Let's write an equation for what David charges.
Let x equal the number of hours he has been moving lawns and y be the charge.
Every hour, you must add another $5.50.
This is the same as multiplying the number of hours by 5.50.
We also need to add that $15 at the beginning. This leaves our equation as
[tex]y=5.50x+15[/tex]
Now, let's write one for Ari.
6.25 per hour...6.25x.
Add the 12 and we get
[tex]y=12x+6.25[/tex]
To find when the two are equal, we can set up this equation:
[tex]5.50x+15=12x+6.25[/tex]
Now, let's simplify. Since all of our decimals are either halves or quarters, you can multiply everything by 4 to get rid of them.
[tex]22x+60=48x+25[/tex]
Subtract 22x from each side to get x on one side.
[tex]60=26x+25[/tex]
Subtract 26 from each side to get x alone.
[tex]35=26x[/tex]
Divide by 26 to find the value of x.
[tex]x=35/26\approx1.346[/tex]
What does this tell us? Well, it tells us that when our number of hours reaches 1.346, their charges are equal.
(The .346 can be estimated as around 20 minutes because it's pretty close to .333, or a third of an hour. This puts 1.346 hours at an hour and 20 minutes.)
We still haven't answered one part of the question, though...after that 1:20 time frame, who is charging more? Well, let's say we had them mow lawns for two hours. Using 2 for x in those earlier equations gives us these values:
[tex]David:\ y=5.50(2)+15=11+15=26[/tex]
[tex]Ari:\ y=6.25(2)+12=13+12=25[/tex]
It seems that Ari charges less than David past that 1:20 point.
Since the question asked "when is Ari's charge greater than or equal to David's" we would respond by saying whenever he mows lawns for less than an hour and 20 minutes.
