First the standard form of a quadratic equation is ax

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Unformatted text preview: 5aR ) = m Finally, divide by the coefficient of b. Recall as well that the “coefficient” is all the stuff that multiplies the b. Doing this gives, b= Example 4 Solve m V + 5aR 111 = + for c. abc Solution First, multiply by the LCD, which is abc for this problem. 1 11 ( abc ) = æ + ö ( abc ) ç ÷ a èb cø bc = ac + ab Next, collect all the c’s on one side (the right will probably be easiest here), factor a c out of the terms and divide by the coefficient. bc - ac = ab c ( b - a ) = ab c= © 2007 Paul Dawkins 83 ab b-a http://tutorial.math.lamar.edu/terms.aspx College Algebra Example 5 Solve y = 4 for x. 5x - 9 Solution First, we’ll need to clear the denominator. To do this we will multiply both sides by 5 x - 9 . We’ll also clear out any parenthesis in the problem after we do the multiplication. y ( 5x - 9 ) = 4 5 xy - 9 y = 4 Now, we want to solve for x so that means that we need to get all terms without a y in them to the other side. So add 9y to both sides and the divide by the coefficient of x. 5 xy = 9 y + 4 9y + 4 x= 5y Example 6 Solve y = 4 - 3x for x. 1 + 8x Soluti...
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This note was uploaded on 06/06/2012 for the course ICT 4 taught by Professor Mrvinh during the Spring '12 term at Hanoi University of Technology.

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