<PRE><B>Question 226538</B>
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*[tex \LARGE 1.\ \ \ \ \ \ \ \ \ \ \left\{2x + y - 3z = 5\cr\ \,x + y - 3z = 1\cr\ \,x\ \ \ \, -\,\ z = 4\right]


Solve the third one for *[tex \LARGE x]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ z\ +\ 4]


Use this expression for *[tex \LARGE x] to substitute into the first two equations:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left\{2(z+4) + y - 3z = 5\cr\ \,(z+4) + y - 3z = 1\right]


Collect terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left\{y -\ z = -3\cr y - 2z = -3\right]


You can go either of two ways from here.  Either solve either of the above equations for one of the variables in terms of the other and make another substitution that will result in a single variable equation, or use the elimination method.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ z\ -\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ z\ -\ 3 - 2z\ =\ -3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -z\ =\ 0]


Which can then be substituted into the original third equation to get *[tex \LARGE x\ =\ 4], which can then be substituted, along with *[tex \LARGE z\ =\ 0] into either the first or second original equation to get *[tex \LARGE y\ =\ -3]


To use the elimination method, you would multiply *[tex \LARGE y\ -\ 2z\ =\ -3] by -1 to get *[tex \LARGE -y\ +\ 2z\ =\ 3] which would then be added to *[tex \LARGE y\ -\ z\ =\ -3] term by term, resulting in *[tex \LARGE 0y\ +\ z\ =\ 0], which value could then be used to proceed as above.


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*[tex \LARGE 2.\ \ \ \ \ \ \ \ \ \ \left\{\ \ \ 9x\ -\ 2y\ =\ \ \ 7\cr-18x\ +\ 4y\ =\ -7\right]


Multiply the first equation by 2:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \left\{\ \ 18x\ -\ 4y\ =\ \ 14\cr-18x\ +\ 4y\ =\ -7\right]


Add the equations, term-by-term:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ 0x\ +\ 0y\ =\ 7]


Which leads to the absurd result:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ 0\ =\ 7]


What this means is that the solution set to the given system is the empty set, that is, there are no solutions.  Graphically speaking, these two equations represent a pair of parallel lines.  The given system is said to be &lt;b&gt;&lt;i&gt;inconsistent&lt;/i&gt;&lt;/b&gt;.


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For your third problem, multiply the first equation by 2 and then add the two equations term by term.  The &lt;b&gt;&lt;i&gt;y&lt;/i&gt;&lt;/b&gt; variable will be eliminated leaving you with a single equation in &lt;b&gt;&lt;i&gt;x&lt;/i&gt;&lt;/b&gt; that can be solved by ordinary means.  Once the value of &lt;b&gt;&lt;i&gt;x&lt;/i&gt;&lt;/b&gt; has been determined, substitute that value back into either of the original equations to create a single variable equation in &lt;b&gt;&lt;i&gt;y&lt;/i&gt;&lt;/b&gt; and solve.  Express your solution in terms of an ordered pair (&lt;b&gt;&lt;i&gt;x&lt;/i&gt;&lt;/b&gt;,&lt;b&gt;&lt;i&gt;y&lt;/i&gt;&lt;/b&gt;), because this represents the point of intersection of the graphs of the two given equations.


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The fourth problem also lends itself to the elimination method.  In this case, I would use a multiplier of -1 on either of the equations, setting up for elimination of the &lt;b&gt;&lt;i&gt;y&lt;/i&gt;&lt;/b&gt; variable.  However, you will be every bit as successful if you multiply the first equation by 2, setting up for the elimination of the &lt;b&gt;&lt;i&gt;x&lt;/i&gt;&lt;/b&gt; variable.  Works both ways just as well.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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