MATH HELP-- I have no idea how to do this

When Joey mowed lawns for 10 hours and walked dogs for 4 hours, he made a total of 62$.When he walked dogs for 5 hours and mowed the lawn for 8, he made a total of 55$. How much does Joey charge per hour for each job?

This is a simultaneous equation:

Let "d" = the amount he charges for walking dogs.
Let "m" = the amount he charges for mowing lawns.

First he mows lawns for 10 hours and walks dogs for 4, this "equals" $62
Therefore you write:

10m + 4d = 62

With the other information you can also write:

8m + 5d = 55.

You now have two equations with two variables. To solve it, you must eliminate one of the variables.

if 10m + 4d = 62,
50m + 20d = 310

8m + 5d = 55
32m + 20d = 220

Now both equations have a common term: "20d".
Now you subtract one equation from the other to eliminate the 20d.
[50m + 20d = 310] - [32m + 20d = 220]
(50m - 32m) + (20d - 20d) = (310 - 220)
18m = 90
m = 5

He charges $5 per hour mowing lawns.

If we return to the original equation: 10m + 4d = 62,
we can substitute "m = 5" into the equation.
10*5+4d = 62
4d = 12
d = 3

He charges $3 per hour for walking dogs.

You need to set up an equation for each scenario.

L- one hour of mowing lawns
D- one hour of walking dogs

62=10L+4D
55=8L+5D

Now you have a system of equations. You can solve by either doing Substitution or Elimination. I'm going to do Substitution.

Solve the first equation for D
62=10L+4D
62-10L=4D
(62-10L)/4=D

You plug what you got for D into the other equation, then solve for L

55=8L+5D
55=8L+5((62-10L)/4)
55=8L+(310-50L)/4
55=8L+77.5-12.5L
-22.5=8L-12.5L
-22.5=-4.5L
L=5

Now plug that into either equation to solve for D.

62=10L+4D
62=10(5)+4D
62=50+4D