Question

select the correct location on the image.
which system of equations has a solution of $(-2, 1, 3)$?
$2x - y + z = -1$
$x - z = -1$
$2y + z = 5$
$2x + y + z = 0$
$y - z = -2$
$2y + 3z = 11$
$x - y - z = -6$
$y + z = 4$
$2y - 3z = 0$
$x + y + z = 2$
$y + 3z = 10$
$2x - 3z = -5$

Explanation:

Step1: Test first system with $x=-2,y=1,z=3$

First equation: $2(-2) - 1 + 3 = -4 -1 +3 = -2
eq -1$

Step2: Test second system with $x=-2,y=1,z=3$

First equation: $2(-2) + 1 + 3 = -4 +1 +3 = 0$
Second equation: $1 - 3 = -2$
Third equation: $2(1) + 3(3) = 2 +9 = 11$

Step3: Verify third system with $x=-2,y=1,z=3$

First equation: $-2 -1 -3 = -6$
Second equation: $1 + 3 = 4$
Third equation: $2(1) - 3(3) = 2 -9 = -7
eq 0$

Step4: Verify fourth system with $x=-2,y=1,z=3$

First equation: $-2 +1 +3 = 2$
Second equation: $1 + 3(3) = 1+9=10$
Third equation: $2(-2) -3(3) = -4 -9 = -13
eq -5$

Answer:

The system of equations:
$2x+y+z=0$
$y-z=-2$
$2y+3z=11$