
The main job of solving word problems is translating them from English to Algebraic. After that, the algebraic manipulation needed to solve the equations is easy.
There are four main steps to solving any word problem:
Before you even begin to solve a word problem, look at the question at the end. Just about every word problem has a question at the end. How many nickels does Jim have? How old is Bob today? That sort of thing. That quantity is the unknown  assign it to a variable. Write down what the variable stands for. Use this sort of language:
Let x be the number of nickels Jim has.
Don't be vague  For example, don't say x=nickels. If you said that, you might get confused between the number of nickels and the value of the nickels.
Sometimes there are two quantities being sought. If so, do they have a simple relationship? For example, is one twice the other? Is one three less than the other? If so, you can assign a variable, such as x, to one of them, and a simple expression, such as 2x or x3 to the other. Write down which is which. It should look like this:
Let x be Bob's age today.
Then x3 is the age, today, of Bob's younger brother.
Now go back through the problem, and look at every English sentence that says anything about this quantity (the quantity that you assigned to a variable). Write down the algebraic equivalent of this quantity. For example, if the problem says "four years ago, Bob was twice as old has his little brother" then you will write,
x4 = 2((x3)4)
If that doesn't come to you right away, take it slower. You can write,
Let x4 be Bob's age, four years ago.
Let (x3)4, which is x7, be Bob's younger brother's age, four years ago.
x4 = 2(x7)
Either way, you'll be able to solve the problem. The rest is easy.
Use the techniques you have learned from the first day of algebra  add or subtract the same value from both sides, multiply or divide both sides by the same number (except never divide by zero).
Don't just write the answer and move on to the next problem. Check your answers! And don't check them using the algebraic expressions you wrote down. Instead, go back to the original wording of the problem, and plug the answers in there, and see if they make sense. That way, you'll catch any errors you made in assigning variables or in translating from English to Algebraic language.
Problem: 
Step 1: Assign variables. Go to the question, which asks how many of each kind of coin there are. Assign a variable, x, to the number of nickels. Then you could assign another variable to the number of dimes, but since there is such a simple relationship between the two, you can save some trouble by using an expression containing x to represent the number of dimes. Here's what you write for step 1.
Let x be the number of nickels Andy has.
Then (20x) is the number of dimes Andy has.
Step 2: Write expressions. The problem says the total value of his coins is $1.10, which is 110 cents. That means the number of nickels multiplied by five plus the number of dimes multiplied by ten equals 110. Here's how you write this:
5x+10(20x)=110
Step 3: Solve it.
5x+10(20x)=110
5x+20010x=110
5x+200=110
5x=90
5x=90
x=90/5
x=18Since x=18, there are 18 nickels. Since (20x)=(2018)=2, there are 2 dimes.
Step 4: Check your work. Go back to the original wording of the problem, and plug in these numbers to see if it still makes sense. Andy has $1.10 in 18 nickels and 2 dimes, 20 coins in all. 18 nickels are 90 cents, and 2 dimes are 20 cents, so the total is 110 cents, or $1.10. Yes, it all checks out.
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