Chapter 10, Problem 28RE

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To determine

To find: the solution of the system

Answer to Problem 28RE

The solution of the system is $\left(2t,t,2-t\right)$

Given:

  $\{\begin{array}{l}x-y+z=2\\ x+y+3z=6\\ 3x-y+5z=10\end{array}$

Explanation:

Let’s begin by labeling our three equations.

  $\{\begin{array}{l}x-y+z=2 Equation1\\ x+y+3z=6 Equation\text{ 2}\\ 3x-y+5z=10 Equation\text{ 3}\end{array}$

Use Operation $1$ to eliminate the $x-term$ from Equation $\text{2}$ . Subtract Equation $1$ from Equation $\text{2}$ .

  $\begin{array}{l}x+y+3z=6 Equation\text{ 2}\\ -[x-y+z=2] \left(-1\right)\times Equation\text{ 2}\\ 2y+2z=4 Equation\text{ 2+}\left(-1\right)\times Equation\text{ 2}=new Equation2\end{array}$ .

Simplify this by dividing both sides by $\text{2}$ .

  $y+z=2$

Our new system, with the new Equation $2$ , is

  $\{\begin{array}{l}x-y+z=2 Equation1\\ y+z=2 Equation\text{ 2}\\ 3x-y+5z=10 Equation\text{ 3}\end{array}$

Now, combine Equation $1$ and $3$ to eliminate the $\text{x-term}$ from Equation $\text{3}$ .

  $\begin{array}{l}3x-y+5z=10 Equation\text{ 3}\\ -3[x-y+z=2] \left(-3\right)\times Equation\text{ 1}\end{array}$

Simplified, this is

  $\begin{array}{l}3x-y+5z=10 Equation\text{ 3}\\ +[-3x+3y-3z=-6] \left(-3\right)\times Equation\text{ 1}\\ \text{2y+2z=4} Equation3+\left(-3\right)\times Equation\text{ 1}=new Equation3\end{array}$

This also simplifier to

  $y+z=2$

One new system, with the new Equation $\text{2}$ , is

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

By the use of Operation $1$ (stated at the beginning), replace Equation $\text{3}$ with the difference of Equation $\text{3}$ and Equation $\text{2}$ . Since Equations $\text{2}$ and $\text{3}$ are the same, this difference is simply $0=0$ .

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

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

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

Now there are two equations with three variables. Use the equations to write $x$ and $z$ in terms of $y$ . However, since $y$ can take on any real value, our system has infinitely many solutions.

To find the complete solution, let’s solve Equation $\text{2}$ for $z$ in terms of $y$ .

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

Now, place this value of $y$ into Equation $1$ .

  $\begin{array}{l}x-y+z=2\\ x-y+\left(2-y\right)=2\\ x-2y+2=2\\ x-2y=0\\ x=2y\end{array}$

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

  $\begin{array}{l}x=2t\\ y=t\\ z=2-t\end{array}$

The ordered triplet is $\left(2t,t,2-t\right)$

The may differ from this if the value of z is $z=t$ rather than $y=t$ . This method is valid because $y$ and $z$ both appear in both Equations $1$ and $2$ . If it is solved in this way, the solution will be $\left(4-2t,2-t,t\right)$ . This solution is also correct.

Conclusion:

Thus, the solution of the system is $\left(2t,t,2-t\right)$

Explanation of Solution

Given:

  $\{\begin{array}{l}x-y+z=2\\ x+y+3z=6\\ 3x-y+5z=10\end{array}$

Let’s begin by labeling our three equations.

  $\{\begin{array}{l}x-y+z=2 Equation1\\ x+y+3z=6 Equation\text{ 2}\\ 3x-y+5z=10 Equation\text{ 3}\end{array}$

Use Operation $1$ to eliminate the $x-term$ from Equation $\text{2}$ . Subtract Equation $1$ from Equation $\text{2}$ .

  $\begin{array}{l}x+y+3z=6 Equation\text{ 2}\\ -[x-y+z=2] \left(-1\right)\times Equation\text{ 2}\\ 2y+2z=4 Equation\text{ 2+}\left(-1\right)\times Equation\text{ 2}=new Equation2\end{array}$ .

Simplify this by dividing both sides by $\text{2}$ .

  $y+z=2$

Our new system, with the new Equation $2$ , is

  $\{\begin{array}{l}x-y+z=2 Equation1\\ y+z=2 Equation\text{ 2}\\ 3x-y+5z=10 Equation\text{ 3}\end{array}$

Now, combine Equation $1$ and $3$ to eliminate the $\text{x-term}$ from Equation $\text{3}$ .

  $\begin{array}{l}3x-y+5z=10 Equation\text{ 3}\\ -3[x-y+z=2] \left(-3\right)\times Equation\text{ 1}\end{array}$

Simplified, this is

  $\begin{array}{l}3x-y+5z=10 Equation\text{ 3}\\ +[-3x+3y-3z=-6] \left(-3\right)\times Equation\text{ 1}\\ \text{2y+2z=4} Equation3+\left(-3\right)\times Equation\text{ 1}=new Equation3\end{array}$

This also simplifier to

  $y+z=2$

One new system, with the new Equation $\text{2}$ , is

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

By the use of Operation $1$ (stated at the beginning), replace Equation $\text{3}$ with the difference of Equation $\text{3}$ and Equation $\text{2}$ . Since Equations $\text{2}$ and $\text{3}$ are the same, this difference is simply $0=0$ .

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

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

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

Now there are two equations with three variables. Use the equations to write $x$ and $z$ in terms of $y$ . However, since $y$ can take on any real value, our system has infinitely many solutions.

To find the complete solution, let’s solve Equation $\text{2}$ for $z$ in terms of $y$ .

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

Now, place this value of $y$ into Equation $1$ .

  $\begin{array}{l}x-y+z=2\\ x-y+\left(2-y\right)=2\\ x-2y+2=2\\ x-2y=0\\ x=2y\end{array}$

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

  $\begin{array}{l}x=2t\\ y=t\\ z=2-t\end{array}$

The ordered triplet is $\left(2t,t,2-t\right)$

The may differ from this if the value of z is $z=t$ rather than $y=t$ . This method is valid because $y$ and $z$ both appear in both Equations $1$ and $2$ . If it is solved in this way, the solution will be $\left(4-2t,2-t,t\right)$ . This solution is also correct.

Conclusion:

Thus, the solution of the system is $\left(2t,t,2-t\right)$