Question 1182223
<br>
Let x be the cost of the ball
Let y be the number of boys<br>
x/y = cost per boy
x/(y-3) = cost per boy if there had been 3 fewer boys
x/(y+4) = cost per boy if there had been 4 more boys<br>
If there had been 3 boys fewer, each would have paid one dollar more...:<br>
{{{x/(y-3)=(x/y)+1}}} [1]<br>
... and if there had been 4 boys more, each would have paid 75 cents less:<br>
{{{x/(y+4)=(x/y)-3/4}}} [2]<br>
Solve each equation by multiplying through by the least common denominator.<br>
[1] {{{x/(y-3)=(x/y)+1}}}<br>
{{{x(y)=x(y-3)+y(y-3)}}}
{{{xy=xy-3x+y^2-3y}}}
{{{3x = y^2-3y}}} [3]<br>
[2] {{{x/(y+4)=(x/y)-3/4}}}<br>
{{{x(4y)=x(4(y+4))-3(y(y+4))}}}
{{{4xy=4xy+16x-3y^2-12y}}}
{{{16x=3y^2+12y}}} [4]<br>
Eliminate x between [3] and [4] by multiplying [3] by 16 and [4] by -3 and adding:<br>
{{{48x=16y^2-48y}}}
{{{-48x=-9y^2-36y}}}
{{{0=7y^2-84y}}}
{{{0=7y(y-12)}}}<br>
y=0 or y=12.  Clearly y=0 makes no sense in the problem, so y=12.<br>
So the number boys is y=12.<br>
Use y=12 in either [1] or [2] to find the cost of the ball.<br>
I leave that much to you....<br>