<PRE><B>Question 1108299</B>
&lt;br&gt;
Let x be the number of children
Let y be the number of parents
Let z be the number of other people&lt;br&gt;
Note that the statement of the problem requires y to be at least 2, since both Gina&#39;s parents attended.
Then...&lt;br&gt;
(1) {{{x+y+z = 100}}}  there were 100 people in attendance
(2) {{{.5x+3y+10z = 100}}}  the total cost of the tickets was $100&lt;br&gt;
We have three unknowns but only two equations; but we can find the answer to the problem (or in some similar problems a small family of solutions) using the fact that the variables have whole number values.&lt;br&gt;
To solve this kind of system of equations, we eliminate one variable and use the requirement that the solutions be whole numbers to find the answer.&lt;br&gt;
Double equation (2) and subtract equation (1) from the result:&lt;br&gt;
{{{x+6y+20z = 200}}}
{{{x+y+z = 100}}}
{{{5y+19z = 100}}}&lt;br&gt;
Solve this equation for one variable in terms of the other:&lt;br&gt;
{{{19z = 100-5y}}}
{{{z = (100-5y)/19 = (5(20-y))/19}}}&lt;br&gt;
The only whole number value of y that makes z also a whole number is y=1.  That makes z=5; and then x=94, because the total number of people was 100.&lt;br&gt;
But that means there were 94 children, 1 parent, and 5 other people in attendance -- and the problem requires the number of parents to be at least 2.&lt;br&gt;
So THERE IS NO SOLUTION to the problem as stated.&lt;br&gt;
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(response amended...)&lt;br&gt;
Tutor ikleyn is right....&lt;br&gt;
The problem as stated requires that there be at least one child and two parents; but it does not require that there be any other people. So in my equation
{{{z = (100-5y)/19 = (5(20-y))/19}}}
y=20 is valid, making z=0 and x=80.&lt;br&gt;
I carelessly read the problem as requiring that there be at least one of each kind of person.
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