1 / 17

Algebraic Equations: Solving Linear and Literal Equations

Learn how to solve linear and literal equations, simplify equations, and solve for specific variables. Practice problems included.

conaway
Download Presentation

Algebraic Equations: Solving Linear and Literal Equations

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. Algebra 2Warm-up: 8/23/2010 Simplify.

  2. Answers • 4x-9 • 86-80x • 38-16x • 54x+14

  3. Quiz Today operations with signed numbers NO CALCULATOR Main Events: Check: p. 14 #28-42 even, 50 **collect signed Honor Code Lesson Quiz PowerPoint Examples 1.3-1.4 Classwork: Practice Problems

  4. Section 1.3-1.4 • Objective: Solve linear and literal equations for a specific variable. linear literal What’s the difference?

  5. 1.3 Solving Linear Equations x – 6 = 9 + 6 +6 x = 15 Example Solve for x: x – 6 = 9 To “undo” subtraction,use addition. Example Solve 112 = –14n 112 = –14n –14 –14 To “undo” multiplication,use division. –8 = n

  6. Two step equation Example 7y – 2 = 19 undo addition (orsubtraction) first. +2 +2 7y = 21 undo multiplication (or division) last. 7y = 21 7 7 y = 3 Be sure to check your solution!

  7. Sometimes, it is necessary to simplify the equation prior to solving it...Example 2x + 3(x – 5) = –10 Distribute first 2x + 3x – 15 = –10 5x – 15 = –10 + 15 +15 5x = 5 5x = 5 5 5 x = 1

  8. Let’s look at an example that has variables on both sides of the equation...Ex. 5x + 6 = 2x – 3 The strategy to solve this type of problem is to get variables on one side, and numbers on the other side. variable side number side 5x + 6 = 2x – 3 – 6 – 6 5x = 2x – 9 –2x –2x 3x = – 9 x = – 3

  9. Practice 1.3Solve…

  10. 1.4 Literal Equations and Formulas Literal Equation: Equation with more than one variable 3x + 2y = 6 This equation can be solved for either x or y.(We will be told which one to solve for). Literal Equation: Equation with more than one variable 3x + 2y = 6 This equation can be solved for either x or y. (We will be told which one to solve for). Here are other Literal Equations that you may have seen before: P = 2L + 2W Here are other Literal Equations that you may have seen before: Literal Equations and Formulas d=rt P = 2L + 2W You can solve for any of the variables in any of these equations.

  11. The formula for the Perimeter of a Rectangle is P = 2L + 2W Example 1: Solve for W: P = 2L + 2W – 2L –2L P – 2L = 2W P – 2L = 2W2 2 P – 2L = W 2

  12. Remember… Distributing vs. factoring Remember that you must distribute the -2 Now do this process in reverse (factor the GCF) What is the common factor? 10

  13. Example 2: Solve for PA = P + Prt Since P is in two terms, we’re going to need to factor it out (think Greatest Common Factor) in order to isolate it. A = P(1 + rt) Now, isolate P by dividing both sides by(1 + rt) A = P(1 + rt)(1 + rt) (1 + rt) P = A 1 + rt

  14. From Geometry, here’s the formula for theVolume of a cylinder: V = πr2h Example 3: Use the formula above to solve for h. h V = πr2h πr2 πr2 r h = V πr2

  15. Here is the formula to find the area of an ellipse :A = πab b a a b Example 4: Find a if A = 65, b = 4 A = πab πb πb a = 65 a = A a ≈ 5.18 πb (3.14)(4)

  16. Practice 1.4 1. Solve for y: 2. Solve for P: 3. Solve for l:

  17. Homework:p. 22 # 35-60 (x5) p. 30 # 4-7(all) Closure: Think of a literal equation that would require factoring a GCF prior to solving

More Related