problem solving involving systems of linear inequalities in two variables

5.10 Systems of Linear Inequalities in Two Variables

Learning objectives.

After completing this section, you should be able to:

• Demonstrate whether an ordered pair is a solution to a system of linear inequalities.
• Solve systems of linear inequalities using graphical methods.
• Graph systems of linear inequalities.
• Interpret and solve applications of linear inequalities.

In this section, we will learn how to solve systems of linear inequalities in two variables. In Systems of Linear Equations in Two Variables , we learned how to solve for systems of linear equations in two variables and found a solution that would work in both equations. We can solve systems of inequalities by graphing each inequality (as discussed in Graphing Linear Equations and Inequalities ) and putting these on the same coordinate system. The double-shaded part will be our solution to the system. There are many real-life examples for solving systems of linear inequalities.

Consider Ming who has two jobs to help her pay for college. She works at a local coffee shop for $7.50 per hour and at a research lab on campus for $12 per hour. Due to her busy class schedule, she cannot work more than 15 hours per week. If she needs to make at least $150 per week, can she work seven hours at the coffee shop and eight hours in the lab?

Determining If an Ordered Pair Is a Solution of a System of Linear Inequalities

The definition of a system of linear inequalities is similar to the definition of a system of linear equations. A system of linear inequalities looks like a system of linear equations, but it has inequalities instead of equations. A system of two linear inequalities is shown here.

To solve a system of linear inequalities, we will find values of the variables that are solutions to both inequalities. We solve the system by using the graphs of each inequality and show the solution as a graph. We will find the region on the plane that contains all ordered pairs ( x , y ) ( x , y ) that make both inequalities true. The solution of a system of linear inequalities is shown as a shaded region in the xy xy -coordinate system that includes all the points whose ordered pairs make the inequalities true.

To determine if an ordered pair is a solution to a system of two inequalities, substitute the values of the variables into each inequality. If the ordered pair makes both inequalities true, it is a solution to the system.

Example 5.89

Determining whether an ordered pair is a solution to a system.

Determine whether the ordered pair is a solution to the system:

• ( − 2 , 4 ) ( − 2 , 4 )
• ( 3 , 1 ) ( 3 , 1 )

Is the ordered pair ( − 2 , 4 ) ( − 2 , 4 ) a solution?

We substitute x = − 2 x = − 2 and y = 4 y = 4 into both inequalities.

The ordered pair ( − 2 , 4 ) ( − 2 , 4 ) made both inequalities true. Therefore ( − 2 , 4 ) ( − 2 , 4 ) is a solution to this system.

Is the ordered pair ( 3 , 1 ) ( 3 , 1 ) a solution?

We substitute x = 3 x = 3 and y = 1 y = 1 into both inequalities.

The ordered pair ( 3 , 1 ) ( 3 , 1 ) made one inequality true, but the other one false. Therefore ( 3 , 1 ) ( 3 , 1 ) is not a solution to this system.

Your Turn 5.89

Solving systems of linear inequalities using graphical methods.

The solution to a single linear inequality was the region on one side of the boundary line that contains all the points that make the inequality true. The solution to a system of two linear inequalities is a region that contains the solutions to both inequalities. We will review graphs of linear inequalities and solve the linear inequality from its graph.

Example 5.90

Solving a system of linear inequalities by graphing.

Use Figure 5.88 to solve the system of linear inequalities:

To solve the system of linear inequalities we look at the graph and find the region that satisfies BOTH inequalities. To do this we pick a test point and check. Let's us pick ( − 1 , − 1 ) ( − 1 , − 1 ) .

Is ( - 1 , - 1 ) ( - 1 , - 1 ) a solution to y ≥ 2 x − 1 ? y ≥ 2 x − 1 ?

Is ( - 1 , - 1 ) ( - 1 , - 1 ) a solution to y < x + 1 ? y < x + 1 ?

The region containing ( − 1 , − 1 ) ( − 1 , − 1 ) is the solution to the system of linear inequalities. Notice that the solution is all the points in the area shaded twice, which appears as the darkest shaded region.

