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=(120)/(y)\\ x+10=(120)/(y-1)\\\\Substitude\ x=(120)/(y)\ into\ the\ second\\\\\ (120)/(y)+10=(120)/(y-1)\ \ |-(120)/(y-1)-10\\\\ (120)/(y)-(120)/(y-1)=-10\\\\Make\ a\ common\ denominator\\\\ (120(y-1))/(y(y-1))-(120y)/((y-1)y)=-10 (120y-120))/(y(y-1))-(120y)/((y-1)y)=-10\\\\ (120y-120-120y)/(y(y-1))=-10`
`Assumptions:\\y \\eq 0\ \ \ and\ \ y \\eq 1\\\\ (-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`