Solve the following system of linear equations in three variables

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 \(y\):

$$ x + y + z = 5 $$

$$ 2x - y + z = 9 $$

(1)
(2)

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$$ 3x + 2z = 14 $$

(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)

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$$ 3x + 5z = 26 $$

(5)

Now we have a new system with two variables. Subtract equation (4) from equation (5) to solve for \(z\):

$$ 3x + 5z = 26 $$

$$ -(3x + 2z = 14) $$

(5)
(4)

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$$ 3z = 12 $$

$$ z = 4 $$

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:

$$ x = 2, \quad y = -1, \quad z = 4 $$