Your Turn 5.90

Graphing systems of linear inequalities.

We learned that the solution to a system of two linear inequalities is a region that contains the solutions to both inequalities. To find this region by graphing, we will graph each inequality separately and then locate the region where they are both true. The solution is always shown as a graph.

Step 1: Graph the first inequality.

Graph the boundary line.

Shade in the side of the boundary line where the inequality is true.

Step 2: On the same grid, graph the second inequality.

Shade in the side of that boundary line where the inequality is true.

Step 3: The solution is the region where the shading overlaps.

Step 4: Check by choosing a test point.

Example 5.91

Solve the system by graphing:

Graph x − y > 3 x − y > 3 by graphing x − y = 3 x − y = 3 and testing a point ( Figure 5.89 ). The intercepts are x = 3 x = 3 and y = − 3 y = − 3 and the boundary line will be dashed. Test ( 0 , 0 ) ( 0 , 0 ) which makes the inequality false so shade the side that does not contain ( 0 , 0 ) ( 0 , 0 ) .

Graph y < − 1 5 x + 4 y < − 1 5 x + 4 by graphing y = − 1 5 x + 4 y = − 1 5 x + 4 using the slope m = − 1 5 m = − 1 5 and y y -intercept b = 4 b = 4 ( Figure 5.90 ). The boundary line will be dashed. Test ( 0 , 0 ) ( 0 , 0 ) which makes the inequality true, so shade the side that contains ( 0 , 0 ) ( 0 , 0 ) .

Choose a test point in the solution and verify that it is a solution to both inequalities. The point of intersection of the two lines is not included as both boundary lines were dashed. The solution is the area shaded twice—which appears as the darkest shaded region.

Your Turn 5.91

Example 5.92, graphing a system of linear inequalities.

Graph x − 2 y < 5 x − 2 y < 5 by graphing x − 2 y = 5 x − 2 y = 5 ( Figure 5.91 ) and testing a point. The intercepts are x = 5 x = 5 and y = − 2.5 y = − 2.5 and the boundary line will be dashed. Test ( 0 , 0 ) ( 0 , 0 ) , which makes the inequality true, so shade the side that contains ( 0 , 0 ) ( 0 , 0 ) .

Graph y > − 4 y > − 4 by graphing y = − 4 y = − 4 and recognizing that it is a horizontal line through y = − 4 y = − 4 ( Figure 5.92 ). The boundary line will be dashed. Test ( 0 , 0 ) ( 0 , 0 ) , which makes the inequality true so shade the side that contains ( 0 , 0 ) ( 0 , 0 ) .

The point ( 0 , 0 ) ( 0 , 0 ) is in the solution, and we have already found it to be a solution of each inequality. The point of intersection of the two lines is not included as both boundary lines were dashed. The solution is the area shaded twice, which appears as the darkest shaded region.

Your Turn 5.92

Systems of linear inequalities where the boundary lines are parallel might have no solution. We will see this in the next example.

Example 5.93

Graphing parallel boundary lines with no solution.

Graph 4 x + 3 y ≥ 12 4 x + 3 y ≥ 12 , by graphing 4 x + 3 y = 12 4 x + 3 y = 12 ( Figure 5.93 ) and testing a point. The intercepts are x = 3 x = 3 and y = 4 y = 4 and the boundary line will be solid. Test ( 0 , 0 ) ( 0 , 0 ) , which makes the inequality false, so shade the side that does not contain ( 0 , 0 ) ( 0 , 0 ) .

Graph y < − 4 3 x + 1 y < − 4 3 x + 1 by graphing y = − 4 3 x + 1 y = − 4 3 x + 1 using the slope m = − 4 3 m = − 4 3 and y y -intercept b = 1 b = 1 ( Figure 5.94 ). The boundary line will be dashed. Test ( 0 , 0 ) ( 0 , 0 ) , which makes the inequality true, so shade the side that contains ( 0 , 0 ) ( 0 , 0 ) .

