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§ 2.3 Solving Linear Equations. Martin-Gay, Beginning and Intermediate Algebra, 4ed 22 Solving Linear Equations Solving Linear Equations in One Variable.

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Presentation on theme: "§ 2.3 Solving Linear Equations. Martin-Gay, Beginning and Intermediate Algebra, 4ed 22 Solving Linear Equations Solving Linear Equations in One Variable."— Presentation transcript:

1 § 2.3 Solving Linear Equations

2 Martin-Gay, Beginning and Intermediate Algebra, 4ed 22 Solving Linear Equations Solving Linear Equations in One Variable 1)Multiply both sides by the LCD to clear the equation of fractions if they occur. 2)Use the distributive property to remove parentheses if they occur. 3)Simplify each side of the equation by combining all like terms. 4)Get all variable terms on one side and all numbers on the other side by using the addition property of equality. 5)Get the variable alone by using the multiplication property of equality. 6)Check the solution by substituting it into original equation.

3 Martin-Gay, Beginning and Intermediate Algebra, 4ed 33 Simplify. Simplify; divide both sides by 7. Simplify both sides. Solving Linear Equations Example: Solve the equation. Add –3y to both sides. Simplify; add –30 to both sides. Multiply both sides by 5.

4 Martin-Gay, Beginning and Intermediate Algebra, 4ed 44 – 0.01(5a + 4) = 0.04 – 0.01(a + 4) Solving Linear Equations 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. – 4a – 4 = 4 – 4 Add 4 to both sides and simplify. – 4a = 4 Divide both sides by – 4. a = – 1

5 Martin-Gay, Beginning and Intermediate Algebra, 4ed 55 Solving Linear 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. Since this equation will be true for every x that is substituted into the equation, the solution is “all real numbers.” Example: Solve the equation.

6 Martin-Gay, Beginning and Intermediate Algebra, 4ed 66 Solving Linear 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.


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