A store owner buys a quantity of balls for $600. If they had each cost $0.25 less, she would have 10 more for the same money. How much did she pay for each ball.

(x-0.25)*(y+10)

is how far I got. Is this even a quadratic equation?

Printable View

- Apr 22nd 2010, 01:14 PMsinjid9Quadratic help
A store owner buys a quantity of balls for $600. If they had each cost $0.25 less, she would have 10 more for the same money. How much did she pay for each ball.

(x-0.25)*(y+10)

is how far I got. Is this even a quadratic equation? - Apr 22nd 2010, 02:15 PMArchie Meade
Hi sinjid9,

the amount of money is the same in both cases and is (price)(quantity).

Hence, if x=price, y=number of balls

$\displaystyle xy=(x-0.25)(y+10)=600$

Now write one variable in terms of the other using $\displaystyle xy=600$

$\displaystyle x=\frac{600}{y},\ or\ y=\frac{600}{x}$

This leads to the quadratic in a single variable.

For example....

$\displaystyle (x-0.25)\left(\frac{600}{x}+10\right)=600$

$\displaystyle \frac{1}{x}(x-0.25)(600+10x)=600$

$\displaystyle (x-0.25)(600+10x)=600x$

$\displaystyle 600x+10x^2-150-2.5x=600x$

$\displaystyle 10x^2-2.5x-150=0$

$\displaystyle 100x^2-25x-1500=0$

$\displaystyle 4x^2-x-60=0$

$\displaystyle (4x+15)(x-4)=0$

$\displaystyle x-4=0$ as x has to be positive.

$\displaystyle x=4\ dollars,\ y=150$

notice $\displaystyle (4-0.25)(150+10)=(3.75)(160)=600$