No shared point exists in both shaded regions, so the system has no solution.

Your Turn 5.93

Some systems of linear inequalities where the boundary lines are parallel will have a solution. We will see this in the next example.

Example 5.94

Graphing parallel boundary lines with a solution.

Graph y > 1 2 x − 4 y > 1 2 x − 4 by graphing y = 1 2 x − 4 y = 1 2 x − 4 using the slope m = 1 2 m = 1 2 and the y y -intercept b = − 4 b = − 4 ( Figure 5.95 ). The boundary line will be dashed. Test ( 0 , 0 ) ( 0 , 0 ) , which makes the inequality true, so shade the side that contains ( 0 , 0 ) ( 0 , 0 ) .

Graph x − 2 y < − 4 x − 2 y < − 4 by graphing x − 2 y = − 4 x − 2 y = − 4 ( Figure 5.96 ) and testing a point. The intercepts are x = − 4 x = − 4 and y = 2 y = 2 and the boundary line will be dashed. Choose a test point in the solution and verify that it is a solution to both inequalities. Test ( 0 , 0 ) ( 0 , 0 ) , which makes the inequality false, so shade the side that does not contain ( 0 , 0 ) ( 0 , 0 ) .

No point on the boundary lines is included in the solution as both lines are dashed. The solution is the region that is shaded twice which is also the solution to x − 2 y < − 4 x − 2 y < − 4 .

Your Turn 5.94

Interpreting and solving applications of linear inequalities.

When solving applications of systems of inequalities, first translate each condition into an inequality. Then graph the system, as we did above, to see the region that contains the solutions. Many situations will be realistic only if both variables are positive, so add inequalities to the system as additional requirements.

Example 5.95

Applying linear inequalities to calculating photo costs.

A photographer sells their prints at a booth at a street fair. At the start of the day, they want to have at least 25 photos to display at their booth. Each small photo they display costs $4 and each large photo costs $10. They do not want to spend more than $200 on photos to display.

• Write a system of inequalities to model this situation.
• Graph the system.
• Could they display 10 small and 20 large photos?
• Could they display 20 large and 10 small photos?

Since x ≥ 0 x ≥ 0 and y ≥ 0 y ≥ 0 (both are greater than or equal to) all solutions will be in the first quadrant. As a result, our graph shows only Quadrant I. To graph x + y ≥ 25 x + y ≥ 25 , graph x + y = 25 x + y = 25 as a solid line. Choose ( 0 , 0 ) ( 0 , 0 ) as a test point. Since it does not make the inequality true, shade the side that does not include the point ( 0 , 0 ) ( 0 , 0 ) .

To graph 4 x + 10 y ≤ 200 4 x + 10 y ≤ 200 , graph 4 x + 10 y = 200 4 x + 10 y = 200 as a solid line. Choose ( 0 , 0 ) ( 0 , 0 ) as a test point. Since it does make the inequality true, shade (bottom left) the side that include the point ( 0 , 0 ) ( 0 , 0 ) .

The solution of the system is the region of Figure 5.97 that is shaded the darkest. The boundary line sections that border the darkly shaded section are included in the solution as are the points on the x x -axis from ( 25 , 0 ) ( 25 , 0 ) to ( 55 , 0 ) ( 55 , 0 ) .

• To determine if 10 small and 20 large photos would work, we look at the graph to see if the point ( 10 , 20 ) ( 10 , 20 ) is in the solution region. We could also test the point to see if it is a solution of both equations. It is not, so the photographer would not display 10 small and 20 large photos.
• To determine if 20 small and 10 large photos would work, we look at the graph to see if the point ( 20 , 10 ) ( 20 , 10 ) is in the solution region. We could also test the point to see if it is a solution of both equations. It is, so the photographer could choose to display 20 small and 10 large photos. Notice that we could also test the possible solutions by substituting the values into each inequality.

