Question

The equations $x+y+z=6, x+2y+3z=10$ and $x+3z=10$ have

• A. a unique solution
• B. no solution
• C. infinitely many solutions
• D. inconsistent

Answer 1

Answer: (A)

a unique solution

Answer 2

We solve the system of equations: $x+y+z=6$ (1) $x+2y+3z=10$ (2) $x+3z=10$ (3)

From (3), we have $x = 10 - 3z$. Substituting this into (1) and (2):

$(10 - 3z) + y + z = 6 \implies y - 2z = -4 \implies y = 2z - 4$ $(10 - 3z) + 2y + 3z = 10 \implies 10 - 3z + 2(2z - 4) + 3z = 10 \implies 10 - 3z + 4z - 8 + 3z = 10 \implies 4z + 2 = 10 \implies 4z = 8 \implies z = 2$

Now we can find $y$ and $x$: $y = 2(2) - 4 = 0$ $x = 10 - 3(2) = 4$

So the unique solution is $(x, y, z) = (4, 0, 2)$.