.25$ and $(x-4)(y+1)=239.25$.

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Use a simple equation involving $x$, the number of students, then solve for $x$. We have that each student (except for those who had birthday) should pay $239.25/x$ plus an additional dollar. $x-4$ did not have birthday so:
Let $A=239.25$. $$ (\frac{A}{x}+1)\cdot (x-4) = A\\ \frac{A+x}{x}\cdot (x-4) = A\\ (A+x)(x-4) = Ax\\ x^2+Ax-4A-4x = Ax\\ x^2-4x-4A = 0 $$ From here on we may use the formula $$ x^2+px+q=0 \Leftrightarrow x = -\frac p 2 \pm \sqrt{\left(\frac p 2\right)^2-q} $$ Here our $-p/2 = 2$. But we know there is more than two students so we have the positive root: $$ x = 2+\sqrt{(-2)^2+4A}\\ = 2+\sqrt{4(1+A)}\\ = 2+2\sqrt{1+A}\\ = 2+2\sqrt{240.25}\\ = 33 $$

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