# Linear Inequalities

An **inequality**, as the name suggests, is a relationship between two quantities that are unequal.

#### Contents

- Numerical Inequalities
- Intervals on the Number Line
- Finding Solutions to Linear Inequalities
- One-step Linear Inequalities
- Two-sided Linear Inequalities
- Multi-step Linear Inequalities
- Linear Inequalities - Multi-step - Intermediate
- Linear Inequalities - Problem Solving
- Linear Inequalities Word Problems
- Linear Inequalities - Word Problems - Intermediate

## Numerical Inequalities

One property of real numbers is that they have order. This order allows us to compare numbers and decide if they are equal to each other or one is greater or less than the other.

It is easiest to understand inequalities in the context of a **number line** (see above). This shows us that the numbers are ordered in a particular way. Those to the left are "less than" those to the right.

To show the inequality of numbers we use a symbolic notation:

**Less than**: The $<$ sign stands for "less than." So, $4 < 10$ is true. Further, $x<10$ means $x$ can be any number less than $10.$**Greater than**: The $>$ sign stands for "greater than". So, $11 > 10$ is true. Further, $x>11$ means $x$ can be any number greater than $11.$**Or equal to**: Sometimes we want to show an inequality that is not strictly greater or less than. We use the same symbol, but with an underline, to show that the number might also be equal to the value we are comparing to. So $4 \geq 4$ is true, and so is $5 \geq 4$. So, $x \leq 3$ means that $x$ can be any number less than**or equal to**$3.$

How many positive integers are there that are $<7?$

Looking at the above number line, we see that the numbers that are less than $7$ and positive are found in between $0$ and $7$.

So the numbers are $1, 2, 3, 4, 5$ and $6$, which make up a total of $6$. $_\square$

What inequality operation should be used in place of the question mark below?

$61\ ?\ 59$

If we look at the extended number line, we would see that the number $61$ is two steps ahead of $59$. It means both "$61$ is greater than $59$" and "$59$ is less than $61.$"

We saw the appropriate sign to show one is greater than the other is "$>$".

So the answer is $61>59$, which is true. $_\square$

How many positive integers are there that are less than $100$ and are multiples of $11?$

We list all the multiples of $11$ that are found between $0$ and $100$:

$\begin{aligned} 11\times1&=11\\ 11\times2&=22\\ 11\times 3&=33\\ 11\times 4& =44\\ 11\times 5& = 55\\ 11\times 6& = 66\\ 11\times 7 &=77\\ 11\times 8 &= 88\\ 11\times 9&=99. \end{aligned}$

Since $11\times 10 =110$ and $110$ is ten steps ahead of $100$, we omit it and end there.

So the numbers are $11, 22, 33, 44, 55, 66, 77, 88, 99$, giving a total of $9$ such numbers. $_\square$

How many integers $x$ satisfy $13<x<21?$

If we split the above inequality, we get two inequalities $13<x$ and $x<21$.

The first inequality $13<x$ means $13$ is less than $x$, which also means $x$ is greater than $13$. So the numbers are found to the right of $13$ on the number line.

The second inequality $x<21$ means $x$ is less than $21$. So the numbers are found to the left of $21$ on the number line.

Combining the two inequalities, we get a list of numbers that are to the right of $13$ and a list of numbers that are to the left of $21$, which means the numbers located between $13$ and $21$.

The numbers are $14, 15, 16, 17 ,18 , 19, 20$, which gives a total of $7$ numbers. $_\square$

## Intervals on the Number Line

In the wiki Representation on the Real Line, we saw that real numbers can be visually represented on a number line, with each point on the line corresponding to a real number and each real number corresponding to a point on the line. How can we use the number line to represent sets of real numbers?

When we draw regions of the number line, we use the following conventions:

- Solid dots and colored lines denote points that are included in the region.
- Open (unfilled) dots denote points that are not included in the region.

Let's consider an example.

Find an inequality representing the above region in the number line.

Since the circle on the real number $-3$ is an open circle, $-3$ is not included in the region. This shows the region includes all real numbers strictly greater than $-3$, i.e., all real numbers $x$ satisfying the inequality $x > -3$. $_\square$

Now, we will consider the same region of the number line and demonstrate that there may be more than one possible inequality to represent the same region.

