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Solving Equations And Formulas
 

Solving Equations And Formulas

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solving multivariable equations for a specific variable (algebra 1)

solving multivariable equations for a specific variable (algebra 1)

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    Solving Equations And Formulas Solving Equations And Formulas Presentation Transcript

    • Solving Equations and Formulas Chapter 3 section 8
    • Solving for a specific Variable
      • Sometimes equations involve multiple variables, or letters that stand for specific things, such as scientific formulas and physics equations
    • Geometry Formulas:
      • The formula for Circumference of a circle is:
      • C =2 πr (2 x pi x radius)
      • We can solve the formula for r: (this means get r by itself!)
    • What do you think we could do?
      • If the equation is set up as C = 2 π r, and you are asked to get r by itself, then you need to first ask yourself: What is attached to r? How can you “undo” the operation?
      • Since 2 π is “stuck” to r by multiplication, you could get rid of it by division……
    • Like this!
      • C = 2 π r
      • Divide both sides by 2 π: 2 π 2 π
      • This leaves you with the following:
      • C = r
      • 2 π
      • That’s it! That is all there is to it!
      • Now r is by itself, and all of the other “stuff” is on the other side!
    • Here’s another one:
      • Try solving the following formula for x:
      • 5x + y = x
    • Follow these steps to get x:
      • 5x + y = x
      • Move your x’s together, to the same side: Which one should you move? The 5x or the x???
      • 5x + y = x
      • - 5x -5x
      • (you should move the 5x because it needs to be with the x on the other side, and it needs to be away from the y!
    • Steps 2 & 3:
      • Now your equation should look like this!
      • y = -4x
      • 2. What would be the next thing you should do to get the x by itself?
      • (If you said DIVIDE BY – 4, THEN YOU’RE RIGHT!)
      • y = -4x
      • -4 -4
      • 3. Your answer will be y = x
      • -4
    • How ‘bout another one?!?
      • Solve 6 – ay = 4( a – b ) for a
    • Just follow the steps you have been using in other types of equations!
      • 6 – ay = 4( a – b )
      • Distributive property:
      • 6 – ay = 4 a – 4 b
      • Get the things with an “a” together, and the things that don’t have an “a” together:
      • 6 – ay = 4 a – 4 b
      • - 4a - 4a
      • Simplify to get:
      • 6 – ay – 4a = - 4b
    • Big Finish:
      • 6 – ay – 4a = - 4b
      • -6 -6
      • Simplify to get: - ay – 4a = - 4b – 6
      • Now, here’s the tricky part: You have two things on the left that have a’s and you can’t combine them because they aren’t like terms!
      • You have to do something called Factoring: it goes like this: Ask yourself what a is being multiplied by in the two terms on the left: there is an a with – y, and an a with – 4, so you group them in parentheses:
    • Factor:
      • - ay – 4a = - 4b – 6
      • Now looks like this:
      • a(- y – 4) = - 4b – 6
      • (Its like the distributive property, backwards!)
      • To get the a by itself, divide both sides by the stuff in the parentheses:
      • a(- y – 4) = - 4b – 6
      • (- y – 4) (- y – 4)
      • Your answer looks REALLY WEIRD!, but that’s OK!
    • Your answer should look like this:
      • a= - 4b – 6
      • - y – 4
      • (you don’t have to keep the () on bottom any more)
      • How’s that
      • for fun!?!
    • Try one more : (if your brain is not already fried!)
      • The perimeter of a square field is given by the equation P = 2 l + 2 w , where P represents the
      • perimeter, l represents the length of the field, and w represents the width of the field.
      • Solve the formula for l .
    • What should you do first?
    • Answer:
      • P = 2 l + 2 w
      • - 2w - 2w
      • P - 2w = 2l
      • 2 2
      • P - 2w = l
      • 2
    • OMG!!!!
      • Yes, these are hard!
      • Yes, you have to do them!
      • Yes, you have an assignment!
      • Book page 168 14 – 32 even
      • (10 problems)
    • You’ll thank me one day…….
    •