Question 1116185
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If a solution without formal algebra is acceptable, then by far the easiest way to find the answer to the problem is by trial and error.<br>
The number of people is certainly going to be a whole number; and the cost per person is very likely to be a whole number.  So look for ways to write 150 as the product of two whole numbers and find two pairs that satisfy the conditions of the problem.<br>
Pairs of whole numbers with a product of 150:<br>
 1 150
 2  75
 3  50
 5  30
 6  25
10  15<br>
The pairs 5*30 and 6*25 satisfy the conditions of the problem: 30 people cost $5 each; 25 people cost $6 each -- 5 fewer people increases the cost for each person by $1.<br>
Here is an algebraic solution....<br>
Let x be the number of people and y the cost per person.  Then
(1) {{{xy = 150}}}
(2) {{{(x-5)(y+1) = 150}}}<br>
Solve (1) for y:
{{{y = 150/x}}}<br>
Substitute into (2):
{{{(x-5)(150/x+1) = 150}}}
{{{150+x-750/x-5 = 150}}}
{{{x-5-750/x = 0}}}
{{{x^2-5x-750 = 0}}}
{{{(x-30)(x+25) = 0}}}
{{{x = 30}}} or {{{x = -25}}}<br>
Of course reject the negative answer; so the number of people is x = 30.<br>
That was a whole lot of work; and in the end you still had to use some trial and error to find the 30 and 25 to factor the quadratic equation....