Question 732040
<pre>
This is an inconsistent system and has no solutions.

3r -  4s =  1 
9r - 12s = -3

Let's try doing it first by substitution.

Solve the first equation for r

3r - 4s = 1
     3r = 1 + 4s
      r = {{{(1+4s)/3}}}

Substitute in

9r - 12s = -3
9{{{((1+4s)/3)}}} - 12s = -3 

3(1+4s) - 12s = -3
3 + 12s - 12s = -3
       3 + 0s = -3
           0s = -6

There is no value of s such that when multiplied by 0
will give -6.  Thus this system has no solution.

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Now let's try doing it by elimination:

3r -  4s =  1 
9r - 12s = -3

Multiply the first equation through by -3

-9s + 12s = -3
 9s - 12s = -3

Add the two equations term by term:

-9s + 12s = -3
 9s - 12s = -3
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 0s +  0s = -6

No matter what values we substitute for r and s,
the left side will be 0 and the right side -6.  So
there can be no solutions.

The system is inconsistent.

Here are the graphs of the two equations:

{{{graph(400,400,-10,10,-10,10,(1-3x)/(-4),(-3-9x)/(-12))}}}

They run parallel and therefore do not intersect and do not have any common
solution.

Edwin</pre>