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.
 Spring '12
 MrVinh

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