Lesson 1: Graphical Solution of System of Linear Inequalities

Video Lesson

Lesson Objective

Dear learner by the end of this lesson, you will be able to:

• describe the definition of a system of linear inequalities.
• draw graph of linear inequalities on a coordinate plane.
• Identify the intersection points of lines as solutions to the system.
• Interpret the graphical representation of different types of solutions

Key Terms

Brainstorming Question

Consider the following system of linear equations:

2x + 3y = 6

4x − y = 5​

Find the values of x and y that satisfy this system.

Solution

To solve the system of equations, we can use the substitution or elimination method. Here, we’ll use the elimination method.

First, let’s manipulate the equations to eliminate one of the variables. We can start with the first equation:

From the first equation, multiply it by 2 to help eliminate xx when we combine with the second equation:

2(2x+3y) = 2(6)  ⟹  4x+6y = 12

Now, we have:

4x+6y=12………(i)

4x−y=5……….(ii)

Next, we’ll subtract equation (ii) from equation (i):

This simplifies to:

7y = 7

Dividing both sides by 7 gives: y =1.

Now, substitute y=1 back into one of the original equations to find x. We’ll use the first equation:

2x+3(1) = 6.

This simplifies to:

2x+3=6

2x=3

x = 3/2.

The solution to the system of equations is x=3/2​ and y=1.

1.1. Linear Inequalities

Definition:

Linear inequalities are the expressions where any two values are compared by the inequality symbols such as, ‘<‘, ‘>’, ‘≤’ or ‘≥’. These values could be numerical or algebraic or a combination of both.

$$\begin{gathered} 3x+5y\le 2 \\ 2x–y\ge 5 \\ x<8\dots \text{etc are Linear inequalities.} \end{gathered}$$

Rules to solve inequalities:

I. Adding or Subtracting the same number to or from both sides of the inequality does not change sign of inequality.

II. Multiplying or dividing both sides of the inequality by the same positive real number does not change the sign of inequality.

III. Multiplying or dividing both sides of the inequality by the same negative real number changes sign of inequality.

Example 1:

$\text{Solve the following}5x+3\le 2x–21$

Solution:

$$\begin{gathered} 5x+3\le 2x–21 \\ \mathrm{\Rightarrow \; 3}x+3\le –21 \\ \mathrm{\Rightarrow \; 3}x\le –24 \\ \mathrm{\Rightarrow }\frac{3x}{3}\le \frac{–24}{3} \\ \text{Therefore,}x\le –8=(–\infty ,–8]\text{is equivalent to the following graphical solution:} \end{gathered}$$

$\text{Fig 1.1. graph of}x\le –8=(–\infty ,–8]$

Example 2:

Solve the following and show their graphical solution.

$\left\{\begin{array}{c}2x-4\le 2\\ x+2\ge 1\end{array}\right.$

Solution:

$$\begin{gathered} \text{From the first inequality}2x\le 2+4\Rightarrow x\le 3. \\ \text{From the second equation}x\ge 1–2\Rightarrow x\ge –1. \end{gathered}$$

Therefore, the solution is [-1, 3]. Graphically, we have

$\text{Fig 1.2. graph of}2x–4\le 2\text{and}x+2\ge 1$

1.2. Linear Inequalities with two Variables and Their Graphical Solution

$$\begin{gathered} \text{For any real number}a,b,c\text{and}dax+by\le 0\text{and}cx+dy\ge bc \\ \text{for}a,c\ne 0\text{are linear inequalities with two variables.} \\ \text{While the given expression:} \\ \left\{\begin{array}{cc}ax+by& \le 0\\ cx+dy& \ge bc\end{array}\right.\text{is the system of linear inequalities with two variables.} \\ \text{Steps to solve Graphically:} \end{gathered}$$

1. Change the inequality sign to an equal sign, and then plot the line.

a) If an inequality is of the form y > mx + c or y < mx + c, then the points on the line y = mx + c are not included in the solution region. So, draw a broken or dotted line.

$$\begin{gathered} \text{b)}\text{if an inequality is of the type}y\ge mx+c\text{or}y\le mx+c,\text{then the points on the line}y=mx+c \\ \text{are also included in the solution region. so draw a solid line.} \end{gathered}$$

2. Test a point created in one half-plane.

a). If the point in the half-plane satisfies the inequality, then the entire half- plane satisfies the inequality.

b). If the point does not satisfy the inequality, then the entire half-plane does not satisfy the inequality. If a point from one half plain fails to satisfy, test a point from the other half- plane.

3. Shade in any half-planes that satisfy the inequality.

4. Take the intersection of regions you draw become a solution or region containing solutions.

5. Find intersection points if any and determine the solution.

Example 3:

1. Determine the solution for the given linear system of inequalities.

$\left\{\begin{array}{cc}2x–3y& \le 5\\ x+2y& \ge 1\end{array}\right.$

Solution:

First draw lines with equation: 2x -3y =5 and x+2y = 1 and identify the region using (0,0) so that (0,0) satisfies the first inequality while it does not satisfy the second inequality.

$$\begin{gathered} x+2y\ge 1. \\ \text{Then shade the region to the opposie side of}(0,0)\text{with respect to the line}x+2y=1 \\ \text{and} \\ \text{Shade the region along the side of}(0,0)\text{with respect to the line}2x–3y=5 \\ \text{From the first equation}2x=3y+5\text{implies}x=\frac{3y+5}{2}\text{and}x+y=1\text{results}\frac{3y+5}{2}+2y=1 \\ \Rightarrow y=–0.43\text{and from the equation}x+2y=1\Rightarrow x=1–2(–0.43)=1.86 \end{gathered}$$

Finally the intersection of the regions become the solution.

