Chapter 14, Problem 51CCSR

Solution

To find: the graph of the solution set of the given system of inequalities.

  

The dotted line represents that the points on the line is not considered. The shaded region represents the solution set.

Hence, solution of system of inequalities is:

For $x>3,y\ge \frac{x-6}{3}$ and for $x\le 3,y>5-2x$ .

Given: $\left\{\begin{array}{l}x-3y\le 6\\ 2x+y>5\end{array}\right.$

Concept Used:

Consider the equation form of any inequality kind of values. Then for some random values of x find the value of y plot the points on a graph and join them to find the graph of a linear equation with equality symbol.

For the region of an inequality equations consider any point that don’t lie on the line and check whether they satisfy the equation or not.

For an equation with $\le \text{ or }\ge$ symbol, the points on the line will also be considered. But of an equation with < or > kind, the points that lie on the line are not considered.

Calculation:

System of inequalities are:

  $\begin{array}{l}x-3y\le 6\mathrm{....}(i)\\ 2x+y>5\mathrm{....}(ii)\end{array}$

Consider

  $\begin{array}{l}x-3y=6\mathrm{....}(i)\\ 2x+y=5\mathrm{....}(ii)\end{array}$

Find and plot the values of x and y for $x-3y=6$

as mentioned below:

  $\begin{array}{cccc}x& 0& 6& 3\\ y& -2& 0& -1\end{array}$

Plot these points and join them as shown in the figure 1.

Find and plot the values of x and y for $2x+y=5$

as mentioned below:

  $\begin{array}{cccc}x& 0& 2& 3\\ y& 5& 1& -1\end{array}$

Plot these points and join them as shown in the figure 1.

Other way of finding the intersecting point:

Multiply equation (ii) by 3 then adding the equation (i) and (ii):

  $\begin{array}{c}x-3y+6x+3y=6+15\\ 7x=21\\ x=3\end{array}$

In equation (ii):

  $\begin{array}{c}3-3y=6\\ -3y=6-3\\ -3y=3\\ y=-1\end{array}$

Solution of system: $\left(x,y\right)\text{ is }\left(3,-1\right)$ .

Graphing the points:

  

Figure !.

For equation $x-3y\le 6$ the points that satisfies $x-3y=6$ will also be considered.

For equation $2x+y>5$ the points that satisfies $2x+y=5$ will not be considered.

  $\begin{array}{l}x-3y\le 6\mathrm{....}(i)\\ 2x+y>5\mathrm{....}(ii)\end{array}$

For an equation with $\le \text{ or }\ge$ symbol, the points on the line will also be considered. But of an equation with < or > kind, the points that lie on the line are not considered.

Now check of the point $\left(0,0\right)$ lies in the region or not, by putting $\left(0,0\right)$ in both the equation.

First check for $x-3y\le 6$

:

  $\begin{array}{c}0-3\left(0\right)\le 6\\ 0\le 6\left(\text{True}\right)\end{array}$

So, the region for $x-3y\le 6$

will be:

  

Now check for $2x+y>5$

:

  $\begin{array}{c}2\left(0\right)+0>5\\ 0>5\left(\text{False}\right)\end{array}$

So, the region for $2x+y>5$

will be:

  

Since the inequality is $2x+y>5$ , the equation of the line will not be considered.

The final graph of the solution set will be:

  

The dotted line represents that the points on the line is not considered. The shaded region represents the solution set.

Conclusion:

Solution of system of inequalities is:

For $x>3,y\ge \frac{x-6}{3}$ and for $x\le 3,y>5-2x$ .