Which of the following inequalities is represented in the above number line?

$\begin{array}{c}&&~(a)\ x+8\le 2x &&~(b)\ 2x-3>x+4 &&~(c)\ -x+3<4x+18 &&~(d)\ 3x+9\le -2x-1\end{array}$

The solutions to the above inequalities are

$\begin{array}{c}&&(a)\ x+8\le 2x &\implies x\ge 8 \\ \\ &(b)\ 2x-3>x+4 &\implies x>7 \\ \\ &(c)\ -x+3<4x+18 &\implies x>-3\\ \\ &(d)\ 3x+9\le -2x-1 &\implies x\le -2. \end{array}$

Therefore, the answer is $(c).$ $_\square$

This example demonstrates that it may be necessary to perform algebraic manipulation to an inequality before being able to determine if the inequality represents a given region in the number line. Possible algebraic manipulations include

- add or subtract a constant to both sides;
- multiply both sides by a positive number;
- multiply both sides by a negative number and switch the inequality sign.

Which of the following number lines represents the inequality

$\\ 2x-11\le -(x+2)?$

We have

$\begin{aligned} 2x-11 &\le -(x+2) \\ 2x-11 &\le -x-2 \\ 3x &\le 9\\ x &\le 3. \end{aligned}$

Therefore, the answer is $(b).$ $_\square$

Which of the following inequalities is represented in the above number line?

$\begin{array}{c}&&~(a)\ \frac{x}{100}<\frac{x}{5}+\frac{19}{50} &&~(b)\ \frac{1}{5}-\frac{2x}{5}>\frac{x}{5}-1 &&~(c)\ \frac{x-1}{2}<\frac{x+3}{4} &&~(d)\ \frac{3x+1}{4}<\frac{x}{3}-1\end{array}$

The solutions to the above inequalities are

$\begin{array}{c}&&(a)\ \frac{x}{100}<\frac{x}{5}+\frac{19}{50} &\implies \frac{19x}{100}>-\frac{38}{100} &\implies x>-2 \\ \\ &(b)\ \frac{1}{5}-\frac{2x}{5}>\frac{x}{5}-1 &\implies \frac{3x}{5}<\frac{6}{5} &\implies x<2 \\ \\ &(c)\ \frac{x-1}{2}<\frac{x+3}{4} &\implies \frac{2x-2}{4}<\frac{x+3}{4} &\implies x<5\\ \\ &(d)\ \frac{3x+1}{4}<\frac{x}{3}-1 &\implies \frac{9x+3}{12}<\frac{4x-12}{12} &\implies x<-3 . \end{array}$

Therefore, the answer is $(c).$ $_\square$

Which of the above number lines represents the solution of the following system of linear inequalities?

$\begin{cases} 2x-1 &\le &x+5\\x+1&>&0\end{cases}$

The first inequality gives $x\le 6.$ The second inequality gives $x>-1.$ Thus, the solution is $-1<x\le 6,$ implying that the appropriate representation of the number line is $(a).$ $_\square$

## Finding Solutions to Linear Inequalities

Linear inequalities are relationships that hold true between two different components. They are usually composed of a $<, >, \leq,$ or $\geq$ symbol.

The notation $x < y$ means that $x$ is less than $y$.

The notation $x \leq y$ means that $x$ is less or equal to $y$.

The notation $x>y$ means that $x$ is greater than $y$.

The notation $x \geq y$ means that $x$ is greater or equal to $y$.

Linear inequalities with one variable can be solved by algebraically manipulating the inequality so that the variable remains on one side and the numerical values on the other. Once this is done, we obtain a relationship that expresses the solution of the inequality.

Linear inequalities can also be solved by graphing and thinking of them visually.

