Chapter 5, Problem 7CT

To determine

To write a system of linear inequalities so that the points $\left(1,2\right)$ and $\left(4,-3\right)$ are solutions of the system, but the point $\left(-2,8\right)$ is not a solution of the system.

Answer to Problem 7CT

  $5x+3y\le 11$

  $10x+6y\le 22$

Explanation of Solution

Given:

Two points $\left(1,2\right)$ and $\left(4,-3\right)$ satisfy the linear inequalities.

Formula used:

Equation of a line passing through two points is given by

  $y-y_1=m\left(x-x_1\right)$ , where $m$ is the slope of the line.

Slope of the line is given by

  $m=\frac{y_2-y_1}{x_2-x_1}$

Calculation:

Since the points $\left(1,2\right)$ and $\left(4,-3\right)$ are the solutions of the linear system of inequalities, they satisfy the inequalities.

The slope of the line $=$

  $m=\frac{y_2-y_1}{x_2-x_1}$

  $\begin{array}{l}=\frac{-3-2}{4-1}\\ =\frac{-5}{3}\end{array}$

The equation of the line passing through the point $\left(1,2\right)$ is

  $y-y_1=m\left(x-x_1\right)$

  $\begin{array}{l}\Rightarrow \left(y-2\right)=\frac{-5}{3}\left(x-1\right)\\ \Rightarrow 3\left(y-2\right)=-5\left(x-1\right)\\ \Rightarrow 3y-6=-5x+5\\ \Rightarrow 5x+3y=5+6\\ \Rightarrow 5x+3y=11\end{array}$

Now, define the inequality such that the point $\left(-2,8\right)$ does not satisfy the inequality.

  $\begin{array}{l}\Rightarrow 5\left(-2\right)+3\left(8\right)\le 11\\ \Rightarrow -10+24\le 11\\ \Rightarrow 14\le 11\end{array}$

Which is a false inequality.

Therefore, the inequality satisfied by the points $\left(1,2\right)$ and $\left(4,-3\right)$ is

  $5x+3y\le 11$

Since, a system of linear inequalities is needed, there should be two inequalities.

There are infinite ways to define another inequality.

The other inequality can be given as $10x+6y\le 22$ .

Conclusion:

Therefore, the system of linear inequalities satisfied by the points $\left(1,2\right)$ and $\left(4,-3\right)$ , but not by the point $\left(-2,8\right)$ is

  $5x+3y\le 11$

  $10x+6y\le 22$