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Basic Algebra

Basic Algebra. Grade: 9 th Gustavo Miranda LRC 320. Vocabulary. Expressions: Don’t contain an “=” sign Contain variables Values are substituted into variables Variables: Represented as letters, i.e. “x” Equations: Also contain variables Want to solve for specified variable

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Basic Algebra

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  1. Basic Algebra Grade: 9th Gustavo Miranda LRC 320

  2. Vocabulary Expressions: • Don’t contain an “=” sign • Contain variables • Values are substituted into variables Variables: • Represented as letters, i.e. “x” Equations: • Also contain variables • Want to solve for specified variable • Contain an “=” sign

  3. Evaluating Basic Expressions • When evaluating, we want to plug in our given value into the variable and simplify by doing the indicated operation. Basic Expression: Ex: x + 5; when x = 2

  4. Evaluating Advanced Expressions • For advanced expressions, we want to combine like terms so that the expression becomes simpler and easier to evaluate. Advanced Expression: Ex:Want this expression to be simplified to: x + 12 – 10x - 30; -9x – 18; when x = 3 When x = 3

  5. Examples for Evaluating Expressions Ex:Evaluate the expression. x – 5; when x = 10 Solution: x – 5 (plug in given value for x) = 10 – 5 = So, the solution to the expression is 5. Ex: Evaluate the expression. 2x + 5 – 3x + 10; when x = 2 Solution: 2x + 5 – 3x + 10 (Simplify terms with x) = -1x + 5 + 10 (Now simplify constants) = -1x + 15 (Plug in value for x) = -1(2) + 15 (Simplify) = -2 + 15 = 13, solution is 13

  6. Solving Basic Equations • With equations, they are expressions that are equal to a number. • The goal with equations is to solve for a variable that contains a number for its solution. Ex: (Basic equation) x + 3 = 10

  7. Solving Advanced Equations • With advanced expressions we wanted to simplify first; the same process is done with equations with more terms. Ex:Solve for x. Want a simplified equation. x + 3 – 3x = 15 -2x + 3 = 15

  8. Examples for Solving Equations Ex: Solve for x. x + 5 = 10 Solution: x + 5 = 10 (Subtract 5 from both sides of equation) - 5 -5 So, x = 5 Ex: Solve for x. x - 5 - 3x - 1 = 9 Solution: x – 5 – 3x – 1 = 9 (combine like terms) -2x – 6 = 9 (Add 6 to both sides of equation) + 6 +6 -2x = 15 (Now divide -2 to both sides of equation) -2 -2 So, x = -15/2

  9. Practice Exercises Evaluate the following expressions: • 2x + 3; when x = 2 • x – 3; when x = 10 • 3x – 5x +1; when x = 3 Solve the following equations for x. a) x + 5 = 2 • 2x – 3 + x = 6 • 3x – 10 = 5 Solutions: 1) 7 2) 7 3) -5 a) x = -3 b) x = 3 c) x = 5

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