Hello janivee!
janviee said:
Hello
Peter takes 15 minutes longer to mow the lawn by himself than charles does. Together they can mow the lawn in 18 minutes. How long does it take charles to mow the lawn by himself?
I'm sorry but what you have done so far is wrong. But I am very happy that you at least showed some work.
Whenever you can make ratios like what I' m going to show you that is what you do.
Time it takes Peter = \(\displaystyle \L c+15\)
Time it takes Charles = \(\displaystyle \L c\)
All time are for one lawn so we can make ratios:
\(\displaystyle \L \frac{1}{c}+\frac{1}{c+15}=\frac{1}{18}\)
Now we need to find an LCM. We can do this by multiplying all the denominators together. And whenever we multiply a term times the denominator we need to to that also to the numerator.
So we multiply are denominators: \(\displaystyle \L c(c+15)18=18c(c+15)\)
So \(\displaystyle \L \frac{number}{18c(c+15)}+\frac{number}{18c(c+15)}=\frac{number}{18c(c+15)}\)
So since our first denominator already has a \(\displaystyle \L c\) in it that means I multiply it by \(\displaystyle \L 18(c+15)\) to get our common denominator \(\displaystyle \L {18c(c+15)}\) so that means we will multiply the first numerator \(\displaystyle \L 1\) by \(\displaystyle \L 18(c+15)\) as well. So for \(\displaystyle \L \frac{1}{c}\) we have: \(\displaystyle \L \frac{18(c+15)}{18(c+15)}\)
Continuing this process you finally reach:
\(\displaystyle \L \frac{18(c+15)}{18(c+15)}+\frac{18c}{18(c+15)}=\frac{c(c+15)}{18(c+15)}\)
Now we can just ignore the denominator.
Solve \(\displaystyle \L 18(c+15)+18c=c(c+15)\)
Can you finish and solve for \(\displaystyle \L c\)?
You will get a negative and a positive numer, toss out the negative number since time is not negative.