Inequalities

Lesson

Before we are introduced to compound inequalities, let's quickly discuss the words "And" and "Or". These are words that you've been using your entire life, but it's important to fully understand what they mean when used in mathematics.

If you heard someone say:

"My teacher is giving us a test on Monday AND an assignment on Tuesday"

Then they mean that both of these things are true, that their teacher is giving them a test on Monday as well as an assignment on Tuesday.

If they said:

"My teacher is giving us a test on Monday OR an assignment on Tuesday"

Then either one of these things could be true on their own.

Consider each of these four different statements.

"$x>1$x>1 AND $x<9$x<9" |

"$x<1$x<1 OR $x>9$x>9" |

"$x>1$x>1 OR $x<9$x<9" |

"$x<1$x<1 AND $x>9$x>9" |

These are examples of compound inequalities, statements involving several inequalities.

Let's go through the compound inequalities one by one.

"$x>1$`x`>1 AND $x<9$`x`<9"

In other words, for any value of $x$`x` to satisfy this statement, it must be both greater than $1$1 and less than $9$9. This means $x$`x` will be a real number in between $1$1 and $9$9.

"$x<1$`x`<1 OR $x>9$`x`>9"

Either one of these inequalities could be true by themselves. $x$`x` could be less than $1$1, and it doesn't matter that it's not greater than $9$9, or vice versa. The only values of $x$`x` that don't satisfy this statement are numbers between $1$1 and $9$9 inclusive.

"$x>1$`x`>1 OR $x<9$`x`<9"

This involves the same inequalities as the first example, except the word "AND" has been replaced with the word "OR". How does this change the statement? Well, now it's okay for just one inequality to be true at a time. $x$`x` can either be greater than $1$1 or it can be less than $9$9. In fact, every possible number satisfies this condition. $x$`x` can be absolutely any value.

"$x<1$`x`<1 AND $x>9$`x`>9"

Is there any number that is both less than $1$1 and greater than $9$9? This is impossible! No number satisfies this compound inequality.

Because the last two aren't very interesting cases, we will mostly encounter examples like the first two.

Previously, we covered how to plot inequalities on a number line.

How do we plot compound inequalities on the number line?

Let's say we have an example like the first one we saw above, say, "$x\ge-5$`x`≥−5 AND $x\le-2$`x`≤−2".

This includes every number in between $-5$−5 and $-2$−2, inclusive. We can plot it like this.

As you can see, while ordinary inequalities are plotted as rays, this kind of compound inequality is plotted as an interval.

Now, say we wanted to plot "$x\le-5$`x`≤−5 OR $x\ge-2$`x`≥−2". As we found before, this means that $x$`x` can either be less than or equal to $-5$−5 OR greater than or equal to $-2$−2. So, we plot both inequalities on the number line, as two rays.

There is another way of writing any compound inequality like "$x\ge-5$`x`≥−5 AND $x\le-2$`x`≤−2".

The values of $x$`x` that satisfy this compound inequality include all numbers between $-5$−5 and $-2$−2, inclusive.

So, another way to write "$x\ge-5$`x`≥−5 AND $x\le-2$`x`≤−2" is:

$-5\le x\le-2$−5≤`x`≤−2

$-5$−5 is less than or equal to $x$`x`, and $x$`x` is less than or equal to $-2$−2. This also makes it easier to see that $x$`x` is in between $-5$−5 and $-2$−2!

Did you know?

In our very first example involving the word "OR":

"My teacher is giving us a test on Monday OR an assignment on Tuesday"

Did you expect that this could also mean that there would be *both* a test and an assignment? Probably not.

What about if a waiter asked you:

"Would you like milk OR sugar with your tea?"

Is it possible to have both? Yes.

There are two interpretations of the word "OR". The first example uses the exclusive "OR" and the second example uses the inclusive "OR".

In mathematics, we will almost always use the inclusive "OR".

