Question 1193490
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Ikleyn's was indeed the easiest possible answer.  But you are to 
find TWO SYSTEMS. Ikelyn has only given you one system.  I'll show 
you how to find a second system.

But instead of doing it for you, I'll do one exactly like it but
change the ordered pair.  I'll do this one instead:

</pre>Find two systems of linear equations that have the ordered pair 
(-3,2) as a solution.<pre>
Make up 4 easy numbers at random for coefficients.
I'll arbitrarily pick 4,3,2, and -5 to use for 
coefficients of x and y for the left sides of two 
equations. Put blanks to be filled in on the right 
side of each equation:

{{{system(4x+3y="___",2x-5y="___")}}}

Now, to the side, we determine what two numbers to put in the blanks
on the right, by substituting the given ordered pair's coordinates 
for x and y in the left sides.

left side of first equation: 4x+3y = 4(-3)+3(2) = -12+6 = -6 

So we put a -6 in the first blank:

{{{system(4x+3y=-6,2x-5y="___")}}}

Substitute in the other left side

left side of second equation = 2x-5y = 2(-3)-5(2) = -6-10 = -16 

So we put a -16 in the second blank:

{{{system(4x+3y=-6,2x-5y=-16)}}}

That's the system. Now do your problem the 
same way.  Maybe you like Ikleyn's simple 
solution, but you might want to make up 
another one like this.

Edwin</pre></font>