Alg_Complete

2 y 3 2 10 y 5 2 2 y 3 2 2 y

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Unformatted text preview: thesis on the left side. Everything inside the parenthesis needs to be multiplied by the 21 before we simplify. At this point we’ve got a problem that is similar the previous problem and we won’t bother with all the explanation this time. 7 ( m - 2 ) + 21 = ( 2m ) ( 3) 7m - 14 + 21 = 6m 7 m + 7 = 6m m = -7 So, it looks like m = -7 is the solution. Let’s verify it to make sure. -7 - 2 ? 2 ( -7 ) + 1= 3 7 -9 14 ? + 1 =3 7 ? - 3 + 1 =- 2 - 2 = -2 OK So, it is the solution. [Return to Problems] (c) 5 10 - y =2 2y - 6 y - 6y + 9 This one is similar to the previous one except now we’ve got variables in the denominator. So, to get the LCD we’ll first need to completely factor the denominators of each rational expression. © 2007 Paul Dawkins 66 http://tutorial.math.lamar.edu/terms.aspx College Algebra 5 10 - y = 2 ( y - 3) ( y - 3) 2 So, it looks like the LCD is 2 ( y - 3) . Also note that we will need to avoid y = 3 since if we 2 plugged that into the equation we would get division by zero. Now, outside of the y’s in the denominator this problem wo...
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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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