What is the smallest integer that satisfies $x > 4?$

Looking at the equality, we observe that it states

all real numbers greater than four. But since we are looking for the smallestintegervalue, we notice that the smallest integer greater than $4$ is five. $_\square$

What is the smallest integer that satisfies $x \geq 4?$

Looking at the equality, we observe that it states

all real numbers greater than or equal to four. Since these numbers are basically all numbers starting from $4$ to $\infty$, including $4$, we know that $4$ is the smallest integer that satisfies the inequality. $_\square$

Use algebra to find the values of $x$ for which $6x + 4 < 2 + x.$

We first gather like terms by adding $-x -4$ to both sides of the inequality and then divide both sides by $5$ to obtain

$\begin{aligned} -x - 4 + 6x + 4 &< 2 + x - x - 4\\\\ 5x &< -2\\\\ x &< \frac{-2}{5} . \end{aligned}$

This means that all values of $x$ less than $\frac{-2}{5}$ are solutions to the relationship above.

In short, we can write the solution as $\left(-\infty,\frac{-2}{5}\right),$ which means all real numbers between but not including negative infinity and $\frac{-2}{5}$. This is essentially a shorter way of rewriting the statement above. $_\square$

Show graphically that the solution obtained for the above example is true.

From our previous equation we know that $6x + 4 < 2 + x$ reduces to $x < -\frac{2}{5}$.

On this graph, we first plotted the line $x = -\frac{2}{5},$ and then shaded the entire region to the left of the line. The shaded area is called the bounded region, and any point within this region satisfies the inequality $x < -\frac{2}{5}.$ Notice also that the line representing the region's boundary is a dashed line; this means that values along the line $x = -\frac{2}{5}$ are not included in the solution set of the inequality. $_\square$

## One-step Linear Inequalities

Solving linear inequalities in one variable is the same as solving linear equations. Given an inequality, what we should do is to isolate the variable on one side.

For example, if we are given the inequality

$x-3>9,$

leave the steps involved for now:

$\begin{aligned} x-3+3&>9+3\\ x+0&>12\\ x&>12. \end{aligned}$

Both $x-3>9$ and $x>12$ have the same solution because performing the basic arithmetic operations $+,-,\times,\div$ to both sides of the inequality doesn't change the inequality. This is known as an *equivalent inequality*. Thus all numbers greater than $12$ satisfy the above inequality.

We have seen that transforming inequalities into an equivalent inequality will lead us to the solution. To change an inequality into an equivalent inequality, we use the four basic arithmetic operations, but what operation to use is entirely dependent on the type of question, so we must keep our eyes sharp to identify what operation to use.

**Addition to form equivalence inequality**

When using addition, we rely on the fact that any number plus zero is the number itself $(x+0=x)$. So, for example, if the inequality is

$x-8<1,$

to create an equivalence inequality, we have to isolate $x$. We can do that by making the left term equal to $x+0,$ which is basically the number itself. So we can add $+8$ to make it zero:

$\begin{aligned} x-8+8&<1\\ x-0&<1\\ x&<1. \end{aligned}$

OK, we have isolated $x$, but we have altered the inequality by simply adding $8$. To correct this, we can add another $8$ to the right side of the inequality to counter the change we made on the left side:

$\begin{aligned} x-8+8&<1\\ x-0&<1+8\\ x&<9. \end{aligned}$

So, all numbers less than $9$ satisfy the inequality.

**Subtraction to from equivalence inequality**

Suppose we are given the inequality $x+3>7.$

If we add $3$ to both sides, $x+3+3>7+3$ or $x+6>10,$ which is no more helpful than the original. But if we subtract $3$ from both sides, we can isolate $x$:

$\begin{aligned} x+3&<7\\ x+3-3&<7-3\\ x&<4. \end{aligned}$

So, all numbers less than $4$ satisfy the inequality.

**Multiplication to form equivalence inequality**

When using multiplication, we rely on the property that any number multiplied by $1$ except for $0$ is the number itself $(m\times1=m, m\neq0)$.

Given the inequality

$\frac { 1 }{ 2 }x <7,$

let's see how we can isolate $x$. We can see that if we multiply $\frac{1}{2}$ by $2$, we get one. So we have

$\begin{aligned} \left( \frac { 1 }{ 2 } x \right) \times 2 &<7\times 2\\ 1\times x&<14\\ x&<14. \end{aligned}$

Another thing not to forget is that when multiplying both sides of the inequality by a negative number, the inequality sign changes. For example, in the inequality $-\frac{1}{3}x<5$, to isolate $x$ we multiply both sides of the inequality by $-3$:

$\begin{aligned} -\left( \frac { 1 }{ 3 } x \right) \times (-3)&<5\times (-3)\\ 1\times x&>-15\\ x&>-15. \end{aligned}$

So, all numbers greater than $-15$ satisfy the inequality.

