Multiply the equation y-3z= 10 by 4 and subtract 4y-z=7 to get-1 1z = 33 and z =-3.

Substituting z=-3 into the equation y-3z= 10 requires y= 1.

X=2y, so x=2.

Therefore, the solutions of ternary linear equations are x = 2, y = 1 and z =-3.

(2) Solution: y=4+2z is obtained from the equation y-2z=4.

Substitute y=4+2z into equations {2x+y+2z= 1, 3x+y=9} and {2x+4z =-3 (1), 3x+2z = 5 (2)}.

Multiply formula (2) by 2 and subtract formula (1) to get 4x = 13 and x = 13/4.

Substitute x= 13/4 into 3x+y=9 to get y =-3/4 and z =- 19/8.

So the solutions of ternary linear equations are x = 13/4, y =-3/4, z =- 19/8.

Solution 4: Substitute 3x-5y=8 into 3x-5y+2z=4 to get 2z =-4 and z =-2.

Substitute z=-2 into the equation 2x+3y-z= 1 to get 2x+3y=- 1.

Multiply the equation 3x-5y=8 by 2, subtract the equation 2x+3y=- 1 and multiply it by 3 to get-19y = 19. ，Y =- 1。 X = 1。

So the solutions of ternary linear equations are x = 1, y =- 1, and z =-2.

Solution 5: Let A be X, B be Y and C be Z. According to the meaning of the question, we can get the equations X+Y+Z = 26, X-Y = 1, 2x+Z-Y = 18.

X=y+ 1 from x-y= 1.

Substituting x=y+ 1 into the equation x+y+z=26, 2x+z-y= 18, we get {2y+z = 25 (1). y+z = 16 (2)}，(。

So x = 10 and z = 7, so the numbers of a, b and c are 10, 9 and 7 respectively.

6 solution: x+y= 16, x =16-y.

Substitute x= 16-y into z+x= 10 to get z-y=-6.

Add z-y=-6 and y+z= 12 to get 2z=6, z=3, y = 9 and x = 7.

So the solutions of ternary linear equations are x=7, y=9 and z=3.

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