Question

Solve the system by elimination.

Answer 1

These are three linear equations in in three variables.


$- 2x + 2y + 3z = 0 - - (1)$

$- 2x - y + z = - 3 - - (2)$
$2x + 3y + 3z = 5 - - (3)$

To solve this by elimination, we are going to add equation (3) to (1) and (2) simultaneously.


First equation (3) + (1)

This implies that

$0+ 5y + 6z = 5$
$5y + 6z = 5 - - (4)$
Next let's add equation (2) and (3)

This implies that

$2y+5z=2--(5)$

Equations (4) and (5) are simultaneous linear equation in two variables.


Equation (4) ×2


This implies that


$10y + 6z = 10 - - -(6)$
Also equation (5) ×5


This implies that

$10y + 25z = 10 - - - (7)$
Now equation (7) - (6) gives

$13z = 0$
$\rightarrow \: z = 0$
Substitute
$z = 0$
in equations (6) or (7) and solve for y.

So using equation (6),
$10y + 6(0) = 10$
$10y + 0 = 10$
$10y = 10$
$\therefore \: y = 1$


Now plug in z=0 and y=1 into (3) or any other containing the three variables and solve for x.


$2x + 3(1) + 3(0) = 5$
$2x + 3 + 0 = 5$
$2x = 2$
$x = 1$

Hence

$x=1, y= 1, z=0$