Applications of Linear Equations Part 1 Lesson

In our Algebra 1 course, we learned how to solve applications of linear equations. These problem types are more commonly known as "word problems". Most students find word problems significantly more challenging than simply solving equations. This is due to the extra work needed to solve a word problem. Instead of just solving an equation, one must read the information presented and create an equation based on the situation given in the word problem. A big key to success is knowing how to correctly translate key phrases for addition, subtraction, multiplication, and division. Let's begin our lesson by looking at some phrases related to addition and subtraction:

Addition & Subtraction Phrases:
Addition	Subtraction
more than	minus
sum of	less than
together	take away
increased by	subtracted from
plus	decreased by
added to	difference

Let's look at a few examples.
Example 1: Solve each word problem.
When a number is increased by 6, the result is 19. Find the number.
The key to solving this problem is being able to translate the phrase "increased by". We should know this involves the addition operation. Let's break this down step by step:
"a number" » x (we can use any variable we want)
"increased by" » +
"6" » 6
So far, we have the algebraic expression:
x + 6
The word for equality is usually indicated by the word "is":
"the result is 19" » = 19
If we put the two together, we end up with an equation that can be used to solve our word problem:
x + 6 = 19
x = 13
The number is 13. If we read back through the problem, we can see this is consistent with what has been given. A number (13) is increased by (+) 6, the result is (=) 19.
(13 + 6 = 19)
Example 2: Solve each word problem.
A number is decreased by 14, the result is -5. Find the number.
The key to solving this problem is being able to translate the phrase "decreased by". We should know this involves the subtraction operation. Let's break this down step by step:
"a number" » x (we can use any variable we want)
"decreased by" » -
"14" » 14
So far, we have the algebraic expression:
x - 14
The word for equality is usually indicated by the word "is":
"the result is -5" » = -5
If we put the two together, we end up with an equation that can be used to solve our word problem:
x - 14 = -5
x = 9
The number is 9. If we read back through the problem, we can see this is consistent with what has been given. A number (9) is decreased by (-) 14, the result is (=) -5.
(9 - 14 = -5)
Note: We must be very careful when translating with subtraction. Remember subtraction is not commutative like addition. This means the order matters. If we see a number decreased by 3:
x - 3 » correct
3 - x » wrong
Additionally, we will have to interpret phrases that involve the multiplication and division operation.

Multiplication & Division Phrases:
Multiplication	Division
times	divided by
tripled	average
product	quotient
multiplied by	ratio of
twice	shared evenly
of	half of

Let's look at a few examples.
Example 3: Solve each word problem.
When a number is multiplied by 7, the result is 91. Find the number.
The key to solving this problem is being able to translate the phrase "multiplied by". We should know this involves the multiplication operation. Let's break this down step by step:
"a number" » x (we can use any variable we want)
"multiplied by" » •
"7" » 7
So far, we have the algebraic expression:
7 • x or 7x
The word for equality is usually indicated by the word "is":
"the result is 91" » = 91
If we put the two together, we end up with an equation that can be used to solve our word problem:
7x = 91
x = 13
The number is 13. If we read back through the problem, we can see this is consistent with what has been given. A number (13) is multiplied by (•) 7, the result is (=) 91.
7 • 13 = 91
Example 4: Solve each word problem.
The quotient of a number and 3 is 31. Find the number.
The key to solving this problem is being able to translate the phrase "quotient of". We should know this involves the division operation. Let's break this down step by step:
"a number" » x (we can use any variable we want)
"quotient of " » ÷
"3" » 3
When we read the "quotient of", the next part is our dividend. The part after the "and" is our divisor. It is very important to translate this correctly as division is not commutative like multiplication.
The quotient of x and 3:
x ÷ 3 » correct
3 ÷ x » wrong
So far, we have the algebraic expression:
x ÷ 3
The word for equality is usually indicated by the word "is":
"is 31" » = 31
If we put the two together, we end up with an equation that can be used to solve our word problem:
x ÷ 3 = 31
x = 93
The number is 93. If we read back through the problem, we can see this is consistent with what has been given. The quotient of a number (93) and 3 is 31.
93 ÷ 3 = 31
Let's put everything together and look at a more challenging problem.
Example 5: Solve each word problem.
3 times the sum of a number and 7 is the same as 9 times the difference of the number and 1.
Let's let x be our unknown number.
"the sum of" refers to addition.
"times" refers to multiplication.
3 times the sum of a number and 7:
3(x + 7)
3 is multiplied by a quantity (the sum of x and 7), so parentheses must be used.
"is the same as" is our equality statement "=".
9 times the difference of the number and 1:
9(x - 1)
Again, we must use parentheses here. 9 is multiplied by a quantity (the difference of x and 1).
Putting the whole picture together gives us:
3(x + 7) = 9(x - 1)
Solve:
3x + 21 = 9x - 9
-6x = -30
x = 5
So our unknown number is 5. Let's read back through the problem and see if this is consistent with what is given.
3 times the sum of a number (5) and 7 is the same as 9 times the difference of the number (5) and 1.
3(5 + 7) = 9(5 - 1)
3(12) = 9(4)
36 = 36

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