An Introduction to Equations Unit 1 Lesson 8
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An Introduction to Equations
Unit 1 Lesson 8
AN INTRODUCTION TO EQUATIONS
Students will be able to:
solve equations and check solutions and
solve equations using tables and mental math.
Key Vocabulary:
Defining a variable
Four-step problem solving plan
Formula
AN INTRODUCTION TO EQUATIONS
EQUATION is a mathematical sentence that uses an equal sign (=). It can be used to represent the relationship between two quantities that have the same value.
TYPES
True equation
If the expressions on either side of the equal sign are equal.
π + π = ππ ππ + π = π + π
AN INTRODUCTION TO EQUATIONS
TYPES
False equation
If the expressions on either side of the equal sign are not equal.
π + π = ππ ππ + π = π + π
Open Sentence
If the equation contains one or more variables, and maybe a true
or a false depending on the values of its variables.
π + π = ππ π + π = ππ
AN INTRODUCTION TO EQUATIONS
Sample Problem 1: Tell whether each equation is true, false, or open. Explain.
A. ππ + ππ = ππ + ππ
B. π β π = ππ
C. ππ + ππ = ππ
AN INTRODUCTION TO EQUATIONS
Sample Problem 1: Tell whether each equation is true, false, or open. Explain.
A. ππ + ππ = ππ + ππ True ππ = ππ
B. π β π = ππ False ππ β ππ
C. ππ + ππ = ππ Open variable π
AN INTRODUCTION TO EQUATIONS
SOLUTION OF AN EQUATION containing a variable is a value of the variable that makes the equation true.
AN INTRODUCTION TO EQUATIONS
Sample Problem 2: Tell whether the given number is a solution of each equation.
A. Is π = π a solution of the equation π β ππ = π?
B. Is π =π
π a solution of the equation ππ + π = ππ?
C. Is π = π a solution of the equation ππ β π = ππ?
AN INTRODUCTION TO EQUATIONS
Sample Problem 2: Tell whether the given number is a solution of each equation.
A. Is π = π a solution of the equation π β ππ = π? π = ππ + π π = ππ π β π
B. Is π =π
π a solution of the equation ππ + π = ππ?
ππ = ππ β π ππ = π π = π π β π
π
C. Is π = π a solution of the equation ππ β π = ππ? ππ = ππ + π ππ = ππ π = π π β π
AN INTRODUCTION TO EQUATIONS
Sample Problem 3: Find the solution of each equation.
A. ππ β π = ππ
B. βππ = ππ β πππ
C. βππ β ππ = βπ
AN INTRODUCTION TO EQUATIONS
Sample Problem 3: Find the solution of each equation.
A. ππ β π = ππ
ππ = ππ + π ππ = ππ π = π
B. βππ = ππ β πππ
βππ β ππ = βπππ βππ = βπππ π = π
C. βππ β ππ = βπ
βππ = βπ + ππ βππ = π π = π
AN INTRODUCTION TO EQUATIONS
Sample Problem 4: Use a table to find the solution of each equation.
A. ππ + ππ = ππ
B. ππ + ππ = ππ
C. ππ = ππ + π
AN INTRODUCTION TO EQUATIONS
Sample Problem 4: Use a table to find the solution of each equation.
A. ππ + ππ = ππ π = π
π ππ + ππ = ππ
π π π + ππ = ππ ππ + ππ = ππ ππ β ππ
π π π + ππ = ππ ππ + ππ = ππ ππ β ππ
π π π + ππ = ππ ππ + ππ = ππ ππ = ππ
π π π + ππ = ππ ππ + ππ = ππ ππ β ππ
AN INTRODUCTION TO EQUATIONS
Sample Problem 4: Use a table to find the solution of each equation.
