Question

A bookstore can purchase several calculators for a total cost of $120. If each calculators cost $1 less, the bookstore could purchase 10 additional calculators at the same total cost. How many calculators can be purchased at the regular price?
How do I put this Into an equation that I can solve?

Answer 1

$x-number\ of\ calculators\ sold\ in\ regular\ price\\\\ y-\ \ first\ price\\\\ (y-1)\ \ -second\ price\\\\x=\frac{120}{y}\\ x+10=\frac{120}{y-1}\\\\Substitude\ x=\frac{120}{y}\ into\ the\ second\\\\\ \frac{120}{y}+10=\frac{120}{y-1}\ \ |-\frac{120}{y-1}-10\\\\ \frac{120}{y}-\frac{120}{y-1}=-10\\\\Make\ a\ common\ denominator\\\\ \frac{120(y-1)}{y(y-1)}-\frac{120y}{(y-1)y}=-10 \frac{120y-120)}{y(y-1)}-\frac{120y}{(y-1)y}=-10\\\\ \frac{120y-120-120y}{y(y-1)}=-10$<u />$Assumptions:\\y \neq 0\ \ \ and\ \ y \neq 1\\\\ \frac{-120}{y(y-1)}=-10\ \ \ | *y(y-1) \\\\ -120=-10y(y-1)\\\\ -120=-10y^2+10y\\\\ 10y^2-10y-120=0\ \ \ |:10\\\\ y^2-y-12=0\\\\Factor\ equation\\\\ y^2+3y-4y-12=0\\\\ (y-4)(y+3)=0\\\\Solutions:\ \ y=4\ \ and\ \ y=-3$