Consider the compound inequality $-2\le1-2x<2$−2≤1−2`x`<2.

Solve the compound inequality for the set of values of $x$

`x`:Which of the following intervals is the correct solution set for $x$

`x`?$\left(-\frac{1}{2},\frac{3}{2}\right]$(−12,32]

A$\left(-\frac{1}{2},\frac{3}{2}\right)$(−12,32)

B$\left[-\frac{1}{2},\frac{3}{2}\right)$[−12,32)

C$\left[-\frac{3}{2},\frac{1}{2}\right)$[−32,12)

D$\left(-\frac{1}{2},\frac{3}{2}\right]$(−12,32]

A$\left(-\frac{1}{2},\frac{3}{2}\right)$(−12,32)

B$\left[-\frac{1}{2},\frac{3}{2}\right)$[−12,32)

C$\left[-\frac{3}{2},\frac{1}{2}\right)$[−32,12)

DWhich of the following graphs represents the correct solution set for $x$

`x`?ABCDABCD

Consider the set described by the inequalities $4x-10<-1.3$4`x`−10<−1.3 OR $4x-10>1.3$4`x`−10>1.3

Solve the first inequality, $4x-10<-1.3$4

`x`−10<−1.3, for $x$`x`.Now solve the second inequality, $4x-10>1.3$4

`x`−10>1.3, for $x$`x`.Which of the following intervals is/are the correct solution set for $x$

`x`?$($($-\infty$−∞$,$,$2.175$2.175$]$]$\cup$∪$[$[$2.825$2.825$,$,$\infty$∞$)$)

A$($($-\infty$−∞$,$,$2.175$2.175$)$)$\cup$∪$($($2.825$2.825$,$,$\infty$∞$)$)

B$($($2.175$2.175$,$,$2.825$2.825$)$)

C$($($-\infty$−∞$,$,$-2.825$−2.825$)$)$\cup$∪$($($-2.175$−2.175$,$,$\infty$∞$)$)

D$($($-\infty$−∞$,$,$2.175$2.175$]$]$\cup$∪$[$[$2.825$2.825$,$,$\infty$∞$)$)

A$($($-\infty$−∞$,$,$2.175$2.175$)$)$\cup$∪$($($2.825$2.825$,$,$\infty$∞$)$)

B$($($2.175$2.175$,$,$2.825$2.825$)$)

C$($($-\infty$−∞$,$,$-2.825$−2.825$)$)$\cup$∪$($($-2.175$−2.175$,$,$\infty$∞$)$)

DWhich of the following graphs represents the correct solution set for $x$

`x`?ABCDABCD

Let $f\left(x\right)=1x-2$`f`(`x`)=1`x`−2 and $g\left(x\right)=4-3x$`g`(`x`)=4−3`x`.

Find all values of $x$`x` for which $f\left(x\right)<-6$`f`(`x`)<−6 or $g\left(x\right)<-8$`g`(`x`)<−8.

$\left(-\infty,-4\right)\cup\left(4,\infty\right)$(−∞,−4)∪(4,∞)

A$\left(-\infty,-4\right)\cap\left(4,\infty\right)$(−∞,−4)∩(4,∞)

B$\left(-\infty,4\right)\cup\left(-4,\infty\right)$(−∞,4)∪(−4,∞)

C$\left(-\infty,4\right)\cap\left(-4,\infty\right)$(−∞,4)∩(−4,∞)

D$\left(-\infty,-4\right)\cup\left(4,\infty\right)$(−∞,−4)∪(4,∞)

A$\left(-\infty,-4\right)\cap\left(4,\infty\right)$(−∞,−4)∩(4,∞)

B$\left(-\infty,4\right)\cup\left(-4,\infty\right)$(−∞,4)∪(−4,∞)

C$\left(-\infty,4\right)\cap\left(-4,\infty\right)$(−∞,4)∩(−4,∞)

D