There are three steps to solving a math problem.
Let's go through a problem in long and excruciating detail. If it gets too bad, you can take a nap.
A cashbox has one-, five-, ten-, and twenty-dollar bills. There are 3 more five-dollar bills than one-dollar bills, half as many ten-dollar bills as one-dollar bills, and two more twenty-dollar bills than one-dollar bills. There is $365 total in the cashbox. That's almost a dollar a day! How many one-dollar bills are in the cashbox? Also, who are the wiseguys who insisted on paying their theater ticket in singles?
After an initial read-through, we can grasp that, in general, we are asked to find the number of one-dollar bills in the cashbox. Plus something about wiseguys that seems significantly less important.
Now we do a second, more careful reading of the problem, translate from English to math, and solve whatever equation or inequality we get.
First we need to translate the problem from English into math. Let x be the number of one-dollar bills we have. Then we have
x + 3 five-dollar bills
x + 2 twenty-dollar bills
We have $365 total, and
$365 = amount in ones + amount in fives + amount in tens + amount in twenties
The amount we have in ones is just x. We don't want to assign variables to any other amount, because that will make things too confusing. We're trying to go toward the light, not away from it.
The amount we have in fives is 5(x + 3) + (5 times the number of 5 dollar bills), and so on. Therefore,
which simplifies to
365 = x + 5x + 15 + 5x + 20x + 40 = 31x + 55
Phew! Our equation still looks long, but we're rid of all parentheses and fractions, and from here it's a matter of shifting all your x terms to one side and everything else to the other. Stop hitting the "shift" key. That won't help you.
After solving this equation, we have x = 10.
If we have 10 one-dollar bills then we have 13 fives, 5 tens, and 12 twenties. The amounts of money do indeed add to
10(1) + 13(5) + 5(10) + 12(20)
which is 365, as it should be. All is right in the universe.