Linear inequalities are simple equations that have a variable on one side of the equals sign and a constant on the other side of the equals sign. Linear inequalities also have an additional constraint: that **one variable cannot** be negative when compared to its corresponding constant.

Linear inequalities can be solved in several ways. One way is to convert the linear inequality to an equation, then solve the equation. Another way is to find all of the solutions to the linear inequality and then check if they are all valid solutions by checking if they are within the bounds of what you are trying to solve for.

Solving *linear inequalities* can be difficult when trying to find all of the solutions to the inequality. There are many ways that this can happen, such as having points that are not in solution sets, having points that are in both solution sets, and having points that are undefined due to being negative.

## Example of a system of linear inequalities

In this example, we will look at solving a system of linear inequalities. We will use the points (3, –2), (1, 1), and (0, 0) as our solutions. We will first need to find all of the solutions to each inequality and then combine them using Boolean algebra.

Solving the *first inequality gives us* a solution of 3 for x. Therefore, all values of x that satisfy this inequality have a value of 3.

Solving the **second inequality gives us** a solution of 1 for y. Therefore, all values of y that satisfy this inequality have a value of 1.

Solving the *third equation gives us* no solution for z = 0. Therefore, there are no values of z that satisfy this equation.

## Solving a system of linear inequalities

Now let’s look at how to solve the system of * linear inequalities*. First, remember that the solution to this problem is a set of values for x and y, so our answer will be a pair of numbers.

We can solve this system by inverting one of the linear inequalities and adding the results. Remember that inverting a *linear inequality means switching* its sign.

So, if we take the solution to *inequality 2* and add it to solution 1, we get the solution to the system as a whole. The point (3, –2) is in the set of solutions because it satisfies both equations.

## The point (3, –2) is in the solution set for the following system of linear inequalities

The system of linear inequalities above has four variables: a, b, c, and d. a=3, b=-2, c=0, and d=1.

This system has two constraints: inequality 1 and inequality 2. Inequality 1 states that b+d>a+c, or -2+1>3+0. Inequality 2 states that ac

Because the point (3, –2) is in the solution set of this system of linear inequalities, we can determine which linear inequality(s) have the point (3,-2) in their solution set. To do this, we must find all of the solutions to each inequality and then see if the point (3,-2) is in any of them.

## Systems of linear inequalities and graphs

Systems of ** linear inequalities** are a little more complex than

**single linear inequalities**, but they are also more versatile.

Systems of linear inequalities are a collection of two or more **linear inequalities combined together**. A system of linear inequalities has a solution set, which is all the values that satisfy all the given inequalities.

Systems of linear inequalities can be graphed as lines on a graph. The points where the lines intersect are the solutions to the system. Points outside of this set do not satisfy the system and thus are not valid solutions.

The number of solutions can be determined by finding how many intersections there are and seeing if that number is greater than one. If it is, then there is only one solution, and none of the other solutions match up with the points on the graph.If it is not, then there are more than one solution and they all match up with the points on the graph.

## Example of a system of quadratic inequalities

In this example, we will look at solving a system of linear inequalities. This example problem will show you how to find the solution set of a system of linear inequalities and which one has the point (3, –2) in its solution set.

First, we must write our system of inequalities as equations. We do this by putting each variable on either the left or right-hand side of the equals sign and adding an equals sign to both sides. Then, we solve each individual equation for the variable and put all of the **solutions together**.

In this problem, there are *two variables*: x and y. So, our equations are going to be x = 3 and y = –2.

## Solve the following quadratic equation by factoring

To solve the equation above, you would have to find the roots of the equation by finding where it **equals zero**. There are two solutions to this equation, which means there are two values that make it *equal zero*.

These solutions are -1 and 2, which mean that x = 1 or x = 2. These solutions show that there are two values of x that make the *original quadratic equation equal zero*.

There is one more solution to this equation that is not mentioned here- 0! 0 is a solution to any non-*zero linear inequality*, because it is false. When solving a linear inequality for a solution, only look for values of x that make the inequality true; do not look for false solutions.

## Solve the following quadratic equation by square rooting

To solve this equation, you would have to find the value of x that makes the * left side equal* to zero. There is only one number that makes the left side of an equation zero, and that is when the value of x is zero.

There are *two possible solutions* for this equation: x could be −3 or it could be −2. These solutions tell you that there are two points on the graph where y equals 2 and **x equals 3**.

These two points are not on the same line, so one more test needs to be done. Check to see if these points are in opposite directions. To do this, take a look at the sign of both numbers and see if they are opposite.

## Solve the following quadratic equation by completing the square

First, find the square root of both sides. Then, add and subtract the square root of the discriminant (the difference between the squares of the coefficients of the linear term and the square of the constant) to both sides. Finally, solve for either variable.

This procedure can be confusing so we will go through it step by step. First, solve:

Square both sides and then simplify:

Add and subtract the discriminant:

Solve for either variable: x= -2 or x= 2

Therefore, y = 3 or y = –1

The point (3, –1) or (–1, 3) is in the solution set.

Now let’s go back to solving by completing the square.

Completing the square is solving an equation containing a quadratic term such as “x” in this problem by finding its ±sign(s), taking its square root, and then adding it to both sides of the equation.

To do this, first solve for “y” by solving either one of these equations: y=3 or y=-1. Then plug these values into one of these equations: x=−2, x=2, or x=-1.

Finally, take the square roots of each side and add them together. The side with a plus (+) sign will have a positive answer while the side with a negative (-) sign will have a negative answer.

### Interpreting Linear Inequalities Using Points-in-the-Solution-Set Intersection

Now we will interpret linear inequalities using points-in-the-solution set intersection.

### Interpreting Linear Inequalities Using Points Within Solution Set Intersection^{[5]}

First let us define what points within solution set intersection means. It means that there is at least one point in each solution set that also lies in another solution set.

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(Jared Pingleton)