Question 163437
Three solutions to the equation Ax-By-Cz-5=0 are (2,4,-1),(3,6,-2), and(-2,-9,-4). Find C.
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Substitute the first solution: (x,y,z)=(2,4,-1) in

Ax-By-Cz-5=0
A(2)-B(4)-C(-1)-5=0
2A-4B+C-5=0
2A-4B+C=5

Substitute the second solution: (x,y,z)=(3,6,-2), in

Ax-By-Cz-5=0
A(3)-B(6)-C(-2)-5=0
3A-6B+2C-5=0
3A-6B+2C=5

Substitute the third solution: (x,y,z)=(-2,-9,-4), in

Ax-By-Cz-5=0
A(-2)-B(-9)-C(-4)-5=0
-2A+9B+4C-5=0
-2A+9B+4C=5

So you have this system of three equations in three
unknowns:

{{{system(2A-4B+C=5, 3A-6B+2C=5, -2A+9B+4C=5)}}}

Do you know how to solve that system?  If not post
again.  The solution to that system is 

{{{matrix(1,5,   A=19 , "," ,  B=7 , ",", C=-5)}}}

All you were asked for was C, which is -5.

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To check, the equation 

{{{Ax-By-Cz-5=0}}}

becomes

{{{(19)x-(7)y-(-5)z-5=0}}}

{{{19x-7y+5z-5=0}}}

Check to see if (2,4,-1) is a solution to
{{{19x-7y+5z-5=0}}}:

{{{19(2)-7(4)+5(-1)-5=0}}}
{{{38-28-5-5=0}}}
{{{0=0}}}

Check to see if (3,6,-2) is a solution to
{{{19x-7y+5z-5=0}}}:

{{{19(3)-7(6)+5(-2)-5=0}}}
{{{57-42-10-5=0}}}
{{{0=0}}}

Check to see if (-2,-9,-4) is a solution to
{{{19x-7y+5z-5=0}}}:

{{{19(-2)-7(-9)+5(-4)-5=0}}}
{{{-38+63-20-5=0}}}
{{{0=0}}}

So the problem checks.

Edwin</pre>