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This question contains no equations to solve, just expressions, so I’ve created three equations:

2x-y+z=a, 

x+y-z=b,

3x-2y+z=c, 

where a, b and c are constants. We can use Cramer’s Rule to find a solution in terms of a, b, c.

The determinant for the coefficients matrix=

| 2 -1  1 |

| 1  1 -1 |

| 3 -2  1 | =

2(1-2)+1(1+3)+1(-2-3)=-2+4-5=-3. This is the common divisor for determining x, y, z.

The constant matrix

( a )

( b )

( c )

Will replace each column in turn for finding x, y, z.

x=

| a -1  1 |

| b  1 -1 | ÷ (-3)

| c -2  1 | =

(a(1-2)+1(b+c)+1(-2b-c))/(-3)=(-a+b+c-2b-c)/(-3)=(-a-b)/(-3)=(a+b)/3.

y=

| 2 a  1 |

| 1 b -1 | ÷ (-3)

| 3 c  1 | =

(2(b+c)-a(1+3)+1(c-3b))/(-3)=(2b+2c-4a+c-3b)/(-3)=(-4a-b+3c)/(-3).

z=

| 2 -1 a |

| 1  1 b | ÷ (-3)

| 3 -2 c | =

(2(c+2b)+1(c-3b)+a(-2-3))/(-3)=(2c+4b+c-3b-5a)/(-3)=(-5a+b+3c)/(-3).

So (x,y,z)=((a+b)/3,(4a+b-3c)/3,(5a-b-3c)/3).

If x, y and z are all integers then we have constraints for a, b, c. For y and z, 3c appears in the numerator and, since 3c is divisible by 3 we can ignore the value of c—any number will do.

5a=3a+2a, so we can replace 5a by 2a for the purposes of division. Similarly we can replace 4a by a.

We now have the fractions (a+b)/3, (a+b)/3, (2a-b)/3. If these are to be integers then a+b and 2a-b must be divisible by 3. We can choose small values for a and b. So (a,b)=(0,3), (3,0), (1,2), (2,1), (3,3) are all possible examples. Let’s pick one: a=2, b=1. We can pick any value for c, so let c=-1.

From these x=1, y=4, z=4.

Another example: a=1, b=2, c=3: x=1, y=-1, z=-2.

 

by Top Rated User (609k points)

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