Chapter 10, Problem 31RE

To determine

To find: the solution of the system

Answer to Problem 31RE

Therefore, the solution to the system of equations is $\left(1-4t,-1-t,t\right)$ .

Given:

  $\{\begin{array}{l}x-3y+z=4\\ 4x-y+15z=5\end{array}$

Explanation:

Consider the system of equations

  $\begin{array}{l}x-3y+z=4 Equation1\\ 4x-y+15z=5 Equation\text{ 2}\end{array}$

Recall the Gauss elimination method,

To solve the system, use the system equivalent operations and reduce the system to triangular form. Then use back substitution to solve for the variables.

And operations that yield equivalent system are

1. Add a non-zero multiple of one equation to another
2. Multiply an equation by a non-zero constant
3. Interchange the positions of two equations.

Use these together to solve the system. Change the system to triangular form by eliminating the $x-term$ from the equation $\text{2}$ by adding equation $\text{2}$ to $-4$ times equation $1$ .

The work is shown as follows.

  $\frac{\begin{array}{l}-4x+12y-4z=-16\\ 4x-y+15z=5\end{array}}{11y+11z=-11}$

So, the equivalent system one step closer to triangular form is

  $\begin{array}{l}x-3y+z=4\\ 11y+11z=-11 \end{array}$

The system is now triangular form, but there are only two equations, use them to solve for $x$ and $y$ in terms of the variable $z$ but $z$ can take any value. So the system has infinite number of solutions.

To find the complete solution of $x$ begin by solving for $y$ in terms of $z$ .

From the second equation,

  $\begin{array}{l}y+z=-1\\ y=-1-z \end{array}$

Then solve for $x$ by substituting $y$ and $z$ in first equation.

  $\begin{array}{l}x-3\left(-1-z\right)+z=4\\ x+3+3z+z=4\\ x+4z+3=4\\ x=4-4z-3\\ =1-4z\end{array}$

To describe the complete solution, let $t$ be any real number. The solution is

  $\begin{array}{l}x=1-4t\\ y=-1-t\\ z=t\end{array}$

Therefore, the solution to the system of equations is $\left(1-4t,-1-t,t\right)$ .

Conclusion:

Therefore, the solution to the system of equations is $\left(1-4t,-1-t,t\right)$ .

Explanation of Solution

Given:

  $\{\begin{array}{l}x-3y+z=4\\ 4x-y+15z=5\end{array}$

Consider the system of equations

  $\begin{array}{l}x-3y+z=4 Equation1\\ 4x-y+15z=5 Equation\text{ 2}\end{array}$

Recall the Gauss elimination method,

To solve the system, use the system equivalent operations and reduce the system to triangular form. Then use back substitution to solve for the variables.

And operations that yield equivalent system are

1. Add a non-zero multiple of one equation to another
2. Multiply an equation by a non-zero constant
3. Interchange the positions of two equations.

Use these together to solve the system. Change the system to triangular form by eliminating the $x-term$ from the equation $\text{2}$ by adding equation $\text{2}$ to $-4$ times equation $1$ .

The work is shown as follows.

  $\frac{\begin{array}{l}-4x+12y-4z=-16\\ 4x-y+15z=5\end{array}}{11y+11z=-11}$

So, the equivalent system one step closer to triangular form is

  $\begin{array}{l}x-3y+z=4\\ 11y+11z=-11 \end{array}$

The system is now triangular form, but there are only two equations, use them to solve for $x$ and $y$ in terms of the variable $z$ but $z$ can take any value. So the system has infinite number of solutions.

To find the complete solution of $x$ begin by solving for $y$ in terms of $z$ .

From the second equation,

  $\begin{array}{l}y+z=-1\\ y=-1-z \end{array}$

Then solve for $x$ by substituting $y$ and $z$ in first equation.

  $\begin{array}{l}x-3\left(-1-z\right)+z=4\\ x+3+3z+z=4\\ x+4z+3=4\\ x=4-4z-3\\ =1-4z\end{array}$

To describe the complete solution, let $t$ be any real number. The solution is

  $\begin{array}{l}x=1-4t\\ y=-1-t\\ z=t\end{array}$

Therefore, the solution to the system of equations is $\left(1-4t,-1-t,t\right)$ .

Conclusion:

Therefore, the solution to the system of equations is $\left(1-4t,-1-t,t\right)$ .