The lines whose equations are \( 2 x+y=3 z \) and \( x+y=6 z \) intersect at which point? \( (3 z,-3 z) \) \( (9 z,-15 z) \) \( (-3 z, 9 z) \)

To find the intersection of the lines given by the equations \( 2x + y = 3z \) and \( x + y = 6z \), we can solve these equations simultaneously. First, we can express \( y \) in terms of \( x \) and \( z \) from one of the equations. Let's rearrange the second equation: 1. From \( x + y = 6z \), we get \( y = 6z - x \). Now, substitute \( y \) into the first equation: 2. Replace \( y \) in \( 2x + y = 3z \): \[ 2x + (6z - x) = 3z \] This simplifies to: \[ 2x - x + 6z = 3z \] \[ x + 6z = 3z \] \[ x = 3z - 6z = -3z \] Now substitute \( x = -3z \) back into the expression for \( y \): 3. \( y = 6z - x \): \[ y = 6z - (-3z) = 6z + 3z = 9z \] Thus, the coordinates of the intersection point are: \[ (x, y) = (-3z, 9z) \] Therefore, the lines intersect at the point \( (-3z, 9z) \).