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§ 1.4 Solving Linear Equations

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Blitzer, Intermediate Algebra, 5e – Slide #2 Section 1.4 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.

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Blitzer, Intermediate Algebra, 5e – Slide #3 Section 1.4 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.

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Blitzer, Intermediate Algebra, 5e – Slide #4 Section 1.4 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.

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Blitzer, Intermediate Algebra, 5e – Slide #5 Section 1.4 Solving Linear EquationsEXAMPLE SOLUTION Solve and check: 5 - 3x + 4x = 1 - 7x ) Simplify the algebraic expressions on each side x + 4x = 1 - 7x x = x Combine like terms: -3x + 4x = x = 13

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Blitzer, Intermediate Algebra, 5e – Slide #6 Section 1.4 Solving Linear Equations 2) Collect variable terms on one side and constant terms on the other side. 5 + x + 7x = x + 7xAdd 7x to both sides CONTINUED 5 + 8x = 13Simplify 5 – 5 + 8x = Subtract 5 from both sides 8x = 8 Simplify

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Blitzer, Intermediate Algebra, 5e – Slide #7 Section 1.4 Solving Linear Equations 3) Isolate the variable and solve. CONTINUED 8x 8Divide both sides by x = 1Simplify

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Blitzer, Intermediate Algebra, 5e – Slide #8 Section 1.4 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 – – Multiply= ? – Add or subtract from left to right = ? Add6 = 6

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Blitzer, Intermediate Algebra, 5e – Slide #9 Section 1.4 Solving Linear EquationsEXAMPLE SOLUTION Solve and check: 1) Simplify the algebraic expressions on each side. Multiply both sides by the LCD: 30 Distributive Property

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Blitzer, Intermediate Algebra, 5e – Slide #10 Section 1.4 Solving Linear Equations Cancel CONTINUED Multiply Distribute

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Blitzer, Intermediate Algebra, 5e – Slide #11 Section 1.4 Solving Linear EquationsCONTINUED Combine like terms14x + 2 = 15x 2) Collect variable terms on one side and constant terms on the other side. 2 = x Subtract 14x from both sides 14x – 14x + 2 = 15x – 14x Simplify 3) Isolate the variable and solve. Already done.

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Blitzer, Intermediate Algebra, 5e – Slide #12 Section 1.4 Solving Linear EquationsCONTINUED 4) Check the proposed solution in the original equation. Replace x with 2 Simplify Original Equation Simplify ? ? ? ?

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Blitzer, Intermediate Algebra, 5e – Slide #13 Section 1.4 Solving Linear EquationsCONTINUED = 1 Simplify 1 = 1 Simplify Since the proposed x value of 2 made a true sentence of 1 = 1 when substituted into the original equation, then 2 is indeed a solution of the original equation.

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Blitzer, Intermediate Algebra, 5e – Slide #14 Section 1.4 Categorizing an Equations Type of EquationsDefinitions IdentityAn equation that is true for all real numbers ConditionalAn equation that is not an identity but is true for at least one real number Inconsistent (contradiction) An equation that is not true for any real number

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Blitzer, Intermediate Algebra, 5e – Slide #15 Section 1.4 Categorizing an EquationEXAMPLE SOLUTION Solve and determine whether the equation is an identity, a conditional equation or an inconsistent equation x = 9x x = 9x x = 9x 4x – 4x = 9x – 4x Subtract 5 from both sides Simplify Subtract 4x from both sides

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Blitzer, Intermediate Algebra, 5e – Slide #16 Section 1.4 Categorizing an Equation 0 = x Divide both sides by 5 Simplify The original equation is only true when x = 0. Therefore, it is a conditional equation. CONTINUED 0 = 5x Simplify

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Blitzer, Intermediate Algebra, 5e – Slide #17 Section 1.4 Categorizing an EquationEXAMPLE SOLUTION Solve and determine whether the equation is an identity, a conditional equation or an inconsistent equation. 5 – (2x – 4) = 4(x +1) - 2x 5 – 2x +4 = 4x x 9 - 2x = 4 - 2x Distribute the -1 and the 4 Simplify Since after simplification we see a contradiction, we know that the original equation is inconsistent and can never be true for any x. 9 = 4 Add 2x to both sides.

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Blitzer, Intermediate Algebra, 5e – Slide #18 Section 1.4 Categorizing an EquationEXAMPLE SOLUTION Solve and determine whether the equation is an identity, a conditional equation or an inconsistent equation x = 3(x +1) - x 3 + 2x = 3x x 3 + 2x = 2x + 3 Distribute the 3 Simplify Since after simplification we can see that the left hand side (LHS) is equal to the RHS of the equation, this is an identity and is always true for all x.

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