1. 1 Equations, Inequalities, and Mathematical Modeling
Copyright © Cengage Learning. All rights reserved.

2. Copyright © Cengage Learning. All rights reserved.
1.2
LINEAR EQUATIONS IN ONE VARIABLE
Copyright © Cengage Learning. All rights reserved.

3. What You Should Learn • Identify different types of equations.
• Solve linear equations in one variable.
• Solve equations that lead to linear equations.
• Find x- and y-intercepts of graphs of equations
algebraically.
• Use linear equations to model and solve real-life problems.

4. Equations and Solutions of Equations

5. Equations and Solutions of Equations
An equation in x is a statement that two algebraic expressions are equal.
For example
3x – 5 = 7, x2 – x – 6 = 0, and
are equations.
To solve an equation in x means to find all values of x for which the equation is true. Such values are solutions.

6. Equations and Solutions of Equations
For instance, x = 4 is a solution of the equation 3x – 5 = 7
because 3(4) – 5 = 7 is a true statement.
The solutions of an equation depend on the kinds of numbers being considered.
For instance, in the set of rational numbers, x2 = 10 has no solution because there is no rational number whose square is 10. However, in the set of real numbers, the equation has the two solutions and

7. Equations and Solutions of Equations
An equation that is true for every real number in the domain of the variable is called an identity. For example
x2 – 9 = (x + 3)(x – 3)
is an identity because it is a true statement for any real value of x.
The equation
where x  0, is an identity because it is true for any nonzero real value of x.
Identity
Identity

8. Equations and Solutions of Equations
An equation that is true for just some (or even none) of the real numbers in the domain of the variable is called a conditional equation.
For example, the equation
x2 – 9 = 0
is conditional because x = 3 and x = –3 are the only values in the domain that satisfy the equation.
The equation 2x – 4 = 2x + 1 is conditional because there are no real values of x for which the equation is true.
Conditional equation

9. Linear Equations in One Variable

10. Linear Equations in One Variable
A linear equation has exactly one solution. To see this, consider the following steps. (Remember that a  0)
ax + b = 0
ax = –b
Write original equation.
Subtract b from each side.

11. Linear Equations in One Variable
To solve a conditional equation in x, isolate x on one side of the equation by a sequence of equivalent (and usually simpler) equations, each having the same solution(s) as the original equation.
The operations that yield equivalent equations come from the Substitution Principle and the Properties of Equality.
Divide each side by a.

12. Linear Equations in One Variable

13. Example 1 – Solving a Linear Equation
a. 3x – 6 = 0
3x = 6
x = 2
b. 5x + 4 = 3x – 8
2x + 4 = –8
2x = –12
x = –6
Original equation
Add 6 to each side.
Divide each side by 3.
Original equation
Subtract 3x from each side.
Subtract 4 from each side.
Divide each side by 2.

14. Linear Equations in One Variable
After solving an equation, you should check each solution in the original equation. For instance, you can check the solution of Example 1(a) as follows.
3x – 6 = 0
3(2) – 6 ≟ 0
0 = 0
Try checking the solution of Example 1(b).
Some equations have no solutions because all the x-terms sum to zero and a contradictory (false) statement such as 0 = 5 or 12 = 7 is obtained. For instance, the Equation x = x + 1 has no solution.
Write original equation.
Substitute 2 for x.
Solution checks.

15. Equations That Lead to Linear Equations

16. Equations That Lead to Linear Equations
To solve an equation involving fractional expressions, find the least common denominator (LCD) of all terms and multiply every term by the LCD.
This process will clear the original equation of fractions and produce a simpler equation to work with.

17. Example 3 – An Equation Involving Fractional Expressions
Solve
Solution:
Write original equation.
Multiply each term by the LCD of 12.
Divide out and multiply.
Combine like terms.

18. Example 3 – Solution
cont’d
The solution is
Divide each side by 13.

19. Equations That Lead to Linear Equations
When multiplying or dividing an equation by a variable quantity, it is possible to introduce an extraneous solution.
An extraneous solution is one that does not satisfy the original equation.
Therefore, it is essential that you check your solutions.

20. Finding Intercepts Algebraically

21. Finding Intercepts Algebraically
We have learned to find x- and y- intercepts using a graphical approach.
Because all the points on the x-axis have a y-coordinate equal to zero, and all the points on the y-axis have an x-coordinate equal to zero, you can use an algebraic approach to find x- and y-intercepts, as follows.

22. Finding Intercepts Algebraically
Here is an example.
y = 4x = 4x –1 = 4x = x
y = 4x y = 4(0) y = 1
So, the x-intercept of y = 4x + 1 is and the y-intercept is (0, 1).

23. Application

24. Example 5 – Female Participants in Athletic Programs
The number y (in millions) of female participants in high school athletic programs in the United States from 1999 through 2008 can be approximated by the linear model
y = 0.042t , –1  t  8
where t = 0 represents 2000.

25. Example 5 – Female Participants in Athletic Programs
cont’d
Find algebraically the y-intercept of the graph of the linear model shown in Figure 1.15.
Assuming that this linear pattern continues, find the year
in which there will be 3.36
million female participants.
(Source: National Federation
of State High School Associations)
Female Participants in
High School Athletics
Figure 1.15

26. Example 5(a) – Solution
To find the y-intercept, let t = 0 and solve for y, as follows.
y = 0.042t
= 0.042(0)
= 2.73
So, the y-intercept is (0, 2.73).
Write original equation.
Substitute 0 for t.
Simplify.

27. Example 5(b) – Solution
cont’d
Let y = 3.36 and solve the equation 3.36 = 0.042t for t.
3.36 = 0.042t
0.63 = 0.042t
15 = t
Because t = 0 represents 2000, t = 15 must represent So, from this model, there will be 3.36 million female participants in 2015.
Write original equation.
Subtract 2.73 from each side.
Divide each side by