Algebra Word Problems

 

Common Terms in Word Problems

 

Some terms can help you figure out which operation needs to be done in a word problem

 

Add

Subtract

Multiply

Divide

Sum

Difference

Product

Quotient

Plus

Less/less than

Times

Per

More/more than

Minus

Of

 

 

When to use an Equation

 

In a word problem, if you can identify what the missing value is, the easiest way to find it may be to set up an equation.  Assign a variable to the missing value, then fill in the other numbers in the equation based on what the problem says is being done to the variable.  If the problem say “10 is 4 less than what number?” the number is something we’ll call x.  Since we’re talking about something that is 4 less than that number, we write x – 4.  ‘Is’ is the same thing as an equal sign, so we could write this equation as x – 4 = 10.

 

Systems of Equations

 

Sometimes you will have more than one value unknown in an equation, but the two values will have a relationship with each other.  For these cases, you will need to use a system of equations.  One value will be assigned a variable.  The others will be assigned an expression containing the first variable.  The following examples are some common problems of this type.

 

Consecutive Number Problem

 

Consecutive numbers that follow each other, such as 4, 5 and 6.  Sometimes you may see consecutive even numbers, like 2, 4, 6 or consecutive odd numbers like 5, 7, 9. 

 

The sum of three consecutive numbers is 36.  What are the numbers?

 

We know that each number is one more than the one before it.  Let’s call the first number x.  The second number is one more than it, so it will be x + 1.  The third number will be one more than that, or x + 2.  If we add up all of those values, we will get 36, so we could write this equation as

 

 x + x + 1 + x + 2 = 36

 

To solve, combine like terms:

 

3x + 3 = 36

 

Addition and subtraction operations

 

3x + 3 = 36

- 3         - 3

 

3x = 33

 

Then, multiplication and division operations:

 

3x = 33

÷3     ÷3

x = 11

 

The first number is 11, so the others must be 12 and 13.

 

Age Problem

 

Alex is 3 years younger than Jamie.  Hannah is 5 years older than Jamie.  The sum of their ages is 23.  How old is Alex?

 

Since we know the how old Hannah and Alex are in relation to Jamie, let’s make Jamie be the variable:  x.  Since Alex is 3 years younger than Jamie, we will represent her age with x – 3.  Since Hanna is 5 years older than Jamie, her age will be x + 5.  All together they will add up to equal 23, so we could write an equation as:

 

x + x – 3 + x + 5 = 23

 

Now combine like terms (remember that numbers after a minus sign are negative):

 

3x + 2 = 23

 

Then addition and subtraction:

 

3x + 2 = 23

- 2         -2

3x = 21

 

Then multiplication and division:

 

3x = 21

÷3    ÷3

x = 7

 

Now we must look back at what the problem asked us to find.  7 is our variable, which we picked to represent Jamie’s age.  We were asked to find Alex’s age.  Since Alex is 3 years younger than Jamie (7-3) we find that Alex’s age is 4.

 

Ticket Problem

 

Tickets for a baseball game have separate prices for kids and adults.  Kids tickets are $3.50 and adults are $6.00.  A family of  9 went to a game and their total was 41.50.  How many of  the tickets were children’s tickets?

 

Let’s call the number of children’s tickets x.  The adult tickets plus the children’s will be 9, so the adult tickets can be represented by 9 – x.  The They paid 3.50 for each child ticket, so that total would be 3.5(x) and 6 for each adult, which would be 6(9 – x).  Add those together for:

 

 3.5(x) + 6(9 – x) = 41.5

 

Distribute:

3.5x + 54 – 6x = 41.5

 

Combine like terms:

 

-2.5x + 54 = 41.5

 

Subtract out the 54:

-2.5x = -12.5

 

Divide out the -2.5:

X = 5

 

There were 5 child tickets