1 / 16

Ch. 6 Equations and Inequalities

Ch. 6 Equations and Inequalities. 6.2 Linear Equations in One Variable and Proportions. Objectives. Solve linear equations . Solve linear equations containing fractions. Solve proportions. Solve problems using proportions.

Download Presentation

Ch. 6 Equations and Inequalities

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Ch. 6Equations and Inequalities 6.2 Linear Equations in One Variable and Proportions

  2. Objectives • Solve linear equations. • Solve linear equations containing fractions. • Solve proportions. • Solve problems using proportions. • Identify equations with no solution or infinitely many solutions.

  3. Linear Equations • linear equation in one variable x is an equation that can be written in the formax+ b = 0, where a and b are real numbers, and a  0. • Solving an equation in x determining all values of x that result in a true statement when substituted into the equation. • Equivalent equationshave the same solution set. 4x + 12 = 0 and x = 3 are equivalent equations.

  4. Addition and Multiplying Properties • The Addition Property of EqualityIf a = b, then a + c = b + c • The Multiplication Property of EqualityIf a = b, then a · c = b · c

  5. Using Properties of Equality to Solve Equations

  6. Solving Linear Equation Solve and check: 2(x – 4) – 5x = 5. Step 1. Simplify the algebraic expression on each side:2(x– 4) – 5x = 52x– 8 – 5x = 5.3x – 8 = 5 Step 2. Collect variable terms on one side and constants on the other side.3x – 8 + 8 = 5 + 8 3x = 3

  7. Solving Linear Equation Step 3. Isolate the variable and solve. 3x = 3 3 3 x =1 Step 4. Check the proposed solution in the original equation by substituting 1 for x. 2(x– 4) – 5x = 5 2(1 – 4) – 5(1) = 5 10 – (5) = 5 5 = 5Because the check results in a true statement, we conclude that the solution set of the given equation is {1}.

  8. Application These graphs indicate that persons with a low sense of humor have higher levels of depression. These graphs can be modeled by the following formulas:

  9. We are interested in the intensity of a negative life event with an average level of depression of 7/2 for the high humor group.

  10. Linear Equation with No Solution Solve: 2x + 6 = 2(x + 4) Solution: 2x + 6 = 2(x + 4) 2x + 6 = 2x + 8 2x + 6 – 2x = 2x + 8 – 2x 6 = 8 The original equation 2x + 6 = 2(x + 4) is equivalent to 6 = 8, which is false for every value of x. The equation has no solution. The solution set is Ø.

  11. Linear Equation with Infinitely Many solutions Solve: 4x + 6 = 6(x + 1) – 2x Solution: 4x + 6 = 6(x + 1) – 2x 4x + 6 = 6x + 6 – 2x 4x + 6 = 4x + 6 The original statement is equivalent to the statement 6 = 6, which is true for every value of x. The solution set is the set of all real numbers, expressed as {x|x is a real number}.

  12. Proportions • The property tax on a house with an assessed value of $480,000 is $5760. Determine the property tax on a house with an assessed value of $600,000, assuming the same tax rate.5760 x ------------ = ------------ 480,000 600,000

  13. 5760 x ------------ = ------------ 480,000 600,000 5760 x ----------- = -------- 48 60 5760 ∙ 60 ------------- = x 48 5 1440 155760 ∙ 60------------ = x 48 12 4 1 7200 = x $7200 = x

  14. Example: Changing Recipe Size A chocolate-chip recipe for five dozen cookies requires ¾ cup of sugar. If you want to make eight dozen cookies, how much sugar is needed? Solution: 3 / 4 (cup) x (cup) ----------------- = -------------- 5 (dozen) 8 (dozen) 8 (3/4) 8 3 6---------- = x --- ∙ --- = --- = x 5 5 4 5

  15. x 60” 240” 960” Find the height x of the tree, when a 60”-man casts a shadow 240” long.

  16. Proportion 60 x ----- = ------- 240 960 60 ∙ 960 ----------- = x x = 240240 x 60” 240” 960”

More Related