Solve the following system of linear equations in three variables
$$ x + 20 = \frac{3y}{2} + 10 $$
Multiply by 2
$$ 2x + 40 = 3y + 20 $$
$$ 2x - 3y = 20 - 40 $$
$$ 2x - 3y = -20 \quad ...(1) $$
$$ \frac{3y}{2} + 10 = 2z + 5 $$
Multiply by 2
$$ 3y + 20 = 4z + 10 $$
$$ 3y - 4z = 10 - 20 $$
$$ 3y - 4z = -10 \quad ...(2) $$
$$ 2z + 5 = 110 - (y+z) $$
$$ 2z + 5 = 110 - y - z $$
$$ y + 2z + z = 110 - 5 $$
$$ y + 3z = 105 \quad ...(3) $$
$$ (3) \times (3) \Rightarrow $$
$$ 3y + 9z = 315 \quad ...(3) $$
$$ (2) \times (1) \Rightarrow $$
$$ 3y - 4z = -10 \quad ...(2) $$
$$ (-) \quad (+) \quad \quad (+) $$
$$ (3) - (2) \Rightarrow $$
Substitute the value of z = 25 in (2)
$$ 3y - 4(25) = -10 $$
$$ 3y - 100 = -10 $$
$$ 3y = -10 + 100 $$
$$ 3y = 90 $$
Substitute the value of y = 30 in (1)
$$ 2x - 3(30) = -20 $$
$$ 2x - 90 = -20 $$
$$ 2x = -20 + 90 $$
$$ 2x = 70 $$
$$ \therefore \text{The value of } x=35, y=30 \text{ and } z=25 $$