Your Turn 5.95

Solving Systems of Linear Inequalities by Graphing

Systems of Linear Inequalities

Check Your Understanding

Section 5.10 exercises.

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4.5 Solving Systems of Linear Inequalities (Two Variables)

Learning objectives.

• Check solutions to systems of linear inequalities with two variables.
• Solve systems of linear inequalities.

Solutions to Systems of Linear Inequalities

A system of linear inequalities A set of two or more linear inequalities that define the conditions to be considered simultaneously. consists of a set of two or more linear inequalities with the same variables. The inequalities define the conditions that are to be considered simultaneously. For example,

We know that each inequality in the set contains infinitely many ordered pair solutions defined by a region in a rectangular coordinate plane. When considering two of these inequalities together, the intersection of these sets defines the set of simultaneous ordered pair solutions. When we graph each of the above inequalities separately, we have

When graphed on the same set of axes, the intersection can be determined.

The intersection is shaded darker and the final graph of the solution set is presented as follows:

The graph suggests that (3, 2) is a solution because it is in the intersection. To verify this, show that it solves both of the original inequalities:

Points on the solid boundary are included in the set of simultaneous solutions and points on the dashed boundary are not. Consider the point (−1, 0) on the solid boundary defined by y = 2 x + 2 and verify that it solves the original system:

Notice that this point satisfies both inequalities and thus is included in the solution set. Now consider the point (2, 0) on the dashed boundary defined by y = x − 2 and verify that it does not solve the original system:

This point does not satisfy both inequalities and thus is not included in the solution set.

Solving Systems of Linear Inequalities

Solutions to a system of linear inequalities are the ordered pairs that solve all the inequalities in the system. Therefore, to solve these systems, graph the solution sets of the inequalities on the same set of axes and determine where they intersect. This intersection, or overlap, defines the region of common ordered pair solutions.

Example 1: Graph the solution set: { − 2 x + y > − 4 3 x − 6 y ≥ 6 .

Solution: To facilitate the graphing process, we first solve for y .

For the first inequality, we use a dashed boundary defined by y = 2 x − 4 and shade all points above the line. For the second inequality, we use a solid boundary defined by y = 1 2 x − 1 and shade all points below. The intersection is darkened.

Now we present the solution with only the intersection shaded.

Example 2: Graph the solution set: { − 2 x + 3 y > 6 4 x − 6 y > 12 .

Solution: Begin by solving both inequalities for y .

Use a dashed line for each boundary. For the first inequality, shade all points above the boundary. For the second inequality, shade all points below the boundary.

As you can see, there is no intersection of these two shaded regions. Therefore, there are no simultaneous solutions.

Answer: No solution, ∅

Example 3: Graph the solution set: { y ≥ − 4 y < x + 3 y ≤ − 3 x + 3 .

Solution: The intersection of all the shaded regions forms the triangular region as pictured darkened below:

After graphing all three inequalities on the same set of axes, we determine that the intersection lies in the triangular region pictured.

The graphic suggests that (−1, 1) is a common point. As a check, substitute that point into the inequalities and verify that it solves all three conditions.

Key Takeaway

• To solve systems of linear inequalities, graph the solution sets of each inequality on the same set of axes and determine where they intersect.

Topic Exercises

Part A: Solving Systems of Linear Inequalities

Determine whether the given point is a solution to the given system of linear equations.

1. (3, 2); { y ≤ x + 3 y ≥ − x + 3

2. (−3, −2); { y < − 3 x + 4 y ≥ 2 x − 1

3. (5, 0); { y > − x + 5 y ≤ 3 4 x − 2

4. (0, 1); { y < 2 3 x + 1 y ≥ 5 2 x − 2

5. ( − 1 ,   8 3 ) ; { − 4 x + 3 y ≥ − 12 2 x + 3 y < 6

6. (−1, −2); { − x + y < 0 x + y < 0 x + y < − 2

Part B: Solving Systems of Linear Inequalities

Graph the solution set.