B. ππ + ππ = ππ π = π
π ππ + ππ = ππ
π π π + ππ = ππ π + ππ = ππ ππ = ππ
π π π + ππ = ππ ππ + ππ = ππ ππ β ππ
π π π + ππ = ππ ππ + ππ = ππ ππ β ππ
AN INTRODUCTION TO EQUATIONS
Sample Problem 4: Use a table to find the solution of each equation.
C. ππ = ππ + π π = π
π ππ = ππ + π
π ππ = π π + π ππ = π + π ππ = ππ
π ππ = π π + π ππ = π + π ππ β ππ
π ππ = π π + π ππ = ππ + π ππ β ππ
AN INTRODUCTION TO EQUATIONS
Sample Problem 5: Use a table to find two consecutive integers between which the solution lies.
A. ππ β ππ = ππ
B. ππ + π = ππ
C. π = π β ππ
AN INTRODUCTION TO EQUATIONS
Sample Problem 5: Use a table to find two consecutive integers between which the solution lies.
A. ππ β ππ = ππ π < π < π
π ππ β ππ
π π π β ππ ππ β ππ ππ
π π π β ππ ππ β ππ ππ
π π π β ππ ππ β ππ ππ
AN INTRODUCTION TO EQUATIONS
Sample Problem 5: Use a table to find two consecutive integers between which the solution lies.
B. ππ + π = ππ ππ < π < ππ
π ππ + π
ππ π ππ + π ππ + π ππ
ππ π ππ + π ππ + π ππ
ππ π ππ + π ππ + π ππ
AN INTRODUCTION TO EQUATIONS
Sample Problem 5: Use a table to find two consecutive integers between which the solution lies.
C. π = π β ππ βπ < π < βπ
π π β ππ
βπ π β π βπ π + π π
βπ π β π βπ π + π π
βπ π β π βπ π + π π
AN INTRODUCTION TO EQUATIONS
Sample Problem 6: Find the solution of each equation using mental math or table. If the solution lies between two consecutive integers, identify those integers.
A. ππ β π = ππ
B. ππ = π + βπ
C. π = ππ β ππ
AN INTRODUCTION TO EQUATIONS
Sample Problem 6: Find the solution of each equation using mental math or table. If the solution lies between two consecutive integers, identify those integers.
A. ππ β π = ππ π < π < π
π ππ β π
π π π β π ππ β π ππ
π π π β π ππ β π ππ
π π π β π ππ β π ππ
AN INTRODUCTION TO EQUATIONS
Sample Problem 6: Find the solution of each equation using mental math or table. If the solution lies between two consecutive integers, identify those integers.
B. ππ = π + βπ π = βπ
π π + βπ
βπ π + β βπ π + π ππ
βπ π + β βπ π + π ππ
AN INTRODUCTION TO EQUATIONS
Sample Problem 6: Find the solution of each equation using mental math or table. If the solution lies between two consecutive integers, identify those integers.
C. π = ππ β ππ π < π < π
π ππ β ππ
π ππ β π π ππ β π ππ
π ππ β π π ππ β ππ π
π ππ β π π ππ β ππ π
AN INTRODUCTION TO EQUATIONS
TRANSLATING SENTENCES TO EQUATIONS:
1. Use variables to represent the unspecified numbers or measures referred to in the sentence or problem.
2. Write the verbal expressions as algebraic expressions.
Verbal Expressions that suggest the equals sign:
is equal to is is as much as
equals is the same as is identical to
AN INTRODUCTION TO EQUATIONS
Sample Problem 7: Write an equation for each sentence.
A. Fifteen times the number π is equal to four times the sum of π and π.
B. Three times π subtracted from 57 equals 29.
C. The difference of 10 and a number is 5.
AN INTRODUCTION TO EQUATIONS
Sample Problem 7: Write an equation for each sentence.
A. Fifteen times the number π is equal to four times the sum of π and π.
ππ β π = π π + π
B. Three times π subtracted from 57 equals 29. ππ β ππ = ππ
C. The difference of 10 and a number is 5. ππ β π = π
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