$\text{Fig 1.3. Graph of}2x–3y\le 5\text{and}x+2y\ge 1$

$\text{2.Find the graphical solution for the system:}\left\{\begin{array}{cc}x–3y& <2\\ x+y& \ge 1\end{array}\right.$

Solution:

$$\begin{gathered} \text{First draw the solid line}y=1–x\text{and the line}x–3y=2\text{withe a dotted line and use the point}(0,0) \\ \text{to test the alignment of the region withe respect to the point.} \\ \text{substituting}(0,0)\text{to the equation}x+y\ge 1\text{implies}0+0\ge 1\text{is false therefore the graph lies against the point}(0,0). \\ \text{while substituting to the equation}x–3y<2\text{implies}0<2\text{is true therefore the graph lies along the point}(0,0). \\ \text{Second you determine the intersection point of the two lines: from the first equation}x=3y+2 \\ \text{and substitute this in to the second equation}x+y=1 \\ \text{it gives}(3y+2)+y=1\text{implies}y=–0.25.\text{and}x=3(–0.25)+2=1.25. \\ \text{The Graph of the two inequalities intersect at a point}(1.25,–0.25) \\ \text{which can be a turning point where the solution may directs the position in which gradients lie.} \end{gathered}$$

$\text{Fig 1.4. Graph of}x–3y<2\text{and}x+y\ge 1$

Examples 2 and 3 showed you that the system of linear inequalities can be solved by graphical method their solution is a region. the region is containing many coordinates that are taken to be solution of the system in the xy -plane and it is said to be the feasible region.

Example 4:

Determine the feasible region of the following system of linear Inequalities using graphical solution method.

$$\begin{gathered} \text{a)}\left\{\begin{array}{cc}2x+3y& \ge 4\\ 5x–2y& \le 2\\ x& \ge 0\\ y& \ge 0\end{array}\right. \\ \text{b)}\left\{\begin{array}{cc}x–3y& \ge 2\\ x+3y& \le 4\\ x& \ge 0\\ y& \ge 0\end{array}\right. \end{gathered}$$

Solution:

a) First draw the graph of the lines 2x+3y =4 and 5x – 2y = 2, x = 0 and y = 0.

Second determine the region using a point (0,2), so that the region lies along the point with respect to the lines 2x+3y =4 and 5x-2y = 2.

Third, shade the region the shaded region is the feasible region. Find the intersection points of the two lines 2x+3y =4 and 5x-2y =2.

$$\begin{gathered} \text{From the first equation}x=\frac{4–3y}{2}\Rightarrow 5\left(\frac{4–3y}{2}\right)–2y=2\Rightarrow y=0.84 \\ \text{then}x=\frac{4–3\left(0.84\right)}{2}\Rightarrow x=0.74\text{mark} \end{gathered}$$

A = (0.74, 0.84) and the y- intercept of 2x+3y =4 which is marked as B = (0, 1.33). these two points are part of the solution of the system called corner points.

Corner points A(0.74, 0.84) and B (0, 1.33)

$\text{Fig 1. 5. Graph of}2x+3y\ge 4,5x–2y\le 2,x\ge 0,y\ge 0$

Solution

b). Draw the graph of lines x -3y = 2, x +3y = 4, x=0 and y = 0 determine the region using a point out of these lines.

Solve the two equations x-3y = 2 and x+3y =4 to get the corner points A(2,0), B(3,0.33) and C(4,0).

$\text{Fig 1.6. Graph of}x–3y\ge 2,x+3y\le 4,x\ge 0,y\ge 0$

1.3. Further on the System of Linear Inequalities

Definition:

A point of intersection of two or more boundary lines of a solution region is called the vertex or the corner point of the region.

Example 5:

Find the solution of the system of linear inequality.

$$\begin{gathered} \left\{\begin{array}{cc}2x+3y& \le 6\\ x–y& \ge 1\\ x+y& \ge 1 \\ x\; \ge \; 0 \\ y\; \ge \; 0\end{array}\right. \end{gathered}$$

Solution:

Draw the graph of the straight lines 2x +3y = 6, x-y = 1 and x+y = 1. Use a point P(0, 0.5), then check the position of the region.

i. 2(0)+3(0.5) = 1.5 is less than 6….. the region lies along the region.

ii. 0 – 0.5= -0.5 is less than 1……. implies the region lies against the point.

iii. 0+0.5 = 0.5 is less than 1 …… implies the region lies against the point.

iv. find intersection point of the lines x-y=1 and x+y = 1 which is (1,0), intersection point of 2x+3y =6 and x – y = 1 which is (1.8,0.8) on the other hand (3,0) is the intersection point of 2x+3y =6 and y = 0.

Therefore, the feasible region is a region bounded by the points (1,0), (3,0) and (1.8, 0.8). The points (1,0), (3,0) and (1.8,0.8) are called corner. points.

Fig 1.7.

Example 6:

A company sells one product for Birr 8 and another for Birr 12. How many of each product must be sold so that revenues are at least Birr 2,400? Let x represent the number of products sold at Birr 8 and let y represent the number f products sold at Birr 12. Write a linear inequality in terms of x and y and ketch the graph of all possible solutions.

Solution:

Considering the first product is x which is sold at Birr 8 per item. the second product is y sold at Birr 12 per item. total revenue is at least 2400.

$$\begin{gathered} \text{Revenue inequality:}8x+12y\ge 2400.\text{while the products are either zero(no product) or counted} \\ \text{(greater than zero) called the non-negativity of variables with inequality :}x\ge 0,y\ge 0. \\ \text{Therefore the system of the inequality:} \\ \left\{\begin{array}{cc}8x+12y& \ge 2400\\ x& \ge 0\\ y& \ge 0\end{array}\right. \end{gathered}$$

then the graphical solution for the above system is given as below with:

corner points A(0,200) and B(300,0).