7. { y ≤ x + 3 y ≥ − x + 3

8. { y < − 3 x + 4 y ≥ 2 x − 1

9. { y > x y < − 1

10. { y < 2 3 x + 1 y ≥ 5 2 x − 2

11. { y > − x + 5 y ≤ 3 4 x − 2

12. { y > 3 5 x + 3 y < 3 5 x − 3

13. { x + 4 y < 12 − 3 x + 12 y ≥ − 12

14. { − x + y ≤ 6 2 x + y ≥ 1

15. { − 2 x + 3 y > 3 4 x − 3 y < 15

16. { − 4 x + 3 y ≥ − 12 2 x + 3 y < 6

17. { 5 x + y ≤ 4 − 4 x + 3 y < − 6

18. { 3 x + 5 y < 15 − x + 2 y ≤ 0

19. { x ≥ 0 5 x + y > 5

20. { x ≥ − 2 y ≥ 1

21. { x − 3 < 0 y + 2 ≥ 0

22. { 5 y ≥ 2 x + 5 − 2 x < − 5 y − 5

23. { x − y ≥ 0 − x + y < 1

24. { − x + y ≥ 0 y − x < 1

25. { x > − 2 x ≤ 2

26. { y > − 1 y < 2

27. { − x + 2 y > 8 3 x − 6 y ≥ 18

28. { − 3 x + 4 y ≤ 4 6 x − 8 y > − 8

29. { 2 x + y < 3 − x ≤ 1 2 y

30. { 2 x + 6 y ≤ 6 − 1 3 x − y ≤ 3

31. { y < 3 y > x x > − 4

32. { y < 1 y ≥ x − 1 y < − 3 x + 3

33. { − 4 x + 3 y > − 12 y ≥ 2 2 x + 3 y > 6

34. { − x + y < 0 x + y ≤ 0 x + y > − 2

35. { x + y < 2 x < 3 − x + y ≤ 2

36. { y + 4 ≥ 0 1 2 x + 1 3 y ≤ 1 − 1 2 x + 1 3 y ≤ 1

37. Construct a system of linear inequalities that describes all points in the first quadrant.

38. Construct a system of linear inequalities that describes all points in the second quadrant.

39. Construct a system of linear inequalities that describes all points in the third quadrant.

40. Construct a system of linear inequalities that describes all points in the fourth quadrant.

27: No solution, ∅

37: { x > 0 y > 0

39: { x < 0 y < 0

Systems of Linear Inequalities, Word Problems - Examples - Expii

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How to Solve Systems of Linear Inequalities?

Linear inequalities are expressions in which two linear expressions are compared using the inequality symbols. In this step-by-step guide, you will learn about solving systems of linear inequalities.

The solution to a system of a linear inequality is the region where the graphs of all linear inequalities in the system overlap.

Related Topics

• How to Solve Linear Inequalities
• How to Solve Quadratic Inequalities

A step-by-step guide to solving systems of linear inequalities

The system of linear inequalities is a set of equations of linear inequality containing the same variables. Several methods of solving systems of linear equations translate to the system of linear inequalities. However, solving a system of linear inequalities is somewhat different from linear equations because the signs of inequality prevent us from solving by the substitution or elimination method. Perhaps the best way to solve systems of linear inequalities is by graphing the inequalities.

To solve a system of inequalities, graph each linear inequality in the system on the same \(x-y\) axis by following the steps below:

• Solve the inequality for \(y\).
• Treat the inequality as a linear equation and graph the line as either a solid line or a dashed line depending on the inequality sign. If the inequality sign does not contain an equals sign \((< or >)\) then draw the line as a dashed line. If the inequality sign does have an equals sign \((≤ or ≥)\) then draw the line as a solid line.
• Shade the region that satisfies the inequality.
• Repeat steps \(1 – 3\) for each inequality.
• The solution set will be the overlapped region of all the inequalities.

Solving Systems of Linear Inequalities – Example 1:

Solve the following system of inequalities.

