Q. 369 - Intermediate Algebra

Question

Determine whether the ordered triple is a solution to the system.

$\left\{\begin{array}{l}y=\frac{2}{3}x-2\\ x+3y-z=15\\ x-3y+z=-2\end{array}\right.$

(a) $(-6,5,\frac{1}{2})$

(b) $(5,\frac{4}{3},-3)$.

Step-by-Step Solution

Answer

(a) The ordered triple $(-6,5,\frac{1}{2})$ is not a solution to the system of linear equations.

(b) The ordered triple $(5,\frac{4}{3},-3)$ is not a solution to the system of linear equations.

Step 1. Given the information.

The system of equations is,

$\left\{\begin{array}{l}y=\frac{2}{3}x-2..........\left(1\right)\\ x+3y-z=15.....\left(2\right)\\ x-3y+z=-2.....\left(3\right)\end{array}\right.$

Part a. Step 1. Finding whether ( - 6 , 5 , 1 2 ) is a solution of the equation (1).

Substituting $$\begin{gathered} x=-6 \\ y=5 \\ z=\frac{1}{2} \end{gathered}$$ in the equation

$y=\frac{2}{3}x-2$,

$$\begin{gathered} 5=\frac{2}{3}(-6)-2 \\ 5=-4-2 \\ 5\ne -6 \end{gathered}$$

Part a. Step 2. Finding whether ( - 6 , 5 , 1 2 ) is a solution for equation (2).

Substituting $$\begin{gathered} x=-6 \\ y=5 \\ z=\frac{1}{2} \end{gathered}$$ in the equation

$x+3y-z=15$,

$$\begin{gathered} -6+3\left(5\right)-\frac{1}{2}=15 \\ -6+15-\frac{1}{2}=15 \\ 9-\frac{1}{2}=15 \\ \frac{17}{2}\ne 15 \end{gathered}$$

Part a. Step 3. Finding whether ( - 6 , 5 , 1 2 ) is solution of the equation (3).

Substituting $$\begin{gathered} x=-6 \\ y=5 \\ z=\frac{1}{2} \end{gathered}$$ in the equation

$x-3y+z=-2$,

$$\begin{gathered} -6-3\left(5\right)+\frac{1}{2}=-2 \\ -6-15+\frac{1}{2}=-2 \\ -21+\frac{1}{2}=-2 \\ \frac{-41}{2}\ne -2 \end{gathered}$$

The ordered triple $(-6,5,\frac{1}{2})$ is not a solution for the system of linear equations.

Part b. Step 1. Finding whether ( 5 , 4 3 , - 3 ) is a solution of equation (1).

Substituting $$\begin{gathered} x=5 \\ y=\frac{4}{3} \\ z=-3 \end{gathered}$$ in the equation

$y=\frac{2}{3}x-2$,

$$\begin{gathered} \frac{4}{3}=\frac{2}{3}\left(5\right)-2 \\ \frac{4}{3}=\frac{10}{3}-2 \\ \frac{4}{3}=\frac{10-6}{3} \\ \frac{4}{3}=\frac{4}{3} \end{gathered}$$

Part b. Step 2. Finding whether ( 5 , 4 3 , - 3 ) is solution of equation (2).

Substituting $$\begin{gathered} x=5 \\ y=\frac{4}{3} \\ z=-3 \end{gathered}$$ in the equation

$x+3y-z=15$,

$$\begin{gathered} 5+3\left(\frac{4}{3}\right)-(-3)=15 \\ 5+4+3=15 \\ 12\ne 15 \end{gathered}$$

Part b. Step 3. Finding whether ( 5 , 4 3 , - 3 ) is a solution for the equation.

Substituting $$\begin{gathered} x=5 \\ y=\frac{4}{3} \\ z=-3 \end{gathered}$$ in the equation

$x-3y+z=-2$,

$$\begin{gathered} 5-3\left(\frac{4}{3}\right)+(-3)=-2 \\ 5-4-3=-2 \\ -2=-2 \end{gathered}$$

The ordered triple $(5,\frac{4}{3},-3)$ is not a solution of the system of linear equations.