Should be a whole number.

Let's call the three equations in order A, B and C. To solve for three variables, we treat one variable as if it were an ordinary number and we use two equations to "solve" for the other two variables. Let's make z our "constant" variable and use A and B to solve for x and y.

2x-y=1-z and x+2y=3+4z; double the first and add to the second: 4x-2y=2-2z plus x+2y=3+4z gives 5x=5+2z so x=(5+2z)/5. Now double the second equation and subtract the first: 2x+4y-2x+y=6+8z-1+z. 5y=5+9z, so y=(5+9z)/5.

We now have y and x in terms of z so we can substitute the values in C:

4(5+2z)/5+3(5+9z)/5-7z=-8. Multiply through by 5: 20+8z+15+27z-35z=-40. But, hey, the z's cancel out and we're left with 35=-40! This is false so the equations are inconsistent. The best we can do is drop C and use the dependencies we found for x and y. To make x and y whole numbers z needs to be a multiple of 5, so let z=5 then x=3 and y=10. These three values will satisfy A and B. We can substitute the values in the left hand side of C:

12+30-35=7, so if C had been 4x+3y-7z=7, we would have had a solution, apparently.

If we write z=5n where n is an integer then x=(5+10n)/5=1+2n, and y=(5+45n)/5=1+9n, then C becomes 4+8n+3+27n-35n=7, because this time the n's cancel out, and revised equation C is not an independent equation, so there are many solutions.

The conclusion is that the original C introduces inconsistency and we either go for a multiple solution by introducing n to provide integer-only solutions, or we create an independent C that gives one unique solution. For example, 4x+3y-10z=-8.

by Top Rated User (1.0m points)