2 Answers | Add Yours
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.
We’ve answered 319,633 questions. We can answer yours, too.Ask a question