Question
Solve the following system of linear equations in three variables:
$$ x + y + z = 5 $$
$$ 2x - y + z = 9 $$
$$ x - 2y + 3z = 16 $$
Solution
First, let's label the given equations to make them easier to reference:
$$ x + y + z = 5 $$ (1)
$$ 2x - y + z = 9 $$ (2)
$$ x - 2y + 3z = 16 $$ (3)
By adding equation (1) and (2), we can eliminate the variable \(y\):
$$ x + y + z = 5 $$
$$ 2x - y + z = 9 $$
(2)
$$ 3x + 2z = 14 $$
(4)
Next, we eliminate \(y\) again by multiplying equation (1) by 2 and adding it to equation (3):
Multiply (1) by 2: $$ 2(x+y+z) = 2(5) \implies 2x + 2y + 2z = 10 $$
$$ 2x + 2y + 2z = 10 $$
$$ x - 2y + 3z = 16 $$
(3)
$$ 3x + 5z = 26 $$
(5)
Now we have a new system with two variables. Subtract equation (4) from equation (5) to solve for \(z\):
$$ 3x + 5z = 26 $$
$$ -(3x + 2z = 14) $$
(4)
$$ 3z = 12 $$
$$ z = 4 $$
Substitute \(z=4\) back into equation (4) to find \(x\):
$$ 3x + 2(4) = 14 $$
$$ 3x + 8 = 14 $$
$$ 3x = 6 $$
$$ x = 2 $$
Finally, substitute \(x=2\) and \(z=4\) into equation (1) to find \(y\):
$$ (2) + y + (4) = 5 $$
$$ y + 6 = 5 $$
$$ y = -1 $$
The final solution is:
$$ x = 2, \quad y = -1, \quad z = 4 $$