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Writing & Solving Equations
SOMETHING YOU MIGHT SAY…
In order to solve application problems, it is necessary to
translate English phrases into algebraic symbols. The following are some common phrases and their mathematic translation.
Applications
Translating from Words to Mathematical Expressions
Verbal Expression
The sum of a number and 2
Mathematical Expression
(where x and y are numbers)
Addition
3 more than a number
7 plus a number
16 added to a number
A number increased by 9
The sum of two numbers
x + 2
x + 3
7 + x
x + 16
x + 9
x + y
Applications
Translating from Words to Mathematical Expressions
Verbal Expression
4 less than a number
Mathematical Expression
(where x and y are numbers)
Subtraction
10 minus a number
A number decreased by 5
A number subtracted from 12
The difference between two
numbers
x – 4
10 – x
x – 5
12 – x
x – y
Applications
Translating from Words to Mathematical Expressions
Verbal Expression
14 times a number
Mathematical Expression
(where x and y are numbers)
Multiplication
A number multiplied by 8
Triple (three times) a number
The product of two numbers
14x
8x
3x
xy
of a number (used with
fractions and percent)
34 x3
4
Applications
Translating from Words to Mathematical Expressions
2
x
Verbal Expression
The quotient of 6 and a number
Mathematical Expression
(where x and y are numbers)
Division
A number divided by 15
half a number
(x ≠ 0)6x
x15
Because subtraction and division are not commutative operations, be carefulto correctly translate expressions involving them. For example, “5 less than anumber” is translated as x – 5, not 5 – x. “A number subtracted from 12” isexpressed as 12 – x, not x – 12. For division, the number by which we are dividing is the denominator, andthe number into which we are dividing is the numerator. For example, “a number divided by 15” and “15 divided into x” both translate as . Similarly,“the quotient of x and y” is translated as .
Applications
Caution
x15x
y
Applications
Indicator Words for Equality
Equality
The symbol for equality, =, is often indicated by the word is. In fact, any
words that indicate the idea of “sameness” translate to =.
Applications
Translating Words into Equations
Verbal Sentence Equation
16x – 25 = 87If the product of a number and 16 is decreased
by 25, the result is 87.
= 48The quotient of a number and the number plus
6 is 48.x + 6
x
+ x = 54The quotient of a number and 8, plus the
number, is 54.8x
Twice a number, decreased by 4, is 32. 2x – 4 = 32
Applications
Distinguishing between Expressions and Equations
(a) 4(6 – x) + 2x – 1
(b) 4(6 – x) + 2x – 1 = –15
There is no equals sign, so this is an expression.
Because of the equals sign, this is an equation.
Decide whether each is an expression or an equation.
Six Steps to Solving Application Problems
Step 1 Read the problem, several times if necessary, until you understandwhat is given and what is to be found.
Step 2 If possible draw a picture or diagram to help visualize the problem.
Step 3 Assign a variable to represent the unknown value, using diagrams or tables as needed. Write down what the variable represents. Express any other unknown values in terms of the variable.
Step 4 Write an equation using the variable expression(s).
Step 5 Solve the equation.
Step 6 Check the answer in the words of the original problem.
Applications
Six Steps to Solving Application Problems
Now Lets Write & Solve Some Equations
Example 1:
Fifteen more than twice a number is – 23.
Now Lets Write & Solve Some Equations
Example 2:
The quotient of a number and 9, increased
by 10 is 11.
Now Lets Write & Solve Some Equations
Example 3:
The difference between 5 times a number
is 4 and 16.
CARTOON
Applications
Solving a Geometry Problem
Step 1 Read the problem. We must find the length and width of the rectangle.
The length is 2 ft more than three times the width and the perimeter is
124 ft.
The length of a rectangle is 2 ft more than three times the width. The perimeter
of the rectangle is 124 ft. Find the length and the width of the rectangle.
Step 2 Assign a variable. Let W = the width; then 2 + 3W = length.
Make a sketch.
W
2 + 3W
Step 3 Write an equation. The perimeter of a rectangle is given by the
formula P = 2L + 2W.
124 = 2(2 + 3W) + 2W Let L = 2 + 3W and P = 124.
Applications
Solving a Geometry Problem
Step 4 Solve the equation obtained in Step 3.
The length of a rectangle is 2 ft more than three times the width. The perimeter
of the rectangle is 124 ft. Find the length and the width of the rectangle.
124 = 2(2 + 3W) + 2W
124 = 4 + 6W + 2W
124 – 4 = 4 + 8W – 4
120 8W8 8
15 = W
Remove parentheses
124 = 4 + 8W Combine like terms.
Subtract 4.
Divide by 8.
120 = 8W
=
Applications
Solving a Geometry Problem
Step 5 State the answer. The width of the rectangle is 15 ft and the length is
2 + 3(15) = 47 ft.
The length of a rectangle is 2 ft more than three times the width. The perimeter
of the rectangle is 124 ft. Find the length and the width of the rectangle.
Step 6 Check the answer by substituting these dimensions into the words of
the original problem.