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Module 1 topic 1 notes Module 1 topic 1 notes Presentation Transcript

  • Module 1 ~ Topic 1 Solving Equations
    • Table of Contents
    • Slides 6-14: Solving Linear Equations
    • Slides 15-29 : Practice Questions
    • Audio/Video and Interactive Sites
    • Slides 2: Algebra Cheat Sheet
    • Slides 3: Graphing Calculator Use Guide
    • Slide 5: Video
    • Slide 7: Gizmos
  • Algebra Cheat Sheet
    • You may want to download and print this
    • sheet for reference throughout the course.
  • Special Instructions
    • This module includes graphing calculator work.
    • Refer to the website, TI-83/84 calculator instructions , for resources and instructions on how to use your calculator to obtain the results we do throughout the lessons. I suggest that you bookmark this site if you haven’t done so already.
    • Take careful notes and following along with every example in each lesson. I encourage you to ask me questions and think deeply as you’re studying these concepts!
  • Topic #1: Solving Linear Equations
    • Many real-life phenomena can be described by linear functions. It is important to learn how to solve equations involving such functions to provide us with information about these phenomena.
    Definition: A linear equation in one variable is an equation that can be written in the form ax + b = c , where a, b and c are real numbers , and a ≠ 0 . (The letter x is often used as the variable, but it is not required to be the variable.)
  • Video Break!!!!! Click on this link to watch videos on solving equations.
  • Recall General Rules Order of Operations: PEMDAS Multiplication/Division is done in order, left to right Addition/Subtraction is done in order, left to right Solving Equations First, you must know what you are solving for so you can isolate it. To do that: Take care of any exponents/FOIL or distribution/simplification Get common denominators if necessary Combine like terms on each side of the equal sign Addition/Subtraction across = is done next to isolate the variable Multiplication/Division across =is done last and the variable should now be isolated
  • Gizmos Gizmo: Modeling 2-Step Equations Gizmo: Modeling 1-Step Equations B Gizmo: Modeling 1-Step Equations A Gizmo: Solving 2-Step Equations
  • How to Solve Equations
    • As with any journey, you need to know where you want to
    • end up before you start off, or you will never get there!
    • Figure out what you are solving for.
    • In this case we want to end up with x = #
    Ex: Solve 2) Move everything that is NOT x to the other side. When solving equations, do all addition/subtraction first. Then all multiplication/ division last. Make sure you are where you wanted to end up, x = some number.
  • Example 1
    • Solving an equation in one variable algebraically means to find the value of the
    • variable that makes the mathematical statement true using appropriate algebraic
    • operations.
    Example 1: Solve the equation 5x - 4 = 0 . 5x – 4 = 0 + 4 +4 5x = 4 5x = 4 5 5 x = This means makes 5x – 4 = 0 true.
    • One of the great things about solving equations is that we can always check our
    • solutions! We do this by plugging the value(s) we get for our variable into the
    • original equation and verify that we have a true statement.
    Example 1: Solve the equation 5x - 4 = 0 . 5x – 4 = 0 + 4 +4 5x = 4 5x = 4 5 5 x = Check: Original Equation Now, plug in the value of x that you just found We simplify and get 4-4, which is 0. This is what our original equation stated. This is true.
  • Example 2 Example 1: Solve the equation 5(h – 2) = -4(3-h). 5(h - 2) = -4(3 – h) 5h – 10 = -12 + 4h +10 +10 5h = -2 + 4h 5h = -2 + 4h -4h - 4h h = -2 Check: 5( h - 2) = -4(3 – h ) 5(( -2 ) – 2) = -4(3 – ( -2 )) 5( - 4) = -4 ( 3 + 2) -20 = - 4 (5) -20 = -20 Since both sides of the equation yield the same result, we know that our answer is correct! Solution: We have a couple of choices as to how to solve this equation. You may notice that it is not in the form introduced earlier , but it still a linear equation in one variable, since it can be written in the form described above.   Our goal is to isolate the variable h , so that it appears on one side of the equation, and its value appears on the other. We will need to distribut e on both sides of the equation before we can do that.
  • Example 1 Example 1: Solve the equation Remember: Solving Equations First, you must know what you are solving for so you can isolate it. To do that: Take care of any exponents/FOIL or distribution/simplification Get common denominators if necessary Combine like terms on each side of the equal sign Addition/Subtraction across = is done next to isolate the variable Multiplication/Division across =is done last and the variable should now be isolated
  • Example 1: Solution -45 -45 -10m -10m 140 140
  • Check: We find that both sides of the equation give us the same result when we plug our answer in, which means that we obtained the correct result!
  • Practice Examples Example 2: Example 3: Solve the equation 3p + 2 = 0 . Solve the equation -7m – 1 = 0. Solutions on next slide. Solve these on your own first. Example 4: Solve the equation 14z – 28 = 0.
  • Practice Examples Answers Example 2: Example 3: Solve the equation 3p + 2 = 0 . Solve the equation -7m – 1 = 0. Example 4: Solve the equation 14z – 28 = 0. Check: Check: Check:
  • More Practice Examples Example 5: Example 6: Solve the equation . Solve the equation . Solve these on your own first. Solutions on next slide.
  • More Practice Examples - Answers Example 5: Example 6: Solve the equation . Solve the equation .
  • More Practice Examples Example 7: Example 8: Solve the equation . Solve the equation . Solve these on your own first. Solutions on next slide.
  • Example 7: Example 8: Solve the equation . Solve the equation . More Practice Examples - Answers
  • More Practice Examples Example 9: Example 10: Solve the equation for m . . Solve the equation for . Solve these on your own first. Solutions on next slide.
  • More Practice Examples Example 9: Example 10: Solve the equation for m . . Solve the equation for .
  • More Practice Examples Example 11: Solve the equation for x. Obviously, , so the answer is No Solution No Solution
  • Word Problem Examples Example: The relationship between ºC and ºF can be represented by the equation where F is the number of degrees Fahrenheit, and C is the number of degrees Celsius.
    • Solve the above equation for C.
    • Convert 98 o F into degrees Celsius
  • Solution a): We want to isolate C on one side of the equation. So, we apply the following operations on our original equation. -32 -32 or Now we have an equation that allows us to compute degrees Celsius if we knew degrees Fahrenheit.
  • Solution b): We plug 98 in for F and solve for C . -32 -32 9 9 So, 98ºF is approximately 36.67ºC.
  • Example: When you buy a new car, they say that the value of the car depreciates as soon as you drive it off the lot! Accountants use the following equation to measure depreciation of assets: where … D is the depreciation of the asset per year, C is the initial cost of the asset, S is the salvage value, and L is the asset’s estimated life.   a. What is the salvage value of a machine that cost a company $40,000 initially, has an annual depreciation of $3000, and an estimated life of 10 years? b. Solve the original equation for S , the salvage value, in general.
  • Solution a): We plug 30,000 in for C , 2000 for D , and 10 for L . We then solve for S . So, the salvage value for the machine is $10,000.
  • Solution b): We want to isolate the variable S , treating all of the other letters in the equation as constants. -C -C This equation allows us to calculate the salvage value for any asset, given the initial cost, estimated life, and depreciation value.