Question

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

6.$$\begin{gathered} \left|x+2y+3z=8 \\ x+3y+3z=10 \\ x+2y+4z=9\right| \end{gathered}$$

Short answer

The solution of the system of equations is$x=1,y=2,z=1$

Step-by-step solution

Step 1:Transforming the system

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

$$\begin{gathered} \left|x+2y+3z=8......\left(1\right) \\ x+3y+3z=10......\left(2\right) \\ x+2y+4z=9......\left(3\right)\right| \end{gathered}$$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 equation2 and then subtract the first equation from equation 3.

$$\begin{gathered} \begin{array}{rcl}\left|x+2y+3z=8 \\ x-x+3y-2y+3z-3z=10-8 \\ x-x+2y-2y+4z-3z=9-8\right|& =& \left|x+2y+3z=8 \\ y=2 \\ z=1\right|\end{array} \end{gathered}$$

Now put the value of y and z which is in second equation and the third equation to the first equation.

$$\begin{gathered} \left|\begin{array}{rcl}x+2\left(2\right)+3\left(1\right)& =& 8\\ y& =& 2\\ z& =& 1\end{array}\right|=\left|x=8-7 \\ y=2 \\ z=1\right|=\left|x=1 \\ y=2 \\ z=1\right| \end{gathered}$$

Hence, we get value of$x=1,y=2,z=1$

Checking the solution

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

$$\begin{gathered} 1+2\left(2\right)+3\left(1\right)=8 \\ 1+3\left(2\right)+3\left(1\right)=10 \\ 1+2\left(2\right)+4\left(1\right)=9 \end{gathered}$$

Hence, the solution of the system of equation is$x=1,y=2,z=1$