# Confusing System of Equations

• Mar 10th 2009, 06:22 PM
bleacher39
Confusing System of Equations
This is probably the hardest system of equations problem I have ever encountered.

"Consider the following system of equations x + y:
Ax+By=C
Dx+Ey=F

Where the constants A,B,C,D,E and F form an arithmetic sequence. How might you characterize the solutions to the system of equations?"
• Mar 11th 2009, 02:26 AM
HallsofIvy
Quote:

Originally Posted by bleacher39
This is probably the hardest system of equations problem I have ever encountered.

"Consider the following system of equations x + y:
Ax+By=C
Dx+Ey=F

Where the constants A,B,C,D,E and F form an arithmetic sequence. How might you characterize the solutions to the system of equations?"

If " A,B,C,D,E and F form an arithmetic sequence" then they have some "common difference" d. That is, B= A+ d, C= A+ 2d, D= A+ 3d, E= A+ 4d, and F= A+ 5d.

That's a cute problem! The answer is surprising.
• Mar 11th 2009, 09:10 AM
Soroban
Hllo, bleacher39!

Quote:

Consider the following system of equations: . $\begin{array}{c}Ax+By\:=\:C \\ Dx+Ey\:=\:F\end{array}$

. . where the constants $A,B,C,D,E,F$ form an arithmetic sequence.

How might you characterize the solutions to the system of equations?

Let the six coefficients be: . $a, \;a+d, \;a+2d, \;a+3d, \;a+4d, \;a+5d$

The system becomes: . $\begin{array}{ccc}\quad ax \quad+\quad (a+d)y &=& a+2d \\ (a+3d)x + (a+4d)y &=& a+5d \end{array}$

Solve the system (use your favorite method): . $x = -1,\;y = 2$

The solution is constant!