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Exercises Systems of equations

Example 4 from Algebra and Trigonometry, 11.2 Systems of Linear Equations: Three Variables

Solving an Inconsistent System of Three Equations in Three Variables

Solve the following system.

\begin{array}{l} x−3y+z=4 & (1) \\ -x+2y−5z=3 & (2) \\ 5 x −13 y + 13 z = 8 & (3) \end{array}

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
Looking at the coefficients of $x,$ we can see that we can eliminate $x$ by adding equation (1) to equation (2).
$\frac{\begin{array}{l} x −3 y + z = 4 (1) \\ - x + 2 y −5 z = 3 (2) \end{array}}{-y−4z=7}$
Next, we multiply equation (1) by $−5$ and add it to equation (3).
$\begin{array}{l} \underset{______________________________________}{\begin{array}{l} - 5 x + 15 y - 5 z = −20 & (1) multiplied by −5 \\ 5x-13y+13z=8 & (3) \end{array}} \\ \begin{array}{l} 2y+8z=−12 & (5) \end{array} \end{array}$
Then, we multiply equation (4) by 2 and add it to equation (5).

$\begin{array}{l} \underset{_______________________________________}{\begin{array}{l} −2 y - 8 z = 14 (4) multiplied by 2 \\ 2 y + 8 z = - 12 (5) \end{array}} \\ 0 = 2 \end{array}$
The final equation $0 = 2$ is a contradiction, so we conclude that the system of equations in inconsistent and, therefore, has no solution.

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

Solve the system of three equations in three variables.

\begin{array}{l} x+y+z=2 \\ y−3z=1 \\ 2 x + y + 5 z = 0 \end{array}

How did it go?

PreviousSolving a Real-World Problem Using a System of Three Equations in Three Variables Next Finding the Solution to a Dependent System of Equations

From Algebra and Trigonometry (1st edition) by OpenStax (Rice University), section 11.2, licensed under CC BY 4.0. Original at openstax.org. We split the solution into steps; the words, numbers and figures are the book's. OpenStax doesn't endorse this site.