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§ 2.3

§ 2.3. Solving Linear Equations. Linear Equations. A step-by-step procedure for solving linear equations:. Simplify the algebraic expression on each side. Collect all the variable terms on one side and all the constant terms on the other side. Isolate the variable and solve.

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§ 2.3

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  1. §2.3 Solving Linear Equations

  2. Linear Equations A step-by-step procedure for solving linear equations: • Simplify the algebraic expression on each side. • Collect all the variable terms on one side and all the constant terms on the other side. • Isolate the variable and solve. • Check the proposed solution in the original equation. Blitzer, Introductory Algebra, 5e – Slide #2 Section 2.3

  3. Solving Linear Equations EXAMPLE Solve and check: 7x = 3(x + 8) 7x = 3(x + 8) 7x = 3x + 24 Use the Distributive Property. 7x – 3x = 3x + 24 – 3x Subtract 3x from both sides. 4x = 24 Simplify. x = 6 Solve. 7(6) =3(6 + 8) Check the solution in the original equation. 42 = 3(14) 42 = 42 The true statement 42 = 42 verifies 6 is the solution. Blitzer, Introductory Algebra, 5e – Slide #3 Section 2.3

  4. Linear Equations – An application The formula describes the pressure of sea water, p, in pounds per square foot, at a depth d feet below the surface. At what depth is the pressure 30 pounds per square foot? Substitute the given pressure into the equation for p, p = 35. The equation becomes: Blitzer, Introductory Algebra, 5e – Slide #4 Section 2.3

  5. Linear Equations The depth where the pressure is 30 pounds per square foot is 33 feet. Blitzer, Introductory Algebra, 5e – Slide #5 Section 2.3

  6. Solving Linear Equations Remember the steps: Blitzer, Introductory Algebra, 5e – Slide #6 Section 2.3

  7. Solving a linear equation involving fractions EXAMPLE Solve and check: SOLUTION 1) Simplify the algebraic expressions on each side. Multiply both sides by the LCD: 30 Distributive Property Blitzer, Introductory Algebra, 5e – Slide #7 Section 2.3

  8. Solving a linear equation involving fractions CONTINUED Cancel Multiply Distribute Blitzer, Introductory Algebra, 5e – Slide #8 Section 2.3

  9. Solving a linear equation involving fractions CONTINUED 14x + 2 = 15x Combine like terms 2) Collect variable terms on one side and constant terms on the other side. Subtract 14x from both sides 14x – 14x + 2 = 15x – 14x 2 = x Simplify 3) Isolate the variable and solve. Already done. Blitzer, Introductory Algebra, 5e – Slide #9 Section 2.3

  10. Solving a linear equation involving fractions CONTINUED 4) Check the proposed solution in the original equation. ? Original Equation ? Replace x with 2 ? Simplify ? Simplify Blitzer, Introductory Algebra, 5e – Slide #10 Section 2.3

  11. Solving a linear equation involving fractions CONTINUED Simplify 1 - 0 = 1 Simplify 1 = 1 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. Blitzer, Introductory Algebra, 5e – Slide #11 Section 2.3

  12. Linear equations having no solutions Some equations are not true for any real number. Such equations are called inconsistent equations or contradictions. Example: Solve: 2x = 2(x +3) 2x = 2x +6 Distributive Property. 2x – 2x = 2x + 6 – 2xSubtract 2x from both sides. 0 = 6 Simplify. False, since 0 ≠ 6. Inconsistent, no solution. Blitzer, Introductory Algebra, 5e – Slide #12 Section 2.3

  13. Linear Equations having many solutions An equation that is true for all real numbers is called an identity. Example: Solve: 5x + 10 =5(x + 2) 5x + 10 = 5x + 10 Distributive Property. 5x + 10 – 5x = 5x + 10 – 5x Subtract 5x from both sides. 10 = 10 Simplify. Always true no matter what x is. This is an Identity, that is, all real numbers are solutions. Blitzer, Introductory Algebra, 5e – Slide #13 Section 2.3

  14. Categorizing an Equations Blitzer, Introductory Algebra, 5e – Slide #14 Section 2.3

  15. Categorizing an Equation EXAMPLE Solve and determine whether the equation is an identity, a conditional equation or an inconsistent equation. 5 + 4x = 9x + 5 SOLUTION 5 + 4x = 9x + 5 5 - 5 + 4x = 9x + 5 - 5 Subtract 5 from both sides 4x = 9x Simplify 4x – 4x = 9x – 4x Subtract 4x from both sides Blitzer, Introductory Algebra, 5e – Slide #15 Section 2.3

  16. Categorizing an Equation CONTINUED 0 = 5x Simplify Divide both sides by 5 Simplify 0 = x The original equation is only true when x = 0. Therefore, it is a conditional equation. Blitzer, Introductory Algebra, 5e – Slide #16 Section 2.3

  17. Categorizing an Equation EXAMPLE Solve and determine whether the equation is an identity, a conditional equation or an inconsistent equation. 5 – (2x – 4) = 4(x +1) - 2x SOLUTION 5 – (2x – 4) = 4(x +1) - 2x Distribute the -1 and the 4 5 – 2x +4 = 4x + 4 -2x Simplify 9 - 2x = 4 - 2x Add 2x to both sides. 9 = 4 Since after simplification we see a contradiction, we know that the original equation is inconsistent and can never be true for any x. Blitzer, Introductory Algebra, 5e – Slide #17 Section 2.3

  18. Categorizing an Equation EXAMPLE Solve and determine whether the equation is an identity, a conditional equation or an inconsistent equation. 3 + 2x = 3(x +1) - x SOLUTION 3 + 2x = 3(x +1) - x 3 + 2x = 3x + 3 - x Distribute the 3 3 + 2x = 2x + 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. Blitzer, Introductory Algebra, 5e – Slide #18 Section 2.3

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