# Equations, Inequalities, and Problem Solving

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Equations, Inequalities, and Problem Solving

3.2 Equations that Reduce to Linear Form

What You Will Learn Solve linear equations containing symbols of grouping Solve linear equations involving fractions Solve linear equations involving decimals

Equations Containing Symbols of Grouping

Equations Containing Symbols of Grouping
In this section you will continue your study of linear equations by looking at more complicated types of linear equations. To solve a linear equation that contains symbols of grouping, use the following guidelines. 1. Remove symbols of grouping from each side by using the Distributive Property. 2. Combine like terms. 3. Isolate the variable using properties of equality. 4. Check your solution in the original equation.

Example 1 – Solving Linear Equations Involving Parenthesis
Solve 4(x – 3) = 8. Then check your solution. Solution: 4(x – 3) = 8 4  x – 4  3 = 8 4x – 12 = 8 4x – = 4x = 20 x = 5 Write original equation. Distributive Property Simplify. Add 12 to each side. Combine like terms. Divide each side by 4. Simplify.

Example 1 – Solving Linear Equations Involving Parenthesis
cont’d Check 4(5 – 3) 8 4(2) 8 8 = 8 The solution is x = 5. Substitute 5 for x in original equation. Simplify. Solution checks.

Equations Containing Symbols of Grouping
The linear equation in the next example involves both brackets and parentheses. Watch out for nested symbols of grouping such as these. The innermost symbols of grouping should be removed first.

Example 2 – Equations Involving Symbols of Grouping (a)
a. Solve 5(x + 2) = 2(x – 1) Solution: 5(x + 2) = 2(x – 1) 5x + 10 = 2x – 2 5x – 2x + 10 = 2x – 2x – 2 3x + 10 = – 2 3x + 10 – 10 = – 2 – 10 3x = –12 x = –4 Original equation Distributive Property Subtract 2x from each side Combine like terms Subtract 10 from each side Combine like terms Divide each side by 3

Example 2 – Equations Involving Symbols of Grouping (b)
b. Solve 2(x – 7) – 3(x + 4) = 4 – (5x – 2) Solution: 2(x – 7) – 3(x + 4) = 4 – (5x – 2) 2x – 14 – 3x – 12 = 4 – 5x – 2 –x – 26 = –5x + 6 –x + 5x – 26 = –5x + 5x + 6 4x – 26 = 6 4x – = 4x = 32 x = 8 Original equation Distributive Property Combine like terms Add 5x to each side Combine like terms Add 26 to each side Combine like terms Divide each side by 4

Example 2 – Equations Involving Symbols of Grouping (c)
c. Solve 5x – 2[4x + 3(x – 1)] = 8 – 3x. Solution: 5x – 2[4x + 3(x – 1)] = 8 – 3x 5x – 2[4x + 3x – 3] = 8 – 3x 5x – 2[7x – 3] = 8 – 3x 5x – 14x + 6 = 8 – 3x –9x + 6 = 8 – 3x –9x + 3x + 6 = 8 – 3x + 3x Write original equation. Distributive Property Combine like terms inside brackets. Distributive Property Combine like terms. Add 3x to each side.

Example 2 – Equations Involving Symbols of Grouping (c)
cont’d –6x + 6 = 8 –6x + 6 – 6 = 8 – 6 –6x = 2 The solution is x = Check this in the original equation. Combine like terms. Subtract 6 from each side. Combine like terms. Divide each side by –6.

Equations Involving Fractions
To solve a linear equation that contains one or more fractions, it is usually best to first clear the equation of fraction by multiplying each side by the least common multiple (LCM) of the denominators.

Example 3 – Solving Linear Equations Involving Fractions (a)
a. Solve Solution: Original Equations Multiply each side by LCM 6 Distributive Property Simplify Add 2 to each side Divide each side by 9

Example 3 – Solving Linear Equations Involving Fractions (b)
b. Solve Solution: Original Equations Multiply each side by LCM 6 Distributive Property Simplify Add 2 to each side

Example 3 – Solving Linear Equations Involving Fractions (c)
c. Solve Solution: Original Equations Distributive Property Multiply each side by LCM 12 Simplify Subtract 2 to each side Divide each side by 8

Equations Involving Fractions
A common type of linear equation is one that equates two fractions. To solve such an equation, consider the fractions to be equivalent and use cross-multiplication. That is, if then a  d = b  c.

Example 4 – Finding a Test Score
To get an A in a course, you must have an average of at least 90 points for 4 tests of 100 points each. For the first 3 tests, your scores were 87, 92, and 94. What must you score on the fourth test to earn a 90% average for the course? Solution: Verbal Model: Labels: Score of 4th test = x (points) Score of first 3 tests: 87, 82, (points)

Example 4 – Finding a Test Score
cont’d Equation: You can solve this equation by multiplying each side by 4 You need a score of 97 on the fourth test to earn a 90% average Write equation Multiply each side by LCM 4 Simplify Combine like terms Subtract 263 from each side

Equations Involving Decimals

Equations Involving Decimals
Many real-life applications of linear equations involve decimal coefficients. To solve such an equation, you can clear it of decimals in much the same way you clear an equation of fractions. Multiply each side by a power of 10 that converts all decimal coefficients to integers.

Example 5 – Solving a Linear Equation Involving Decimals
Solve 0.3x + 0.2(10 – x) = 0.15(30). Then check your solution. Solution: 0.3x + 0.2(10 – x) = 0.15(30) 0.3x + 2 – 0.2x = 4.5 0.1x + 2 = 4.5 10(0.1x + 2) = 10(4.5) x + 20 = 45 x = 25 Write original equation. Distributive Property Combine like terms. Multiply each side by 10. Simplify. Subtract 20 from each side.

Example 5 – Solving a Linear Equation Involving Decimals
cont’d Check 0.3(25) + 0.2(10 – 25) (30) 0.3(25) + 0.2(–15) (30) 7.5 – 4.5 = 4.5 The solution is x = 25. Substitute 25 for x in original equation. Perform subtraction within parentheses. Multiply. Solution checks.

Example 7 – Finding Your Gross Pay per Paycheck
The enrollment y (in millions) at postsecondary schools from 2000 through 2009 can be approximated by the linear model y = , where t represents the year, with t = 0 corresponding to Use the model to predict the year in which the enrollment will be 14 million students. (Source: U.S. Department of Education) Solution: To find the year in which the enrollment will be 14 million students, substitute 14 for y in the original equation and solve the equation for t. 14 = 0.35t 4.9 = 0.35t 14 = t Because t = 0 corresponds to 2000, the enrollment at postsecondary schools will be 14 million during 2014. Substitute 14 for y in original equation Subtract 9.1 from each side Divide each side by 0.35