\(\begin{cases}x\:-\:5y\ge \:6\\ \:3x\:+\:2y>1\end{cases}\)

First, isolate the variable \(y\) to the left in each inequality:

\(x -5y≥ 6\)

\(x≥6 + 5y\)

\(5y≤ x- 6\)

\(y≤0.2 x -1.2\)

\(3x+ 2y> 1\)

\(2y>1-3x\)

\(y> 0.5-1.5x\)

Now, graph \(y≤ 0.2x-1.2\) and \(y > 0.5 -1.5x\) using a solid line and a broken one, respectively.

The solution of the system of inequality is the darker shaded area which is the overlap of the two individual solution regions.

Exercises for Solving Systems of Linear Inequalities

Determine the solution to the following system of inequalities..

• \(\color{blue}{\begin{cases}5x-2y\le 10 \\ \:3x+2y>\:6\end{cases}}\)
• \(\color{blue}{\begin{cases}-2x-y<\:-1 \\ \:4x+\:2y\:\le -6\end{cases}}\)

by: Effortless Math Team about 2 years ago (category: Articles )

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Linear Inequalities In Two Variables

Linear inequalities in two variables represent the inequalities between two algebraic expressions where two distinct variables are included. In linear inequalities in two variables, we use greater than (>), less than (<), greater than or equal (≥) and less than or equal (≤) symbols, instead of using equal to a symbol (=).

What is Linear Inequalities?

Any two real numbers or two algebraic expressions associated with the symbol ‘<’, ‘>’, ‘≤’ or ‘≥’ form a linear inequality. For example, 9<11, 18>17 are examples of numerical inequalities and x+7>y, y<10-x, x ≥ y > 11 are examples of algebraic inequalities.

The symbols ‘<‘ and ‘>’ represent the strict inequalities and the symbols ‘≤’ and ‘≥’ represent slack inequalities. To represent linear inequalities in one variable in a number line is a visual representation and is a convenient way to represent the solutions of the inequality. Now, we will discuss the graph of a linear inequality in two variables.

What are Linear Inequalities in Two Variables?

Linear inequalities in two variables represent the unequal relation between two algebraic expressions that includes two distinct variables. Hence, the symbols used between the expression in two variables will be ‘<’, ‘>’, ‘≤’ or ‘≥’, but we cannot use equal to ‘=’ symbol here.

The examples of linear inequalities in two variables are:

• 3x < 2y + 5
• 8y – 9x > 10

Note:   4x 2 + 2x + 5 < 0 is not an example of linear inequality in one variable, because the exponent of x is 2 in the first term. It is a quadratic inequality.

How to Solve Linear Inequalities in Two Variables?

The solution for linear inequalities in two variables is an ordered pair that is true for the inequality statement. Let us say if Ax + By > C is a linear inequality where x and y are two variables, then an ordered pair (x, y) satisfying the statement will be the required solution.

The method of solving linear inequalities in two variables is the same as solving linear equations .

For example, if 2x + 3y > 4 is a linear inequality, then we can check the solution, by putting the values of x and y here.

Let x = 1 and y = 2

Taking LHS, we have;

2 (1) + 3 (2) = 2 + 6 = 8

Since, 8 > 4, therefore, the ordered pair (1, 2) satisfy the inequality 2x + 3y > 4. Hence, (1, 2) is the solution.

We can also put different values of x and y to find different solutions here.

Graphical Solution of Linear Inequalities in Two Variables

The statements involving symbols like ‘<’(less than), ‘>’ (greater than), ‘≤’’(less than or equal to), ‘≥’ (greater than or equal to) and two distinct variables are called linear inequalities in two variables. Let us see here, how to find the solution of such expressions, graphically.

Below are the two examples of linear inequalities shown in the figure. The graph of y > x – 2 and y ≤ 2x + 2 are:

Real-life Examples:

Following example validates the difference between equation and inequality:

Statement 1: The distance between your house and school is exactly 4.5 kilometres,

The mathematical expression of the above statement is,

x = 4.5 km, where ‘x’ is the distance between house and the school.

