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§ 2.2 The Multiplication Property of Equality. Blitzer, Introductory Algebra, 5e – Slide #2 Section 2.2 Properties of Equality PropertyDefinition Addition.

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Presentation on theme: "§ 2.2 The Multiplication Property of Equality. Blitzer, Introductory Algebra, 5e – Slide #2 Section 2.2 Properties of Equality PropertyDefinition Addition."— Presentation transcript:

1 § 2.2 The Multiplication Property of Equality

2 Blitzer, Introductory Algebra, 5e – Slide #2 Section 2.2 Properties of Equality PropertyDefinition Addition Property of Equality The same real number or algebraic expression may be added to both sides of an equation without changing the equation’s solution set. Multiplication Property of Equality The same nonzero real number may multiply both sides of an equation without changing the equation’s solution set. In this section, we introduce an additional property of equality for use in solving equations. This new property is the Multiplication Property of Equality.

3 Multiplicative property of equalityEXAMPLE Blitzer, Introductory Algebra, 5e – Slide #3 Section 2.2

4 Since we know that division is the multiplication by the multiplicative inverse, we can divide both sides of an equation by the same number without changing the equation’s solution. Dividing both sides by the same nonzero number Blitzer, Introductory Algebra, 5e – Slide #4 Section 2.2

5 Fractional Coefficients Blitzer, Introductory Algebra, 5e – Slide #5 Section 2.2

6 Solve: 3x – 10 = - 4x + 18 3x – 10 + 4x = - 4x + 18 + 4x Add 4x to both sides. 7x – 10 = 18 Simplify. 7x – 10 + 10 = 18 + 10 Add 10 to both sides. 7x = 28Simplify. Divide both sides by 7. x = 4Solve. Using both properties of Equality Blitzer, Introductory Algebra, 5e – Slide #6 Section 2.2 EXAMPLE

7 Blitzer, Introductory Algebra, 5e – Slide #7 Section 2.2 Solving Linear Equations Solving a Linear Equation 1) Simplify the algebraic expressions on each side. 2) Collect all the variable terms on one side and all the numbers, or constant terms, on the other side 3) Isolate the variable and solve. 4) Check the proposed solution in the original equation. Remember the steps for solving linear equations.

8 Blitzer, Introductory Algebra, 5e – Slide #8 Section 2.2 Solving Linear EquationsEXAMPLE SOLUTION Solve and check: 5 - 3x + 4x = 1 - 7x + 12. 1) Simplify the algebraic expressions on each side. 5 - 3x + 4x = 1 - 7x + 12 5 + x = 13 - 7x Combine like terms: -3x + 4x = x 1 + 12 = 13

9 Blitzer, Introductory Algebra, 5e – Slide #9 Section 2.2 Solving Linear Equations 2) Collect variable terms on one side and constant terms on the other side. 5 + x + 7x = 13 - 7x + 7xAdd 7x to both sides CONTINUED 5 + 8x = 13Simplify 5 – 5 + 8x = 13 - 5 Subtract 5 from both sides 8x = 8 Simplify

10 Blitzer, Introductory Algebra, 5e – Slide #10 Section 2.2 Solving Linear Equations 3) Isolate the variable and solve. CONTINUED 8x 8Divide both sides by 8 8 8 x = 1Simplify

11 Blitzer, Introductory Algebra, 5e – Slide #11 Section 2.2 Solving Linear Equations 4) Check the proposed solution in the original equation. CONTINUED 5 - 3x + 4x = 2 - 7x + 6Original equation 5 – 3(1) + 4(1) 1 – 7(1) + 12Replace x with 1 ? = 5 – 3 + 4 1 – 7 + 12Multiply= ? 2 + 4 – 6 + 12Add or subtract from left to right = ? Add6 = 6 It checks, and the solution set is {1}. The original sentence is true when x is 1.


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