# simultaneous equations (?) power company question.

• Aug 28th 2006, 01:56 PM
Paulotron
simultaneous equations (?) power company question.
Hello.

I have this question.

one company charges a standing charge of 15 per quarter, with a cost of 9 per unit used, whereas another companycharges 10.5 per quarter, with a cost of 11 per unit.

find out which number of units makes the cost of each company the same.
• Aug 28th 2006, 02:15 PM
ThePerfectHacker
Quote:

Originally Posted by Paulotron
Hello.

I have this question.

one company charges a standing charge of 15 per quarter, with a cost of 9 per unit used, whereas another companycharges 10.5 per quarter, with a cost of 11 per unit.

find out which number of units makes the cost of each company the same.

$15+9x=10.5+11x$
Thus,
$15+9x-11x=10.5$
$15-2x=10.5$
$-2x=-4.5$
$x=2.25$
• Aug 28th 2006, 02:19 PM
topsquark
Quote:

Originally Posted by Paulotron
Hello.

I have this question.

one company charges a standing charge of 15 per quarter, with a cost of 9 per unit used, whereas another companycharges 10.5 per quarter, with a cost of 11 per unit.

find out which number of units makes the cost of each company the same.

The first company charges
$y = 9x + 15$ for x units per quarter.

The second company charges
$y = 11x + 10.5$ for x units per quarter.

We need a number of units x that makes both of these expressions equal.

You can do this in several ways. You can graph both cost functions and see for what value of x they intersect.

The other way is algebraic. We have two equations for which we are setting the x and y values to be equal. Since the y values are equal we have:
$y = 9x + 15 = 11x + 10.5$
for some value of x. Solve this equation for x and you have your answer.

-Dan