Chapter 15
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| (a1 | ): 5x | +2y | +2z | = | +27 |
| (a2 | ): 11x | +8y | +2z | = | +57 |
| (a3 | ): -3x | +1y | -2z | = | -16 |
A Solution: First subtract equation (a1) from
(a2). This implies yields equation (b2) below.
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(b1 |
:) 5x |
+2y |
+2z |
= |
+27 |
|
(b2 |
:) 6x |
+6y |
= |
+30 |
|
|
(b3 |
:) -3x |
+1y |
-2z |
= |
-16 |
Second, add equation (b1) to equation (b3). This gives
| (c1 | ): 5x | +2y | +2z | = | +27 |
| (c2 | ): 6x | +6y | = | +30 | |
| (c3 | ): 2x | +3y | = | +11 |
The above steps have eliminated z from the last two equations.
Third, from equation (c2) subtract two times equation (c3). This implies
| (d1 | ): 5x | +2y | +2z | = | +27 |
| (d2 | ): 2x | = | +8 | ||
| (d3 | ): 2x | +3y | = | +11 |
Finally, we may change the order of equations. This yields the more suggestive system of equations:
| (d2 | ): 2x | = | +8 | ||
| (d3 | ): 2x | +3y | = | +11 | |
| (d1 | ): 5x | +2y | +2z | = | +27 |
This last step was optional. Now we can do the following.
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Exercise: Solve
| (a1 | :) x | +y | + 3z | = | +10 |
| (a2 | :) x | -y | +2z | = | +5 |
| (a3 | :) 2x | +4y | -5z | = | +9 |
Note that you can and should check your answer (values for x, y and z) satisfy each equation. If one is not satisfied then there is an error somewhere in your work, the solution or the check.
Exercise: Solve
| (a1 | :) x | +y | + 3z | = | +10 |
| (a2 | :) x | -y | +2z | = | +5 |
| (a3 | :) 2x | +4y | +5z | = | +9 |
See the difference a "small" change in the problem makes.
More Chapter Sections:
Three Skills
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Chapters and Appendices
Words Before Symbols:
What is a Variable?
Introduction
Variation between Examples
Variation of Letters
A letter denotes a variable
Cases of Double Variation
Three Notions of a Variable
Constants, Parameters
& Variables
Talking about numbers
Dependent
or Independent
Variable, a Matter of Choice
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