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

Linear Equations in One Variable 1.2 Copyright © Cengage Learning. All rights reserved.

Objectives Identify different types of equations. Solve linear equations in one variable. Solve rational 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.

Equations and Solutions of Equations

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.

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

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

Equations and Solutions of Equations An equation that is true for just some (but not all) 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. Conditional equation

Equations and Solutions of Equations The equation 2x + 4 = 6 is conditional because x = 1 is the only value in the domain that satisfies the equation. A contradiction is an equation that is false for every real number in the domain of the variable. For example, the equation 2x – 4 = 2x + 1 is a contradiction because there are no real values of x for which the equation is true. Contradiction

Linear Equations in One Variable

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. Divide each side by a.

Linear Equations in One Variable So, the equation ax + b = 0 has exactly one solution, To solve a conditional equation in x, isolate x on one side of the equation by a sequence of equivalent equations, each having the same solution(s) as the original equation. The operations that yield equivalent equations come from the properties of equality.

Linear Equations in One Variable

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.

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). Write original equation. Substitute 2 for x. Solution checks.

Rational Equations That Lead to Linear Equations

Rational Equations That Lead to Linear Equations A rational equation is an equation that involves one or more fractional expressions. To solve a rational equation, 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.

Example 3 – Solving a Rational Equation Solve Solution: Original equation Multiply each term by the LCD. Simplify. Combine like terms.

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

Rational Equations That Lead to Linear Equations When multiplying or dividing an equation by a variable expression, it is possible to introduce an extraneous solution that does not satisfy the original equation.

Finding Intercepts Algebraically

Finding Intercepts Algebraically You 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.

Example 5 – Finding Intercepts Algebraically Find the x- and y-intercepts of the graph of each equation algebraically. a. y = 4x + 1 b. 3x + 2y = 6 Solution: a. To find the x-intercept, set y equal to zero and solve for x, as follows. y = 4x + 1 0 = 4x + 1 –1 = 4x = x Write original equation. Substitute 0 for y. Subtract 1 from each side. Divide each side by 4.

Example 5 – Solution cont’d So, the x-intercept is . To find the y-intercept, set x equal to zero and solve for y, as follows. y = 4x + 1 y = 4(0) + 1 y = 1 So, the y-intercept is (0, 1). Write original equation. Substitute 0 for x. Simplify.

Example 5 – Solution cont’d b. To find the x-intercept, set y equal to zero and solve for x, as follows. 3x + 2y = 6 3x + 2(0) = 6 3x = 6 x = 2 So, the x-intercept is (2, 0). Write original equation. Substitute 0 for y. Simplify. Divide each side by 3.

Example 5 – Solution cont’d To find the y-intercept, set x equal to zero and solve for y, as follows. 3x + 2y = 6 3(0) + 2y = 6 2y = 6 y = 3 So, the y-intercept is (0, 3). Write original equation. Substitute 0 for x. Simplify. Divide each side by 2.

Application

Example 6 – Female Participants in Athletic Programs The number y (in millions) of female participants in high school athletic programs in the United States from 2000 through 2010 can be approximated by the linear model y = 0.045t + 2.70, 0  t  10 where t = 0 represents 2000.

Example 6 – Female Participants in Athletic Programs cont’d (a) Find algebraically and interpret the y-intercept of the graph of the linear model shown in Figure 1.10. (b) Use the linear model to predict the year in which there will be 3.42 million female participants. Female Participants in High School Athletics Figure 1.10

Example 6(a) – Solution To find the y-intercept, let t = 0 and solve for y, as follows. y = 0.045t + 2.70 = 0.045(0) + 2.70 = 2.70 So, the y-intercept is (0, 2.70). This means that there were about 2.70 million female participants in 2000. Write original equation. Substitute 0 for t. Simplify.

Example 6(b) – Solution cont’d Let y = 3.42 and solve for t, as follows y = 0.045t + 2.70 3.42 = 0.045t + 2.70 0.72 = 0.045t 16 = t Because t = 0 represents 2000, t = 16 must represent 2016. So, from this model, there will be 3.42 million female participants in 2016. Write original equation. Substitute 3.42 for y. Subtract 2.70 from each side. Divide each side by 0.045.