1 / 7

Solving Algebraic Equations Using Opposite Operations

Solving Algebraic Equations Using Opposite Operations. First Degree Equations (One Step Equations). Solving Equations. An equation is like a balance scale As long as you do the “same thing” to both sides of the “scale” or equation, you do not change the value of the equation

reece
Télécharger la présentation

Solving Algebraic Equations Using Opposite Operations

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. Solving Algebraic Equations Using Opposite Operations First Degree Equations (One Step Equations)

  2. Solving Equations An equation is like a balance scale As long as you do the “same thing” to both sides of the “scale” or equation, you do not change the value of the equation Therefore we can use opposite operations, to balance out an equation, to solve of unknown variables

  3. Solving Equations by Subtraction Given the following equation, use opposite operations to solve for the unknown variable n + 6 = 14 n + 6 – 6 = 14 -6 n = 8 p + 3.4 = 22 p + 3.4 – 3.4 = 22 – 3.4 p = 18.6 CHECK: p + 3.4 = 22 18.6 + 3.4 = 22 22 = 22

  4. Solving Equations by Addition Given the following equation, use opposite operations to solve for the unknown variable r – 5 = 12 r – 5 + 5 = 12 + 5 r = 17 p – 9 = 0.9 p – 9 + 9 = 0.9 + 9 p = 9.9 CHECK: p – 9 = 0.9 9.9 – 9 = 0.9 0.9= 0.9

  5. Solving Equations by Dividing Given the following equation, use opposite operations to solve for the unknown variable 5n = 25 5n = 25 5 5 n = 5 100 x n = 2000 100 x n ÷ 100 = 2000 ÷ 100 n = 20 CHECK: 100 x n = 2000 100 x 20 = 2000 2000 = 2000

  6. Solving Equations by Multilpying Given the following equation, use opposite operations to solve for the unknown variable n / 7 = 7 n / 7 x 7 = 7 x 7 n = 49 n ÷ 100 = 5 n ÷ 100 x 100 = 5 x 100 n = 500 CHECK: n ÷ 100 = 5 500 ÷ 100 = 5 5 = 5

  7. Homework http://studyjams.scholastic.com/studyjams/jams/math/algebra/add-sub-equations.htm http://studyjams.scholastic.com/studyjams/jams/math/algebra/mult-div-equations.htm

More Related