Question

In exercises, 1 through 10, find all solutions of the linear systems using elimination.Then check your solutions.

9.$$\begin{gathered} \left|x+2y+3z=1 \\ 3x+2y+z=1 \\ 7x+2y-3z=1\right| \end{gathered}$$

Short answer

The system of equation has infinite number of solutions. One of them is$x=\frac{-1}{4},y=1,z=\frac{-1}{4}$.

Step-by-step solution

Step 1:Transforming the equation. 

To get the solution, we will transform the value of x,y,z.

$$\begin{gathered} \left|x+2y+3z=1......\left(1\right) \\ 3x+2y+z=0......\left(2\right) \\ 7x+2y-3z=0......\left(3\right)\right| \end{gathered}$$Into the form$$\begin{gathered} \left|x=.... \\ y=.... \\ z=... \\ \right| \end{gathered}$$

Eliminating the variables

In the given system of equations, we can eliminate the variables by adding or subtracting the equations.

In this system, we can eliminate the variable x from equation 2 and 3. First of all subtract the 3 times of equation 1 from equation2 and then subtract 7 times of first equation from equation 3.

$$\begin{gathered} \begin{array}{rcl}\left|x+2y+3z=1 \\ 3x-3x+2y-6y+z-9z=-2 \\ 7x-7x+2y-14y-3z-21z=-6\right|& =& \left|x+2y+3z=1 \\ -4y-8z=-2 \\ -12y-24z=-6\right|\end{array} \end{gathered}$$

Now divide the equation 2 by-2 and equation 3 by -6.

$$\begin{gathered} \begin{array}{rcl}& & \left|x+2y+3z=1 \\ 2y+4z=1 \\ 2y+4z=1\right|\end{array} \end{gathered}$$

Now subtract the equation 3 from equation 2.

$$\begin{gathered} \begin{array}{rcl}& & \left|x+2y+3z=1 \\ 0=0 \\ 2y+4z=1\right|\end{array} \end{gathered}$$

Now we are left with only two equations. This means the system of equation has infinite number of solutions.

Choosing the value of variable y.

Let the value of y be 1. Now put this in equation 1 and equation3.

$$\begin{gathered} \begin{array}{rcl}\left|x+2\left(1\right)+3z=1 \\ 2\left(1\right)+4z=1\right|& =& \left|x+3z=-1 \\ z=\frac{-1}{4}\right|=\left|x=\frac{-1}{4} \\ z=\frac{-1}{4}\right|\end{array} \end{gathered}$$

Hence, we get value of $x=\frac{-1}{4},y=1,z=\frac{-1}{4}$. Thus the system has only trivial solution.

Checking the solution

Now check the solution by putting the value of x, y and z in the given system of equation.

$$\begin{gathered} \frac{-1}{4}+2\left(1\right)+3\left(\frac{-1}{4}\right)=1 \\ 3\left(\frac{-1}{4}\right)+2\left(1\right)+\left(\frac{-1}{4}\right)=1 \\ 7\left(\frac{-1}{4}\right)+2\left(1\right)-3\left(\frac{-1}{4}\right)=1 \end{gathered}$$

Hence, the system of equations has infinite number of solutions. One of them is $x=\frac{-1}{4},y=1,z=\frac{-1}{4}$