2.1 – Linear Inequalities in One Variable

Linear Inequalities in One Variable

We have dealt with equalities in some of our problems up to this point. Now we will deal with another expression used often in math called an inequality. We have the following symbols and their meanings:

$\begin{tabular}{l l} \raisebox{8pt}{\phantom{M}}$<$ & Less than\\ \raisebox{8pt}{\phantom{M}}$>$ & Greater than\\ \raisebox{8pt}{\phantom{M}}$\le$ & Less than or equal to\\ \raisebox{8pt}{\phantom{M}}$\ge$ & Greater than or equal to\\ \end{tabular}$

A Solution of an inequality is any value that makes the inequality a true statement. The solution set of an inequality is all numbers that make the statement true.

Interval Notation is a way of writing a set of real numbers. For example, suppose $a$ is a real number less than $b$ , another real number. The interval (a,b) is the set of all real numbers between the two (but not including them). To include $a$ and $b$ in the interval, square brackets are used: $[a,b]$

Set Builder Notation is a way of writing the set in a way that populates (usually an infinite number of) solutions. For example, to indicate the set of all numbers greater than 4, the set builder notation is $\{x|x>4\}$ . This is read as “The set of all $x$ values such that $x$ is greater than 4.”

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Another way of expressing solutions is by Graphing. This is simply a visual representation of the interval(s). For example, the interval $(-2,1)$ might be graphed

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Note the open circles that mean the graph goes up to, but does not include the numbers (at -2 and 1). This is called “open”. If the end points are to be included, the circles are filled in. This is called “closed” The interval $[-2,1]$ is graphed:

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Many combinations can be made by having an interval be open, closed, or open on one end and closed on the other. Let’s view some examples.

Express the solution set of the inequality $z\ge 4$ in interval notation
Since the solution set is all numbers 4 or greater, the interval for this is $[4,\infty)$ . We use the square bracket [ on the left to indicate that we include 4 in the solution. On the right side of the interval, we have a round parentheses ). This is to indicate that we don’t actually include the value as a valid solution.

Infinity is not a real number, it is the idea that the numbers simply never end in that direction. Intervals including an infinity (either positive or negative) will have always a parentheses next to the infinity.

Express the solution set of the inequality $x < 5$ in interval notation
Any number less than (but not equal to) 5 solves this, so we will use a parentheses around the 5, so we have: $(-\infty,5)$ .

Addition Principle of Inequalities

Recall the lever from section 1.2, but instead of starting in a balanced state, it begins with one side heavier than the other:

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If we add equal amounts to each side, will this have any effect on the inbalance?

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NO! It should be easy to see that adding the same amount to the left side and the right side leaves the lever in the same state of imbalance. That is, it hasn’t effected the inequality. This gives us the Addition Principle of Inequalites:

The Addition Principle of Inequalities
If $a<b$ Then $a+c<b+c$ likewise, If $a>b$ Then $a+c>b+c$

The above principle is valid for strict inequalities (those involving a < or > symbol) as well as inclusive inequalities (those including a $\le$ or $\ge$ symbol).

Express the solution set of the inequality $x+4<8$ by graphing as well as in interval notation.
Here we must isolate the variable $x$ first, using the Addition Principle of Inequalities So we can add (or subtract) from each side of the inequality with no problem: $x+4 {\color{red} \ -\ 4}<8{\color{red} \ -\ 4}$ $x<4$ We can then graph the solutions. Because this is a strict inequality, we use an open circle at 4, and then highlight the numbers lower than 4: the interval notation is given by $(-\infty,4)$

Express the solution set of the inequality $x+4<8$ by graphing as well as in interval notation.

Here we must isolate the variable $x$ first, using the Addition Principle of Inequalities
So we can add (or subtract) from each side of the inequality with no problem:
$x+4 {\color{red} \ -\ 4}<8{\color{red} \ -\ 4}$
$x<4$
We can then graph the solutions. Because this is a strict inequality, we use an open circle at 4, and then highlight the numbers lower than 4:

the interval notation is given by $(-\infty,4)$

Graph the solution set of the inequality $x-3\ge-2$
Again we can isolate the $x$ term by adding 3 to each side. Doing so yields: $x\ge1$ And we graph this using a closed circle to indicate that 1 is included in the solution set: The interval notation of this set is $[1,\infty)$

Graph the solution set of the inequality $x-3\ge-2$

Again we can isolate the $x$ term by adding 3 to each side. Doing so yields:
$x\ge1$
And we graph this using a closed circle to indicate that 1 is included in the solution set:

The interval notation of this set is $[1,\infty)$

Write the solution set to the inequality $x-5<6$ in set builder notation.
First lets isolate the $x$ . $x-5<6$ $x<11$ So this is our condition. We then write the set builder as: $\{x \| x < 11\}$ . The $\|$ symbol can be read as “such that” so the entire set builder notation can be read “The set of $x$ values, such that that $x$ is less than eleven.”

Multiplication Principle of Inequalities

We saw that the addition principle works for inequalities just as it does for equalities, but can we say the same is true for the multiplication principle? Let’s see. Consider the following example. Take a true inequality:

$4<7$
This is simply true by the nature of the inequality. If we multiply both sides by a value:

$4{\color{red} \cdot 4}<7{\color{red} \cdot4}$

$16<28$
We see that the inequality is still true. – This will work for any positive number that we multiply by! But what if we were to multiply a negative number? Consider this example. Take the same true inequality from above:

$4<7$

$4{\color{red} \cdot (-6)}<7{\color{red} \cdot(-6)}$

$-24<-42$
We see that this inequality has become false!. (Recall that a number is greater than another if it is to the right on a number line.) So we see that multiplying (and subsequently, dividing) by a negative number causes our inequality to become false. We can combat this by turning the inequality from greater than to less than, (or vice versa).

Multiplication Principle of Inequalities
Let $c$ be a positive number. If $a<b$ Then $ac<bc$ and $a(-b)>c(-b)$

Write the solution set to the inequality in set builder notation $3x-6\le15$
First, we use the addition principle: $3x-6\le15$ $3x\le21$ And then the multiplication principle: $x\le7$ And out set builder notation is: $\{x\|x\le7\}$

Solve the inequality. Graph and write solutions in interval notation and set builder notation.
$3x+2\le 5x-10$ $\underline{{\color{red}-5x-2}}\phantom{=}\underline{{\color{red}-5x-2}}$ $-2x\le-12$ $x\ge6$ Interval notation: $[6,\infty)$ Set Builder notation: $\{x\|x\ge6\}$

Solve the inequality. Graph and write solutions in interval notation and set builder notation.