Statement 2: The distance between your house and the school is at least 4.5 kilometers.

Here, the distance can be 4.5 km or more than that. Therefore the mathematical expression for the above statement is,

x ≥ 4.5 km, where ‘x’ is a variable that is equal to the distance between the house and the school.

Important Facts

• We can add, subtract, multiply and divide by the same number to solve the inequalities
• While multiplying and dividing by negative number, the inequality sign get reversed
• In graphical solution, the ordered pair outside the shaded portion does not solve the inequality
• Numerical inequalities: If only numbers are involved in the expression, then it is a numerical inequality. Example:  10 > 8, 5 < 7
• Literal inequalities: x < 2, y > 5, z < 10 are the examples for literal inequalities.
• Double inequalities: 5 < 7 < 9 read as 7 less than 9 and greater than 5 is an example of double inequality.
• Strict inequality: Mathematical expressions involve only ‘<‘ or ‘>’  are called strict inequalities. Example: 2x + 3 < 6, 2x + 3y > 6
• Slack inequality: Mathematical expressions involve only ‘≤′ or ‘≥’ are called slack inequalities. Example: 2x + 3 ≤ 6, 2x + 3y ≥ 6

Related Articles

• Linear Inequalities
• Solving Linear Inequalities
• Represent Linear Inequalities In One Variable On Number Line
• Linear Equations In Two Variables
• Cross Multiplication- Pair Of Linear Equations In Two Variables

Solved Examples on Linear Inequalities in Two Variables

1.) Classify the following expressions into:

• Linear inequality in one variable.
• Linear inequality in two variables.
• Slack inequality.

5x < 6, 8x + 3y ≤ 5, 2x – 5 < 9 , 2x ≤ 9 , 2x + 3y < 10.

2.) Solve y < 2 graphically.

Solution: Graph of y = 2. So we can show it graphically as given below:

Let us select a point, (0, 0) in the lower half-plane I and putting y = 0 in the given inequality, we see that: 1 × 0 < 2 or 0 < 2 which is true. Thus, the solution region is the shaded region below the line y = 2.

Hence, every point below the line (excluding all the points on the line) determines the solution of the given inequality.

Linear Inequalities in Two Variables Word Problem

In an experiment, a solution of hydrochloric acid is to be kept between 25° and 30° Celsius. What is the range of temperature in degree Fahrenheit if the conversion formula is given by C = 5/9 (F – 32), where C and F represent the temperature in degree Celsius and degree Fahrenheit, respectively.

Solution: As per the question it is given:

25<C<30

Now if we put C = 5/9 (F – 32), we get;

25 < 5/9 (F – 32) < 30

9/5 x 25 < F – 32 < 30 x 9/5

45 < F -32< 54

77 < F < 86

Thus, the required range of temperature is between 77° F and 86° F.

Frequently Asked Questions – FAQs

What is a system of linear inequalities in two variables, what is an example of linear inequality in two variables, what are the symbols used in linear inequalities in two variables, is y≥2x−3 a linear inequality.

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Linear inequalities in two variables

• Linear inequalities with two variables I
• Linear inequalities with two variables II
• Linear inequalities with two variables III

The solution of a linear inequality in two variables like Ax + By > C is an ordered pair (x, y) that produces a true statement when the values of x and y are substituted into the inequality.

Is (1, 2) a solution to the inequality

$$2x+3y>1$$

$$2\cdot 1+3\cdot 2\overset{?}{>}1$$

$$2+5\overset{?}{>}1$$

The graph of an inequality in two variables is the set of points that represents all solutions to the inequality. A linear inequality divides the coordinate plane into two halves by a boundary line where one half represents the solutions of the inequality. The boundary line is dashed for > and < and solid for ≤ and ≥. The half-plane that is a solution to the inequality is usually shaded.

Graph the inequality

$$y\geq -x+1$$

Video lesson

Graph the linear inequality

$$y \geq 2x -3$$

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