Solve the following system of linear equations: 2x + 2y - z = -7 x − y + 2z = 4 3x + 3y − z = 0

Subtract equation (1) from equation (3):

$$(3x + 3y - z) - (2x + 2y - z) = 0 - (-7)$$ $$3x - 2x + 3y - 2y - z + z = 7$$ $$x + y = 7 \quad (*)$$

Multiply equation (1) by 2:

$$2(2x + 2y - z) = 2(-7)$$ $$4x + 4y - 2z = -14 \quad (1')$$

Add equation (1') to equation (2):

$$(4x + 4y - 2z) + (x - y + 2z) = -14 + 4$$ $$4x + x + 4y - y - 2z + 2z = -10$$ $$5x + 3y = -10 \quad (**)$$

From $(*)$, we can express $x$ in terms of $y$:

$$x = 7 - y$$

Substitute this into $(**)$:

$$5(7 - y) + 3y = -10$$ $$35 - 5y + 3y = -10$$ $$35 - 2y = -10$$ $$-2y = -10 - 35$$ $$-2y = -45$$ $$y = \frac{-45}{-2}$$ $$y = \frac{45}{2}$$

Now substitute the value of $y$ back into $x = 7 - y$:

$$x = 7 - \frac{45}{2}$$ $$x = \frac{14}{2} - \frac{45}{2}$$ $$x = \frac{14 - 45}{2}$$ $$x = \frac{-31}{2}$$

Using equation (1):

$$2x + 2y - z = -7$$ $$2\left(\frac{-31}{2}\right) + 2\left(\frac{45}{2}\right) - z = -7$$ $$-31 + 45 - z = -7$$ $$14 - z = -7$$ $$-z = -7 - 14$$ $$-z = -21$$ $$z = 21$$

The solution is:

$$x = -\frac{31}{2}, \quad y = \frac{45}{2}, \quad z = 21$$