Q. 164 - Intermediate Algebra

Question

In the following exercises, determine whether the ordered triple is a solution to the system.

$$\begin{gathered} y-10 z=-8 \\ 2 x-y=2 \\ x-5 z=3 \\ \left(a\right) (7,12,2) \\ \left(C\right) (2,2,1) \end{gathered}$$

Step-by-Step Solution

Answer

Part (a) The ordered triple $(7,12,2)$ is not a solution of the system of linear equations.

Part (b) The ordered triple$(2,2,1)$ is not a solution of the system of linear equations.

Part (a) Step 1. Given information

We have been given system of linear equations:

$\begin{array}{c}y-10z=-8\\ 2z-y=2\\ x-5z=3\end{array}$

And coordinates$(7,12,2)$

Part (a) Step 2. Testing

Substitute coordinate in each equation and check if left hand side is equal to right hand side.

We will have to check if

$$\begin{gathered} x=7, \\ y=12, \\ z=2 \end{gathered}$$Satisfy equations.

Consider equation one.

$\begin{array}{r}y-10z=-8\\ 12-10\left(2\right)=-8\\ 12-20=-8\\ -8=-8\end{array}$

This is true.

Now substitute this values in equation two.

$\begin{array}{r}2z-y=2\\ 2\left(2\right)-12=2\\ 4-12=2\\ -8=2\end{array}$

This is not true.

The given coordinate is not solution of given equations.

Part (b) Step 1. Given information

We have been given system of linear equations:

$\begin{array}{c}y-10z=-8\\ 2z-y=2\\ x-5z=3\end{array}$

and coordinates: $(2,2,1)$

Part (b) Step2. Testing

First substitute 

$$\begin{gathered} x=2, \\ y=2, \\ z=1 \end{gathered}$$Into the equation one

$\begin{array}{r}y-10z=-8\\ 2-10\left(1\right)=-8\\ 2-10=-8\\ -8=-8\end{array}$

This is true.

Also substitute the values into the equation two.

$\begin{array}{r}2z-y=2\\ 2\left(1\right)-2=2\\ 2-2=2\\ 0=2\end{array}$

This is not true.

So, the ordered triple$(2,2,1)$ is not a solution of the system of linear equations.