Chapter 10.9, Problem 37E

To determine

The given system of equation has to be solved and graphed for its solutions, co-ordinates of all vertices and tells that the set of solution is bounded or not.

  $\{\begin{array}{l}x+2y\le 14\\ 3x-y\ge 0\\ x-y\ge 2\end{array}$

Answer to Problem 37E

The set of solution is:

  $\begin{array}{l}\left(x_1,y_1\right)=\left(6,4\right)\\ \left(x_2,y_2\right)=\left(-1,-3\right)\\ \left(x_3,y_3\right)=\left(2,0\right)\end{array}$

  

All solutions, or solution set is not in the shaded plane so, the solution of the given inequalities is unbounded.

Explanation of Solution

Given:

  $\begin{array}{l}x+2y\le 14\\ 3x-y\ge 0\\ x-y\ge 2\end{array}$

Calculations:

The given eqns. are:

  $x+2y\le 14$.......(1)

  $3x-y\ge 0$.......(2)

  $x-y\ge 2$.......(3)

Use the test point $\left(0,0\right)$ to determine whether which region satisfies the inequalities equation:

Substitute $\left(0,0\right)$ in eq. (1), (2) and (3):

For eq. (1):

  $\begin{array}{l}x+2y\le 14\\ 0+0\le 14\\ 0\le 14\end{array}$

It satisfies the given equation. Need to shade the region on or below the solid line which does have the point $\left(0,0\right)$ .

For eq. (2):

  $\begin{array}{l}3x-y\ge 0\\ 0\ge 0\end{array}$ ;

It does not satisfy the given equation. Need to shade the region on or below the solid line which does have the point $\left(0,0\right)$ .

For eq. (3):

  $\begin{array}{l}x-y\ge 2\\ x-y\ge 2\\ 0\cancel{\ge }2\end{array}$ ;

It does not satisfy the given equation. Need to shade the region on or below the solid line which does not have the point $\left(0,0\right)$ .

So, the graph of the given inequality equations is:

  

Now, the determine the vertex:

  $x+2y=14$.......(4)

  $3x-y=0$.......(5)

  $x-y=2$.......(6) The first vertex is the intersection point of two lines shown by taking the subtraction of eq. (4) and (6):

  $\begin{array}{l}x+2y-x+y=14-2\\ 3y=12\\ y=4\end{array}$

Put $y=4$ in eq. (6):

  $\begin{array}{l}x-y=2\\ x=2+4\\ x=6\end{array}$

So, the first vertex point is $\left(x_1,y_1\right)=\left(6,4\right)$ .

The second vertex is the intersection point of two lines shown by subtraction of eq. (5) and (6):

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

Put $x=-1$ in eq. (6):

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

So, the second vertex is $\left(x_2,y_2\right)=\left(-1,-3\right)$ .

The other vertex is the value of the $x$ intercept with the line of equation $x-y=2$ :

  $\begin{array}{l}x-y=2\\ x-0=2\\ x=2\end{array}$

So, $x$ intercept is $\left(2,0\right)$ .

So, the set of solution is:

  $\begin{array}{l}\left(x_1,y_1\right)=\left(6,4\right)\\ \left(x_2,y_2\right)=\left(-1,-3\right)\\ \left(x_3,y_3\right)=\left(2,0\right)\end{array}$

  

All solutions, or solution set is not in the shaded plane so, the solution of the given inequalities is unbounded.

Conclusion:

Hence, the given set of equations is derived as above.