Question

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

7. $$\begin{gathered} \left|x+2y+3z=1 \\ x+3y+4z=3 \\ x+4y+5z=4\right| \end{gathered}$$

Short answer

The system of equation has no solution.

Step-by-step solution

Step 1:Transforming the system

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

$$\begin{gathered} \left|x+2y+3z=1......\left(1\right) \\ x+3y+4z=3......\left(2\right) \\ x+4y+5z=4......\left(3\right)\right| \end{gathered}$$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 equation 1 from equation 2 and then subtract the first equation from equation 3.

$$\begin{gathered} \begin{array}{rcl}\left|x+2y+3z=1 \\ x-x+3y-2y+4z-3z=3-1 \\ x-x+4y-2y+5z-3z=4-1\right|& =& \left|x+2y+3z=1 \\ y+z=2 \\ 2y+2z=3\right|\end{array} \end{gathered}$$

Now multiply equation 2 by 2 and then subtract equation 2 from equation 3.

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

For whatever the value of the value 0 will never equal to 1. Thus, the given system of equations is inconsistent.

Hence, the given system of the equation has no solution.