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Exercises Systems of equations

Example 5 from Algebra and Trigonometry, 11.2 Systems of Linear Equations: Three Variables

Finding the Solution to a Dependent System of Equations

Find the solution to the given system of three equations in three variables.

\begin{array}{r} 2x+y−3z=0 & (1) \\ 4 x + 2 y −6 z = 0 & (2) \\ x - y + z = 0 & (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
First, we can multiply equation (1) by $−2$ and add it to equation (2).
$\begin{array}{l} \underset{____________________________________________}{\begin{array}{l} −4 x −2 y + 6 z = 0 equation (1) multiplied by −2 \\ 4 x + 2 y −6 z = 0 \end{array}} \\ 0 = 0 \end{array}$
We do not need to proceed any further. The result we get is an identity, $0 = 0,$ which tells us that this system has an infinite number of solutions. There are other ways to begin to solve this system, such as multiplying equation (3) by $−2,$ and adding it to equation (1). We then perform the same steps as above and find the same result, $0 = 0.$
When a system is dependent, we can find general expressions for the solutions. Adding equations (1) and (3), we have

$\begin{array}{l} \underset{_____________}{\begin{array}{l} \begin{array}{l} 2 x + y −3 z = 0 \end{array} \\ x-y+z=0 \end{array}} \end{array}$
We then solve the resulting equation for $z .$

$\begin{array}{l} 3 x −2 z = 0 \\ z=\frac{3}{2}x \end{array}$
We back-substitute the expression for $z$ into one of the equations and solve for $y .$

$\begin{array}{l} 2 x + y - 3 (\frac{3}{2} x) = 0 \\ 2 x + y - \frac{9}{2} x = 0 \\ y=\frac{9}{2}x-2x \\ y=\frac{5}{2}x \end{array}$
So the general solution is $(x, \frac{5}{2} x, \frac{3}{2} x) .$ In this solution, $x$ can be any real number. The values of $y$ and $z$ are dependent on the value selected for $x .$

Your turn

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

Solve the following system.

\begin{array}{l} 3 x - 2 y - z = 4 \\ x + 6 y + 5 z = 24 \end{array}

How did it go?

PreviousSolving an Inconsistent System of Three Equations in Three Variables Next Solving a System of Nonlinear Equations Representing a Parabola and a Line

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.