Chapter 10.9, Problem 41E

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.

  $\{\begin{array}{l}y>x+1\\ x+2y\ge 12\\ x+1>0\end{array}$

Answer to Problem 41E

The set of solution is:

  $\begin{array}{l}\left(x_1,y_1\right)=\left(\frac{10}{3},\frac{13}{3}\right)\\ \left(x_2,y_2\right)=\left(-1,\frac{13}{2}\right)\\ \left(x_3,y_3\right)=\left(-1,0\right)\end{array}$

  

All solutions, or solution set is in the shaded plane so, the solution of the given inequalities is bounded.

Explanation of Solution

Given:

  $\begin{array}{l}y>x+1\\ x+2y\le 12\\ x+1>0\end{array}$

Calculations:

The given eqns. are:

  $y>x+1$.......(1)

  $x+2y\le 12$.......(2)

  $x+1>0$.......(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) and (2), (3):

For eq. (1):

  $\begin{array}{l}y>x+1\\ 0>0+1\\ 0\cancel{>}1\end{array}$ ;

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

For eq. (2):

  $\begin{array}{l}x+2y\le 12\\ 0+0\le 12\\ 0\le 12\end{array}$ ;

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

For eq. (3):

  $\begin{array}{l}x+1>0\\ 0+1>0\\ 1>0\end{array}$ ;

It satisfies the given equation.

Clearly, need to shade the region on or below the solid line which does have the point $\left(0,0\right)$ .

So, the graph of the given inequality equations is:

  

Now, the determine the vertex:

  $y=x+1$.......(4)

  $x+2y=12$.......(5)

  $x+1=0$.......(6)

Solve first two eqns:

Put $y=x+1$ in eq. (5):

  $\begin{array}{l}x+2y=12\\ x+2\left(x+1\right)=12\\ x+2x+2=12\\ 3x=10\\ x=\frac{10}{3}\end{array}$

Now, $y=x+1=\frac{10}{3}+1=\frac{13}{3}$

So, the first vertex point is $\left(x_1,y_1\right)=\left(\frac{10}{3},\frac{13}{3}\right)$ .

Solve two eqns by taking subtraction of (5) and (6):

  $\begin{array}{l}x+2y-x-1=12-0\\ 2y=13\\ y=\frac{13}{2}\end{array}$

From eq. (6):

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

So, the second vertex point is $\left(x_2,y_2\right)=\left(-1,\frac{13}{2}\right)$ .

Solve (4) and (6):

From eq. (6):

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

Put in eq. (4):

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

So, the third vertex point is $\left(x_3,y_3\right)=\left(-1,0\right)$ .

So, the set of solution is:

  $\begin{array}{l}\left(x_1,y_1\right)=\left(\frac{10}{3},\frac{13}{3}\right)\\ \left(x_2,y_2\right)=\left(-1,\frac{13}{2}\right)\\ \left(x_3,y_3\right)=\left(-1,0\right)\end{array}$

  

All solutions, or solution set is in the shaded plane so, the solution of the given inequalities is bounded.

Conclusion:

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