**Division to form equivalence inequality**

Division relies on the same principle as multiplication since division is the reciprocal of multiplication: $m\div n= m \times \frac{1}{n}$.

Consider the inequality $6x<12$.

If we wanted to isolate $x$, we can divide the left term by $6$. Since any number divided by itself except for $0$ is equal to $1$, it follows that $\frac{6}{6}=1$, implying

$\begin{aligned} 6x&<12\\ 6x\div 6&<12\div 6\\ 6x\times \frac { 1 }{ 6 } &<12\times \frac { 1 }{ 6 } \\ 1\times x&<2\\ x&<2. \end{aligned}$

So, all numbers less than $2$ satisfy the inequality.

Just like what we did with multiplication, dividing both sides of an inequality with a negative number changes the sign of the inequality.

Let's look at some more difficult questions to strengthen our understanding.

Solve $6x-3<9$.

This looks like we have to do both addition and division to isolate $x$.

Step 1:$\begin{aligned} 6x-3&<9\\ 6x-3+3&<9+3\\ 6x&<12. \end{aligned}$

Step 2:$\begin{aligned} 6x\times \frac { 1 }{ 6 } &<12\times \frac { 1 }{ 6 } \\ x&<2. \ _\square \end{aligned}$

Note: We didn't necessarily have to do the addition process before the division process, but it is usually easier to do addition and subtraction process before multiplication and division process.

Given that $x$ is an element of the set $\left\{ -2,-1,-7,-4,-5,0 \right\}$, how many values of $x$ satisfy the inequality $-6x+13<37?$

Instead of plugging in all the numbers and checking, let's isolate $x$ to form equivalence inequality.

Step 1: Remove $13$ from the left-hand side:$\begin{aligned} -6x+13-13&<37-13\\ -6x&<24. \end{aligned}$

Step 2: To remove $-6$, divide both sides by $-6$ and change the sign of the inequality:$\begin{aligned} (-6x)\times \left( -\frac { 1 }{ 6 } \right) &<24\times \left( -\frac { 1 }{ 6 } \right) \\ x&>-4. \end{aligned}$

Since $x$ is greater than $-4$, the numbers $-2,-1,0$ satisfy the inequality. Thus, a total $3$ values satisfy the inequality. $_\square$

Solve $\frac { 2 }{ 3 } x-\frac { 1 }{ 2 } >\frac { 3 }{ 2 }.$

Step 1: Remove $\frac{1}{2}$ from the left-hand side:$\begin{aligned} \frac { 2 }{ 3 } x-\frac { 1 }{ 2 } &>\frac { 3 }{ 2 } \\ \frac { 2 }{ 3 } x-\frac { 1 }{ 2 } +\frac { 1 }{ 2 } &>\frac { 3 }{ 2 } +\frac { 1 }{ 2 } \\ \frac { 2 }{ 3 } x&>2. \end{aligned}$

Step 2: Here we could remove the $2$ and $3$ in $\frac{2}{3}$ separately by dividing by $2$ and multiplying by $3$. But if we realize that the reciprocal of a fraction is $1$, i.e. $\frac{m}{n}\times \frac{n}{m}=1$, we can multiply it by its reciprocal:$\begin{aligned} \left( \frac { 2 }{ 3 } x\times \frac { 3 }{ 2 } \right) &>2 \times \left( \frac { 3 }{ 2 } \right) \\ 1\times x&>3\\ x&>3. \end{aligned}$

