Question

In exercises, 1 through 10, findall solutions of the linear systems using elimination.T hen check your solutions.

10.$\begin{array}{rcl}& & \left|\begin{array}{c}x+2y+3z=1\\ 2x+4y+7z=2\\ 3x+7y+11z=8\end{array}\right|\end{array}$

Short answer

The solution of system of equation is $x=-9,y=5,z=0$

Step-by-step solution

Step 1:Transforming the system

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

$$\begin{gathered} \left|x+2y+3z=1......\left(1\right) \\ 2x+4y+7z=2......\left(2\right) \\ 3x+7y+11z=8......\left(3\right)\right| \end{gathered}$$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 2 times of equation 1 from equation 2 and then subtract the 3 times of first equation from equation 3.

$$\begin{gathered} \begin{array}{rcl}\left|x+2y+3z=1 \\ 2x-2x+4y-4y+7z-6z=0 \\ 3x-3x+7y-6y+11z-9z=5\right|& =& \left|x+2y+3z=1 \\ z=0 \\ y+2z=5\right|\end{array} \end{gathered}$$

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

$$\begin{gathered} \begin{array}{rcl}\left|x+2\left(y\right)+3\left(0\right)=1 \\ z=0 \\ y+2\left(0\right)=5\right|& =& \left|x=1-2y \\ z=0 \\ y=5\right|=\left|x=-9 \\ z=0 \\ y=5\right|\end{array} \end{gathered}$$

Hence, we get value of$x=-9,y=5,z=0$

Checking the solution

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

$$\begin{gathered} -9+2\left(5\right)+3\left(0\right)=1 \\ 2(-9)+4\left(5\right)+7\left(0\right)=2 \\ 3(-9)+7\left(5\right)+11\left(0\right)=8 \end{gathered}$$

Which is true.

Hence, the solution of system of equation is $x=-9,y=5,z=0$