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Martin-Gay, Beginning Algebra, 5ed 22 33 Using Both Properties Divide both sides by 3. Example: 3z – 1 = 26 3z = 27 Simplify both sides. z = 9 Simplify.

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Presentation on theme: "Martin-Gay, Beginning Algebra, 5ed 22 33 Using Both Properties Divide both sides by 3. Example: 3z – 1 = 26 3z = 27 Simplify both sides. z = 9 Simplify."— Presentation transcript:

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2 Martin-Gay, Beginning Algebra, 5ed 22

3 33 Using Both Properties Divide both sides by 3. Example: 3z – 1 = 26 3z = 27 Simplify both sides. z = 9 Simplify both sides. 3z – 1 + 1 = 26 + 1 Add 1 to both sides.

4 Martin-Gay, Beginning Algebra, 5ed 44 Using Both Properties Example: 12x + 30 + 8x – 6 = 10 20x + 24 = 10 Simplify the left side. 20x = – 14 Simplify both sides. Divide both sides by 20. 20x + 24 + ( – 24) = 10 + ( – 24) Add –24 to both sides. Simplify both sides.

5 Martin-Gay, Beginning Algebra, 5ed 55 Solving Linear Equations Simplify both sides. Divide both sides by 7. Example: Solve the equation. Add –3y to both sides. Add –30 to both sides. Multiply both sides by 5. Simplify both sides.

6 Martin-Gay, Beginning Algebra, 5ed 66 – 0.01(5a + 4) = 0.04 – 0.01(a + 4) Multiply both sides by 100. Example: Solve the equation. – 1(5a + 4) = 4 – 1(a + 4) Apply the distributive property. – 5a – 4 = 4 – a – 4 Add a to both sides and simplify. – 4a – 4 = 0 Add 4 to both sides and simplify. – 4a = 4 Divide both sides by  4. a = – 1 – 5a – 4 = – aSimplify both sides

7 Martin-Gay, Beginning Algebra, 5ed 77 – 0.01(5a + 4) = 0.04 – 0.01(a + 4) Example: Solve the equation. – 0.05a – 0.04 = 0.04 – 0.01a – 0.04 Apply the distributive property. – 0.05a – 0.04 = – 0.01a Add 0.01a to both sides. – 0.04a – 0.04 = 0 Add 0.04 to both sides and simplify. – 0.04a = 0.04 Divide both sides by  0.04. a = – 1 Simplify both sides.

8 Martin-Gay, Beginning Algebra, 5ed 88 Identity Equations 5x – 5 = 2(x + 1) + 3x – 7 5x – 5 = 2x + 2 + 3x – 7Use distributive property. 5x – 5 = 5x – 5 Simplify the right side. Both sides of the equation are identical. This equation will be true for every x that is substituted into the equation, the solution is “all real numbers.” Example: Solve the equation. 0 = 0 Add 5 to both sides. – 5 = – 5 Add  5x to both sides. Identity Equation.

9 Martin-Gay, Beginning Algebra, 5ed 99 Contradiction Equations Since no value for the variable x can be substituted into this equation that will make this a true statement, there is “no solution.” 3x – 7 = 3(x + 1) 3x – 7 = 3x + 3 Use distributive property. – 7 = 3 Simplify both sides. 3x + ( – 3x) – 7 = 3x + ( – 3x) + 3 Add –3x to both sides. Example: Solve the equation. Contradiction Equation.

10 Martin-Gay, Beginning Algebra, 5ed 10 Solving Linear Equations

11 Martin-Gay, Beginning Algebra, 5ed 11 Examples c) a) d) b)


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