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Math Project

Math Project. By: Jake Covey Nathan Henke Zach Bassuener. Math it’s a beautiful Thing!!!. http://www.youtube.com/watch?v=52CzD31SqaM. Example 1 - solve by clearing fractions. Instructions. Work shown. Find LCD in the equation. LCD is x

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Math Project

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  1. Math Project By: Jake Covey Nathan Henke Zach Bassuener

  2. Math it’s a beautiful Thing!!! http://www.youtube.com/watch?v=52CzD31SqaM

  3. Example 1- solve by clearing fractions Instructions Work shown Find LCD in the equation. LCD is x Take the LCD out of the equation by multiplying the whole equation by it. Move all variables on to one side of the equation Reverse foil the equation to find the zeros. x + 3/x = 4 X(x+3/x = 4) X2 +3= 4x X2 -4x +3= 0 (x-1) or (x-3)

  4. Example 2- Solve a Rational Equation Instructions Work shown Multiply the equation by x-4 (LCD) Distributive property Quadratic formula Simplify the equation Simplify even further

  5. Example 3- Eliminating Extraneous Solutions Instructions Work Shown • To the right is the original equation • Next you are going to multiply the equation by the LCD to get rid of the fractions • Then you are going to use the distributive property • Finally you are going to factor your equation. or x=3

  6. Example 3- Eliminating Extraneous Solutions Cont. Work shown Now replace x by your answers in the equations and see if they work. When plugging the -1/2 into the equation and simplifying you find out that -1/2 works. Making it a valid solution, but when you test 3 you find out that it doesn’t work making it an extraneous solution

  7. Example 4 – Eliminating Extraneous Solutions The LCD of the equation on the right is x(x+2) First you are going to multiply the equation by the LCD to cancel the fraction This will expand the equation Next simplyand Factor You are going to come up with the answers x=0 and x=-2 once you substitute these answers into the original equation you find out that it results in a division by 0 making both answers extraneous solutions. This equation has no solution

  8. Example 5- Calculating Acid Mixture How much pure acid must be added to 50 mL of a 35% acid solution to produce a mixture that is 75% acid? 100% 35% 75% + = 50+x x 50 1x + .35(50) = .75(50+x) This is the set up for the equation. View next slide.

  9. Example 5- Calculating Acid Mixtures Work Shown 1x + .35(50) = .75(50+x) 1x + 17.5= 37.5 + .75x - 17.5 -17.5 1x = 20+.75x -.75x -.75x .25x = 20 x= 80mL

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