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Ch 2 Sec 4: Slide #1 Columbus State Community College Chapter 2 Section 4 Solving Equations Using Division.

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Presentation on theme: "Ch 2 Sec 4: Slide #1 Columbus State Community College Chapter 2 Section 4 Solving Equations Using Division."— Presentation transcript:

1 Ch 2 Sec 4: Slide #1 Columbus State Community College Chapter 2 Section 4 Solving Equations Using Division

2 Ch 2 Sec 4: Slide #2 Solving Equations Using Division 1.Solve equations using the division property of equality. 2.Simplify equations before using the division property of equality. 3.Solve equations such as –x = 3.

3 Ch 2 Sec 4: Slide #3 Division Property of Equality If a = b, then as long as c is not 0. In other words, you may divide both sides of an equation by the same nonzero number and still keep it balanced. a c b c =

4 Ch 2 Sec 4: Slide #4 Equality Principle for Solving an Equation As long as you do the same thing to both sides of an equation, the balance is maintained and you still have a true equation. (The only exception is that you cannot divide by 0.)

5 Ch 2 Sec 4: Slide #5 Using the Division Property of Equality EXAMPLE 1 Using the Division Property of Equality (a)5n = 30 Solve each equation and check each solution. 5n = 30 5 n = 6 5n = 30 = 30 30 5 · 6 = 30 Check: Balance statement Solution 5

6 Ch 2 Sec 4: Slide #6 Using the Division Property of Equality EXAMPLE 1 Using the Division Property of Equality (b)48 = – 8x Solve each equation and check each solution. 48 = – 8x –8–8 – 6 = x 48 = – 8x = – 8 · – 6 48 = 48 Check: Balance statement Solution –8–8

7 Ch 2 Sec 4: Slide #7 CAUTION Be careful to divide both sides by the same number as the coefficient of the variable term. In Example 1 (b), the coefficient of – 8x is – 8, so divide both sides by – 8. (Do not divide by the opposite of – 8, which is 8. Use the opposite only when you’re eliminating a term.) Divide both sides by the coefficient – 8. 48 = – 8x –8–8 – 6 = x –8–8

8 Ch 2 Sec 4: Slide #8 Simplifying before Solving Equations EXAMPLE 2 Simplifying before Solving Equations (a)2k – 6k = – 14 – 6 Solve each equation and check each solution. 2k – 6k = – 14 – 6 – 4k = – 20 –4–4 k = 5 2k – 6k = – 14 – 6 = – 20 10 – 30 2(5) – 6(5) = – 20 Check: Balance statement Solution –4–4 = – 20 – 20

9 Ch 2 Sec 4: Slide #9 Simplifying before Solving Equations EXAMPLE 2 Simplifying before Solving Equations (b)2 – 8 = 11m – 9m Solve each equation and check each solution. 2 – 8 = 11m – 9m – 6 = 2m 2 – 3 = m 2 – 8 = 11m – 9m = 11( – 3) – 9( – 3) –6–6 –6–6 = – 33 + 27 Check: Balance statement Solution 2 = – 6 –6–6

10 Ch 2 Sec 4: Slide #10 Solving an Equation of the Type – x = 7 EXAMPLE 3 Solving an Equation of the Type – x = 7 Solve – x = 7 and check the solution. – 1 x = 7 –1–1 x = – 7 – 1 x = 7 = 7 7 – 1( – 7) = 7 Check: Balance statement Solution –1–1

11 Ch 2 Sec 4: Slide #11 CAUTION As the last step in solving an equation, do not leave a negative sign in front of a variable. For example, do not leave – x = – 9. Write the understood – 1 as the coefficient so that – x = – 9 is written as – 1x = – 9. Then divide both sides by – 1 to get x = 9. The solution is 9.

12 Ch 2 Sec 4: Slide #12 Solving Equations Using Division Chapter 2 Section 4 – Completed Written by John T. Wallace


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