2007 paul dawkins 68

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Unformatted text preview: if it is still there after we finish the simplification work. So, let’s finish the problem. © 2007 Paul Dawkins 67 http://tutorial.math.lamar.edu/terms.aspx College Algebra 2 z 2 - 20 z = 3 z + 9 + 2 z 2 - 14 z - 60 -20 z = -11z - 51 51 = 9 z 51 =z 9 17 =z 3 Notice that the z2 did in fact cancel out. Now, if we did our work correctly z = 17 should be the 3 solution since it is not either of the two values that will give division by zero. Let’s verify this. æ 17 ö 2ç ÷ ? è 3 ø= 3 +2 17 17 +3 - 10 3 3 34 ? 3 = 3 +2 26 13 3 3 34 æ 3 ö ? æ 3 ö ç ÷ = 3ç - ÷ + 2 3 è 26 ø è 13 ø 17 17 = 13 13 OK The checking can be a little messy at times, but it does mean that we KNOW the solution is correct. [Return to Problems] Okay, in the last couple of parts of the previous example we kept going on about watching out for division by zero problems and yet we never did get a solution where that was an issue. So, we should now do a couple of those problems to see how they work. Example 2 Solve each of the following equations. 2 -x...
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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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