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How to solve for a variable in a given equation. Lesson Plan. Table of Contents. First Step Applying the First Step Second Step Now You Try ! Part 2: Solving for x in a Quadratic Equation Finding the Zeroes/Roots Solve for x once again Practice Problem Quiz – Question #1 Question #2

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table of contents
Table of Contents
  • First Step
  • Applying the First Step
  • Second Step
  • Now You Try!
  • Part 2: Solving for x in a Quadratic Equation
  • Finding the Zeroes/Roots
  • Solve for x once again
  • Practice Problem
  • Quiz – Question #1
  • Question #2
  • Question #3
  • Question #4
  • Question #5
  • Question #6
  • Question #7
  • Question #8
first step
First Step
  • Lets use the example x + 12 = 4x. In this problem we are solving for the variable x.
  • The first thing we want to do is get all the like terms together by putting all the plain numbers on one side of the equal sign, and put all the numbers with an x attached to them on the other side of the equal sign.
applying the first step
Applying the First Step
  • We will get all the like terms together through adding and subtracting.
  • In our example of x + 12 = 5x we will subtract the x from the left side of the equal sign to cross out the x.
  • Then we subtract the x from the right side of the equal sign as well.
  • Whatever we do to the left side of the equal sign, we do the same operation to the right side.
second step
Second Step
  • We now have 12 = 4x after subtracting x from both sides.
  • The next step is to get x alone in this equation.
  • We do this by dividing the right and left sides of the equation by whatever number is multiplied by x.
  • In our example we would divide both sides by 4 to get x = 3 because 4 divided by 4 equals x and 12 divided by 4 equals 3.
  • x = 3 is our answer!
now you try
Now You Try!

If 6x - 12 = 4x + 6 then solve for x.

  • Click on the correct answer!

a) x = 3 c) x = 9 b) x = -3 d) x = -6/10

correct
Correct!
  • Go to Part 2 of the Lesson

Quiz:

  • Go to Question #2
  • Go to Question #3
  • Go to Question #4
  • Go to Question #5
  • Go to Question #6
  • Go to Question #7
  • Go to Question #8
try again
Try Again!
  • If you need help, go back to the Table of Contents to find what you need to review.
  • Back to You Try!
  • Back to Practice Problem

Quiz:

  • Back to Question #1
  • Back to Question #2
  • Back to Question #3
  • Back to Question #4
  • Back to Question #5
  • Back to Question #6
  • Back to Question #7
  • Back to Question #8
part 2 solving for x in a quadratic equation
Part 2: Solving for x in a quadratic equation
  • This is called finding the “zeroes” or “roots” of the equation. In exact terms, we are finding where the graph any given equation crosses the x-axis.
  • Here we will work with the example: x² + 5x + 6 = 0
finding the zeroes roots
Finding the zeroes/roots
  • First, we need to find 2 numbers that add to the middle term in the equation and also multiply to the third term in the equation.
  • In our example of x² + 5x - 6 = 0 these two numbers are 6 and -1 because 6 + -1 = 5 (the middle term) and 6 x -1 = -6 (the third term).
  • Now we insert the two numbers we found into

(x±) (x±) = 0

solve for x once again
Solve for x once again
  • We just “unfoiled” the equation x² + 5x - 6 = 0 to be (x + 6) (x – 1) = 0
  • Now we simply solve for x treating the contents of both parenthesis as separate equations equaling zero.
  • Therefore, x + 6 = 0 and x – 1 = 0.
  • When we solve for x we get zeroes of x = -6 and x = 1. That’s our answer! Now lets practice!
practice problem
practice problem

If x² + 5x + 6 = 0 then what are the zeroes of this equation?

a) x = 2, x = 3 c) x = -2, x = -3

b) x = 6, x = -1 d) x = -6, x = 1

quiz question 1
Quiz! – Question #1

If 20x + 35 = 2+ 31x then solve for x.

a) x = 3 c) x = 37/51

b) x = 51/37 d) x = 1/3

question 2
Question #2

If 50x + 200 = 300 + 25x then solve for x.

a) x = 20 c) x = 4

b) x = 1/4 d) x = 5

question 3
Question #3

If x² + 3x - 28 = 0 then what are the zeroes of this equation?

a) x = 4, x = 7 c) x = -4, x = 7

b) x = 4, x = -7 d) x = -4, x = -7

question 4
Question #4

If x² - 8x + 12 = 0 then what are the zeroes of this equation?

a) x = 6, x = 2 c) x = -6, x = 2

b) x = 6, x = -2 d) x = -6, x = -2

question 5
Question #5

If 30x + 75 = 125 + 5x then solve for x.

a) x = 8 c) x = 200/35

b) x = 2 d) x = 50/35

question 6
Question #6

If x² + 8x + 40 = 0 then what are the zeroes of this equation?

a) x = -8, x = -5 c) x = 8, x = -1

b) x = 8, x = 5 d) x = -8, x = 1

question 7
Question #7

If 8x + 7 – 4x = 27 – 12x + 5 then solve for x.

a) x = 25/39 c) x = -8/39

b) x = 2 d) x = 25/16

question 8
Question #8

If x² + 8x - 48 = 0 then what are the zeroes of this equation?

a) x = 12, x = 4 c) x = 12, x = -4

b) x = -12, x = -4 d) x = -12, x = 4

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CONGRATULATIONS!!!

YOU FINISHED THE LESSON!!!