Thus, all numbers greater than $3$ satisfy the inequality. $_\square$

## Two-sided Linear Inequalities

## Solve $2x+3<x<3x+16.$

The first inequality gives

$\begin{aligned} 2x+3 &<x \\ x&<-3. \qquad (1) \end{aligned}$

The second inequality gives

$\begin{aligned} x &<3x+16 \\ -16 &<2x \\ -8 &<x. \qquad (2) \end{aligned}$

Combining $(1)$ and $(2)$ gives $-8<x<-3. \ _\square$

Solve $3x-8<2x<5x-33.$

The first inequality gives

$\begin{aligned} 3x-8 &<2x \\ x&<8. \qquad (1) \end{aligned}$

The second inequality gives

$\begin{aligned} 2x &<5x-33 \\ 33 &<3x \\ 11 &<x. \qquad (2) \end{aligned}$

Since there is no value of $x$ that satisfies both $(1)$ and $(2),$ there is no solution. $_\square$

If the solution to the following inequalities is all the non-negative numbers, what is the value of $a?$

$x-9 \le 3x+1 \le 7x-a$

The first inequality gives

$\begin{aligned} x-9 &\le 3x+1 \\ -10 &\le 2x \\ x &\ge -5. \qquad (1) \end{aligned}$

The second inequality gives

$\begin{aligned} 3x+1 &\le 7x-a \\ a+1 &\le 4x \\ x &\ge \frac{a+1}{4}. \qquad (2) \end{aligned}$

As the result of combining $(1)$ and $(2),$ we should have the required solution to this problem: $x\ge 0. \qquad (3)$

Since $(1)$ contains $(3),$ it must be true that $(2)$ is equivalent to $(3):$

$x \ge \frac{a+1}{4} \Leftrightarrow x\ge 0.$

This implies

$a+1=0 \implies a=-1. \ _\square$

## Multi-step Linear Inequalities

To solve inequalities involving the expression $mx+b$, we need to consider the properties of inequalities.

Properties of Inequalities:

- The sense of inequality is unchanged if the same real number is added to both sides.
- The sense of inequality is unchanged if both sides are multiplied by the same
*positive*real number. - The sense of inequality is reversed if both sides are multiplied by the same
*negative*real number. - If $a>b$ and $c>d ,$ then $a+c> b+d.$
- $a>b>0$ and $c>d>0,$ then $ac>bd.$

## Solve the solution set of $3-4x \leq 2x +9.$

By Property $1$ above, $-6x \leq 6.$

Then by Property $3$ above, $x \geq -1. \ _\square$

## Linear Inequalities - Multi-step - Intermediate

## Linear Inequalities - Problem Solving

When solving a problem (as opposed to merely using a given formula), it is generally good to proceed along the following lines:

- First, you have to understand the problem.
- After understanding, make a plan.
- Carry out the plan.
- Look back on your work. How could it be better?

For more details on each of these, see How to Solve Problems on Brilliant.

Prove that for all positive reals $x$, the following inequality is always true:

$x + \frac1x \geq 2.$

Because the square of a real number is always non-zero, we have $\Big(\sqrt x - \frac1{\sqrt x}\Big)^2 \geq 0.$ Perform some algebraic manipulation, and you will get the desired inequality. $_\square$

Let $a,b,c$ be positive reals. Also, let $k$ be the largest possible real such that

$\dfrac{a}{1}+\dfrac{b}{1}+\dfrac{c}{1}+\dfrac{a+b}{1}+\dfrac{b+c}{1}+\dfrac{c+a}{1}\le \dfrac{a+b+c}{k}.$

If $k$ can be expressed as $\frac{p}{q}$ for relatively prime positive integers $p$ and $q$, then what is $p+q?$

- Inequality of Denominators: the harder version of this inequality

## Linear Inequalities Word Problems

If that doughnut weighs heavier than the flip-flop by 100 grams, but that flip-flop is lighter than the dry cement by 1000 kg. Which is heavier, the doughnut or the dry cement? And by how much?

To be continued

The diagram shows how a mobile will be balanced when left to hang, and the rods are all tilted to the maximum degree.

Assuming that the fulcrum is at the center of each rod, what are the relative weights of these shapes?

## Linear Inequalities - Word Problems - Intermediate

A classroom has $n\times n$ number of tables. If the teacher removes the least number of squares such that there is still a perfect square number of tables, then there will be 4 students who don't have a sit. How many extra tables are there if the teacher did not remove the tables?

$(n-1)^2 = m + 4$ or something. To be continued.

**Cite as:**Linear Inequalities.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/linear-inequalities/