Chapter 10, Problem 36RE

To determine

To find: the solution of the system.

Answer to Problem 36RE

The solution is $\left(t,t-1,2-t\right)$

Explanation of Solution

Given:

  $\{\begin{array}{l}x-y=1\\ x+y+2z=3\\ x-3y-2z=-1\end{array}$

Calculation:

Let's begin by labeling our three equations.

  $\{\begin{array}{l}x-y=1 Equation1\\ x+y+2z=3 Equation\text{ 2}\\ x-3y-2z=-1 Equation\text{ 3}\end{array}$

Use Operation $1$ to eliminate $z-term$ from Equation $\text{3}$ . Add Equation $\text{2}$ to Equation $\text{3}$ .

  $2x-2y=2$

This reduces to

  $x-y=1$

Our new system, with the new Equation $\text{3}$ , is

  $\{\begin{array}{l}x-y=1 Equation1\\ x+y+2z=3 Equation\text{ 2}\\ x-3y-2z=-1 Equation\text{ 3}\end{array}$

By the use of Operation $1$ (stated at the beginning), replace Equation $\text{3}$ with the difference of Equation $\text{3}$ and Equation $1$ . Since Equations $1$ and $\text{3}$ are the same. This difference is simply $0=0.$

  $\{\begin{array}{l}x-y=1 Equation1\\ x+y+2z=3 Equation\text{ 2}\\ 0=0 Equation\text{ 3}\end{array}$

Drop Equation $\text{3}$ from our system because it is true, but provides no extra information about the variables.

  $\{\begin{array}{l}x-y=1 Equation1\\ x+y+2z=3 Equation\text{ 2}\end{array}$

Now there are two equations with three variables. Use them to write $xandz$ in terms of

  $y$ . However, since $y$ can take on any real value, our system have infinitely many solutions.

Let's solve Equation $1$ for $x$ in terms of $y$ .

  $x-y=1$

  $x=y+1$

Now, place this value of $x$ into Equation $2$ .

  $\begin{array}{l}x+y+2z=3\\ \left(y+1\right)+y+2z=3\\ 2y+2z+1=3\\ 2y+2z=2\\ y+z=1\\ z=1-y\end{array}$

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

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

Write this as the ordered triplet

  $\left(t+1,t,1-t\right)$

The solution you find may differ from this if you chose to use $x=t$ rather than $y=t$ .

This method is valid because $x$ and $y$ both appear in both Equations $1and2$ .

If you solved it this way, you get the solution $\left(t,t-1,2-t\right)$ . This solution is also correct.

Conclusion:

Therefore, the solution is $\left(t,t-1,2-t\right)$