# Introduction to Linear Inequalities

When one expression is given to be greater than or less than another expression, we have an **inequation**.For example, consider:

\[2x + 3 > 7\]

This is an example of an *inequation in one variable*. The *solution to this inequation* will be the set of all values of *x* for which this inequation** ***is satisfied*, that is, the left side is greater than the right side. The solution in this case is simple to evaluate:

\[2x > 7 - 3\,\,\,\Rightarrow \;\;2x > 4\,\,\, \Rightarrow \;\;x > 2\]

Thus, all values of *x *greater than 2 will satisfy this inequality, because for all values of *x *greater than 2, the term \(2x + 3\) will be greater than 7.

Now, consider the inequation

\[x + 2y > 3\]

This is an example of an inequation in two variables. The *solution to this inequation* will be the set of all *pairs* of values for *x* and *y* such that the expression \(x + 2y\) is greater than 3.For example, one possible solution is \(x =3,\;y = 1,\) because

\[\left( 3 \right) +2\left( 1 \right) = 5 > 3\]

Note that this is *just one possible solution*. The solution to the inequation as a whole will be *all* pairs of values which satisfy the inequation.

The two examples considered above have **strict** inequalities: this means that the two sides can never be equal. However, we can also have inequations which don’t have strict inequalities. For example,

\[\begin{array}{l} -3x + 2 \ge 5\\x - 4y \le \sqrt 2 \end{array}\]

are linear inequations which are not strict, because the two sides in each inequation can also be equal.

As we have seen, *a linear inequation in one variable* involves an inequality between two linear expressions, or between a linear expression and a constant, where there is only one variable involved. Some examples:

\[\begin{array}{l}x +7 < \sqrt 2 \\ - 3 - \sqrt 3 x \ge 10\\3x + 1 \le \pi x + 2\\1 - 10x > 2+ 11x\end{array}\]

Every linear inequation in one variable can be simplified to the following general forms:

\[\begin{array}{l}ax +b > 0 & {\rm{or}} & ax + b \ge 0\\ax + b < 0 & {\rm{or}}& ax + b \le 0\end{array}\]

*a* and *b* are arbitrary real constants, and *a* is not equal to 0.

How do we solve a linear inequation, that is, how do we find the solution to a linear inequation? Consider the following inequation:

\[3x + 1 \le 7\]

We proceed as follows:

\[3x \le 7 - 1\,\,\,\Rightarrow \;\;\;3x \le 6\,\,\, \Rightarrow \;\;\;x \le 2\]

Now, we want to represent this *solution set* on a number line. Thus, we simply highlight that part of the number line lying to the left of 2:

We see that any number lying on the red part of the number line will satisfy this inequation, and so it is a part of the solution set for this inequation. Note that we have drawn a *solid dot* exactly at the point 2. This is to indicate that 2 is also a part of the solution set.

Now, consider the following inequation:\[4x > - 3x +21\]

We first solve this inequation algebraically:

\[4x + 3x >21\,\,\, \Rightarrow \;\;\;7x > 21\,\,\, \Rightarrow \;\;\;x > 3\]

Now, we plot the solution set on a number line:

Any point lying on the red part of the number line will satisfy this inequation. Note that in this case, we have drawn a hollow dot at the point 3. This is to indicate that 3 is *not *a part of the solution set (this is because the given inequation had a strict inequality).

Consider the following inequation:

\[ - 2x + 3 > 5\]

Suppose that we solve this inequation as follows:

\[ - 2x > 5 -3\,\,\, \Rightarrow \;\;\; - 2x > 2\,\,\, \Rightarrow \;\;\;x > - 1\]

Is this solution correct? *No, *it is not! Let us try to understand why. Let us pick an arbitrary number in this solution set, say \(x = 2,\) and see whether it satisfies the original inequation. We have

\[ - 2\left( 2 \right)+ 3 = - 4 + 3 = - 1\]

This is *not* greater than 5, and thus *x* equal to 2 does not satisfy the original inequation, which means that the solution set \(x > - 1\) is incorrect. Where did we go wrong?

To understand that, consider the following inequation:

\[ - x > 1\]

Can we multiply –1 on both sides and say the following?

\[\begin{array}{l}\left({ - 1} \right) \times \left( { - x} \right) > \left( { - 1} \right) \times1\\ \Rightarrow \;\;\;x > - 1\end{array}\]

No! This is incorrect. If *–x*is greater than 1, this cannot mean that *x* is greater than –1. In fact,what this will *actually* mean is that *x* is less than –1. Only then can the negative of *x* be greater than 1. Take your time and reflect on this argument in detail.

Thus, when we multiplied the inequation by –1 on both sides, we should actually have *reversed the direction of the inequality*:

\[\begin{array}{l} - x> 1\\ \Rightarrow \;\;\;\left( { - 1} \right) \times \left( { - x} \right)< \left( { - 1} \right) \times 1\\ \Rightarrow \;\;\;x < -1\end{array}\]

In fact, extending this line of reasoning, whenever you multiply two sides of an inequation by a negative number, the direction of the inequality will get reversed.

Let us summarize our discussion into the following rules:

**(1)** You can add or subtract any real number on both sides of an inequation:

\[\begin{array}{l}x> a\\ \Rightarrow \;\;\;\left\{ \begin{array}{l}x + k > a + k\\x - k >a - k\end{array} \right.\end{array}\]

**(2)** You can multiply any *positive *real number on both sides of an inequation:

\[\begin{array}{l}x> a,\,\,\,k > 0\\ \Rightarrow \;\;kx > ka\end{array}\]

**(3)** If you multiply a *negative *real number on both sides of an inequation, the *direction of the inequality will get reversed*:

\[\begin{array}{l}x> a,\,\,\,k < 0\\ \Rightarrow \;\;\;kx < ka\end{array}\]

**Example 1: **Solve the following inequation, and plot the solution set on a number line:

\[2x - 5 > 3 - 7x\]

**Solution:** We have:

\[2x - 5 > 3 -7x\,\,\, \Rightarrow \;\;9x > 8\,\,\, \Rightarrow \;\;x > \frac{8}{9}\]

The solution set is plotted below. Note the hollow dot carefully:

**Example 2: **Solve the following inequation, and plot the solution set on a number line:

\[ - 2x - 39 \ge -15\]

**Solution:** We have:

\[\begin{array}{l} -2x - 39 \ge - 15\,\,\, \Rightarrow \;\; - 2x \ge 24\\ \Rightarrow \;\;2x \le - 24\,\,\, \Rightarrow \;\;x \le - 12\end{array}\]

The solution set is plotted below. Note the solid dot at *x* equal to –12.

**Example 3: **A number *x *satisfies both the following linear inequations *simultaneously*:

\[\begin{array}{l}2x +1 > 7\\ - 3x > - 18\end{array}\]

How many integer values can *x *take?

**Solution:** We will find the solution sets for the two inequations separately, and then find those values of *x *which are *common to* both the solution sets.

### Inequality - 1

We have:

\[2x + 1 > 7\,\,\,\Rightarrow \;\;\;2x > 6\,\,\, \Rightarrow \;\;\;x > 3\]

### Inequality - 2

We have

\[ - 3x > -18\,\,\, \Rightarrow \;\;\;3x < 18\,\,\, \Rightarrow \;\;\;x < 6\]

Now, for *x* to simultaneously satisfy both inequations, it must be greater than 3 *as well as* less than 6. Thus, the solution set for the two inequations taken together will be as follows:

We see that *x* can take two integer values: 4 and 5. Note that the solution set itself is infinite. *x *can take any value between 3 and 6. However, there are only two integer values between 3 and 6.