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Exercises Systems of equations

Example 5 from Algebra and Trigonometry, 11.8 Solving Systems with Cramer's Rule

Using Cramer’s Rule to Solve an Inconsistent System

Solve the system of equations using Cramer’s Rule.

\begin{array}{l} 3 x - 2 y = 4 (1) \\ 6 x - 4 y = 0 (2) \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.

1
Solution
We begin by finding the determinants $D, D_{x}, and D_{y} .$

$D = | \begin{matrix} 3 & - 2 \\ 6 & - 4 \end{matrix} | = 3 (- 4) - 6 (- 2) = 0$

We know that a determinant of zero means that either the system has no solution or it has an infinite number of solutions. To see which one, we use the process of elimination. Our goal is to eliminate one of the variables.
1. Multiply equation (1) by $−2.$
2. Add the result to equation $(2) .$
$\begin{array}{l} \underset{_______________}{\begin{array}{l} \begin{array}{l} - 6 x + 4 y = −8 \end{array} \\ 6 x - 4 y = 0 \end{array}} \\ 0 = −8 \end{array}$
We obtain the equation $0 = −8,$ which is false. Therefore, the system has no solution. Graphing the system reveals two parallel lines. See Figure 1.

Figure 1

How did it go?

PreviousSolving a 3 × 3 System Using Cramer’s Rule Next Use Cramer’s Rule to Solve a Dependent System

From Algebra and Trigonometry (1st edition) by OpenStax (Rice University), section 11.8, 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.