# Solve 3x + 2y + z = 10, z + 2y + 3z = 18 and x + y + z = 3 by elimination.

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The set of equations 3x + 2y + z = 10, x + 2y + 3z = 18 and x + y + z = 3 has to be solved by elimination. This involves adding the equations appropriately multiplied by a constant to eliminate all but one variable.

3x + 2y + z = 10 ...(1)

x + 2y + 3z = 18 ...(2)

x + y + z = 3 ...(3)

Looking at the given equations notice that (1) + (2) gives:

3x + x + 2y + 2y + z + 3z = 10 + 18

=> 4x + 4y + 4z = 28

=> x + y + z = 7 ...(4)

But (3) gives x + y + z = 3

It is not possible for both (3) and (4) to hold.

The set of equations has no solution. The three equations cannot hold for any value of x, y and z.

Oops. Sorry educators, the second equation was x+2y+3z=18. I wanted to edit the question but there was no option to do that.