1. Equations, Inequalities, and Problem Solving
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2. 3.2 Equations that Reduce to Linear Form

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

4. Equations Containing Symbols of Grouping

5. 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.

6. 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.

7. 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.

8. 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.

9. 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

10. 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

11. 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.

12. 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.

13. 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.

14. 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

15. 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

16. 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

17. 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.

18. 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)

19. 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

20. Equations Involving Decimals

21. 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.

22. 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.

23. 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.

24. 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