1. Chapter 2 Linear Equations and Inequalities in One Variable

2. § 2.1 The Addition Property of Equality

3. The same real number (or algebraic expression) may be added to both sides of an equation without changing the equation’s solution. This can be expressed symbolically as follows: If a = b, then a + c = b + c The Addition Property of Equality Blitzer, Introductory Algebra, 5e – Slide #3 Section 2.1

4. Blitzer, Introductory Algebra, 5e – Slide #4 Section 2.1 Linear Equations Definition of a Linear Equation A linear equation in one variable x is an equation that can be written in the form ax + b = 0, where a and b are real numbers and a is not equal to 0. An example of a linear equation in x is 4x + 2 = 6. Linear equations in x are first degree equations in the variable x.

5. Blitzer, Introductory Algebra, 5e – Slide #5 Section 2.1 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. Subtraction Property of EqualityThe same real number or algebraic expression may be subtracted from both sides of an equation without changing the equation’s solution set.

6. Blitzer, Introductory Algebra, 5e – Slide #6 Section 2.1 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.

7. Blitzer, Introductory Algebra, 5e – Slide #7 Section 2.1 Solving an equation using the addition property

8. Since we know that subtraction is just addition of an opposite or an additive inverse, we can also subtract the same number from both sides of an equation without changing the equation’s solution. Solve: x + 6 = 9 x + 6 – 6 = 9 – 6 Subtract 6 from both sides. x = 3 Blitzer, Introductory Algebra, 5e – Slide #8 Section 2.1 Subtracting from both sides of an equation

9. Now consider an equation in which we would need to subtract variable terms from both sides. Remember that our goal is to isolate all the variable terms on one side. To do this in the equation below, we must get the 4x term off the RHS by adding its opposite, -4x, to both sides. Solve: 5x = 4x + 3 5x – 4x = 4x + 3 – 4x Subtract 4x from both sides. This simplifies to: x = 3 Blitzer, Introductory Algebra, 5e – Slide #9 Section 2.1 Adding and Subtracting Variable Terms in an Equation

10. Blitzer, Introductory Algebra, 5e – Slide #10 Section 2.1 Solving Linear Equations EXAMPLE: Solve for x 14x + 2 = 15x 2) Collect variable terms on one side and constant terms on the other side. 2 = x Subtract 14x from both sides14x – 14x + 2 = 15x – 14x Simplify Check the proposed solution of 2 in the original equation. When we insert the 2 for x, we Get the sentence 14(2) + 2 = 15(2) or we get 28 + 2 = 30, which is true. Then x = 2 is the solution.

11. Blitzer, Introductory Algebra, 5e – Slide #11 Section 2.1 Solving Linear EquationsEXAMPLE SOLUTION Solve and check: 5 + 3x - 4x = 1 - 2x + 12. 1) Simplify the algebraic expressions on each side. 5 + 3x - 4x = 1 - 2x + 12 5 - x = 13 - 2x Combine like terms: +3x - 4x = -x 1 + 12 = 13

12. Blitzer, Introductory Algebra, 5e – Slide #12 Section 2.1 Solving Linear Equations 2) Collect variable terms on one side and constant terms on the other side. 5 - x + 2x = 13 - 2x + 2xAdd 2x to both sides CONTINUED 5 + 1x = 13Simplify 5 – 5 + 1x = 13 - 5 Subtract 5 from both sides x = 8 Simplify

13. Blitzer, Introductory Algebra, 5e – Slide #13 Section 2.1 Solving Linear Equations 4) Check the proposed solution in the original equation. CONTINUED 5 + 3x - 4x = 1 - 2x + 12Original equation 5 + 3(8) - 4(8) 1 – 2(8) + 12Replace x with 8 ? = 5 +24 – 32 1 – 16 + 12Multiply= ? 29 -32 – 16 + 13Add or subtract from left to right = ? True - It checks. The solution set is {8}. -3 = -3