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Section 1.1 Algebra

Subsection 1.1.1 Sets and Number Systems

A set can be thought of as any collection of distinct objects considered as a whole. Typically, sets are represented using set-builder notation and are surrounded by braces. Recall that \((,)\) are called parentheses or round brackets; \([,]\) are called square brackets; and \(\{,\}\) are called braces or curly brackets.

Example 1.1. Sets.

The collection \(\{a,b,1,2\}\) is a set. It consists of the collection of four distinct objects, namely, \(a\text{,}\) \(b\text{,}\) \(1\) and \(2\text{.}\)

Let \(S\) be any set. We use the notation \(x\in S\) to mean that \(x\) is an element inside of the set \(S\text{,}\) and the notation \(x\not\in S\) to mean that \(x\) is not an element of the set \(S\text{.}\)

Example 1.2. Set Membership.

If \(S=\{a,b,c\}\text{,}\) then \(a\in S\) but \(d\not\in S\text{.}\)

The intersection between two sets \(S\) and \(T\) is denoted by \(S\cap T\) and is the collection of all elements that belong to both \(S\) and \(T\text{.}\) The union between two sets \(S\) and \(T\) is denoted by \(S\cup T\) and is the collection of all elements that belong to either \(S\) or \(T\) (or both).

Example 1.3. Union and Intersection.

Let \(S=\{a,b,c\}\) and \(T=\{b,d\}\text{.}\) Then \(S\cap T=\{b\}\) and \(S\cup T=\{a,b,c,d\}\text{.}\) Note that we do not write the element \(b\) twice in \(S\cup T\) even though \(b\) is in both \(S\) and \(T\text{.}\)

Numbers can be classified into sets called number systems.

Symbol Description Set notation
\(\mathbb{N} \) the natural numbers \(\{1, 2, 3,\ldots\}\)
\(\mathbb{Z} \) the integers \(\{\dots,-3,-2, -1, 0, 1, 2, 3,\dots\} \)
\(\mathbb{Q} \) the rational numbers Ratios of integers: \(\left\{\frac{p}{q}\,:\,p,q\in\mathbb{Z},q\not=0\right\} \)
\(\mathbb{R} \) the real numbers an be written using a finite or infinite decimal expansion
\(\mathbb{C} \) the complex numbers These allow us to solve equations such as \(x^2+1=0 \)

In the table, the set of rational numbers is written using set-builder notation. The colon, \(:\text{,}\) used in this manner means such that. Often times, a vertical bar \(|\) may also be used to mean such that. The expression \(\left\{\frac{p}{q}\,:\,p,q\in\mathbb{Z},q\not=0\right\}\) can be read out loud as "the set of all fractions \(p\) over \(q\) such that \(p\) and \(q\) are both integers and \(q\) is not equal to zero".

Example 1.4. Rational Numbers.

The numbers \(-\frac{3}{4}\text{,}\) \(2.647\text{,}\) \(17\text{,}\) \(0.\bar{7}\) are all rational numbers. You can think of rational numbers as fractions of one integer over another. Note that 2.647 can be written as a fraction:

\begin{equation*} 2.647=2.647\times\frac{1000}{1000}=\frac{2647}{1000}\text{.} \end{equation*}

Also note that in the expression \(0.\bar{7}\text{,}\) the bar over the \(7\) indicates that the \(7\) is repeated forever:

\begin{equation*} 0.77777777\ldots=\frac{7}{9}\text{.} \end{equation*}

All rational numbers are real numbers with the property that their decimal expansion either terminates after a finite number of digits or begins to repeat the same finite sequence of digits over and over. Real numbers that are not rational are called irrational.

Example 1.5. Irrational Numbers.

Some of the most common irrational numbers include:

  • \(\sqrt 2\text{.}\) Can you prove this is irrational? (The proof uses a technique called contradiction.)

  • \(\pi\text{.}\) Recall that \(\pi\) (pi) is defined as the ratio of the circumference of a circle to its diametre and can be approximated by \(3.14159265\text{.}\)

  • \(e\text{.}\) Sometimes called Euler's number, \(e\) can be approximated by \(2.718281828459\text{.}\) We will review the definition of \(e\) in a later chapter.

Let \(S\) and \(T\) be two sets. If every element of \(S\) is also an element of \(T\text{,}\) then we say \(S\) is a subset of \(T\) and write \(S\subseteq T\text{.}\) Furthermore, if \(S\) is a subset of \(T\) but not equal to \(T\text{,}\) we often write \(S\subset T\text{.}\) The five sets of numbers in the table give an increasing sequence of sets:

\begin{equation*} \mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R} \subset \mathbb{C}\text{.} \end{equation*}

That is, all natural numbers are also integers, all integers are also rational numbers, all rational numbers are also real numbers, and all real numbers are also complex numbers.

Subsection 1.1.2 Law of Exponents

The Law of Exponents is a set of rules for simplifying expressions that governs the combination of exponents (powers). Recall that \(\sqrt[n]{~}\) denotes the \(n\)-th root. For example \(\sqrt[3]{8}=2\) represents that the cube root of \(8\) is equal to \(2\text{.}\)

Definition 1.6. Law of Exponents.


If \(m,n\) are positive integers, then:

  1. \(x^n=x\cdot x\cdot\ldots\cdot x\) (\(n\) times)

  2. \(x^0=1\text{,}\) for \(x\neq 0\)

  3. \(\ds{x^{-n}=\frac{1}{x^n}}\text{,}\) for \(x\neq 0\)

  4. \(x^{m/n}=\sqrt[n]{x^m}\) or \(\left(\sqrt[n]{x}\right)^m\text{,}\) for \(x\geq 0\)


  1. \(x^ax^b=x^{a+b}\)

  2. \(\ds{\frac{x^a}{x^b}=x^{a-b}}\text{,}\) for \(x\neq 0\)

  3. \(\left(x^a\right)^b=x^{ab}=x^{ba}=\left(x^b\right)^a\)


  1. \((xy)^a=x^ay^a\text{,}\) for \(x\geq 0\text{,}\) \(y\geq 0\)

  2. \(\ds{\left(\frac{x}{y}\right)^a=\frac{x^a}{y^a}}\text{,}\) for \(x\geq 0\text{,}\) \(y>0\)

In the next example, the word simplify means to make simpler or to write the expression more compactly.

Example 1.7. Laws of Exponents.

Simplify the following expression as much as possible assuming \(x,y>0\text{:}\)

\begin{equation*} \frac{3x^{-2}y^3x}{y^2\sqrt x}\text{.} \end{equation*}

Using the Law of Exponents, we have:

\begin{equation*} \begin{array}{>{\displaystyle}r>{\displaystyle}c>{\displaystyle}l>{\displaystyle}l} \frac{3x^{-2}y^3x}{y^2\sqrt x} \amp = \amp \frac{3x^{-2}y^3x}{y^2x^{\frac{1}{2}}}, \amp \mbox{since \(\ds\sqrt x = x^{\frac{1}{2}}\)} ,\\ ~ \amp = \amp \frac{3x^{-2}yx}{x^{\frac{1}{2}}}, \amp \mbox{since \(\ds\frac{y^3}{y^2}=y\)} ,\\ ~ \amp = \amp \frac{3y}{x^{\frac{3}{2}}}, \amp \mbox{since \(\ds\frac{x^{-2}x}{x^{\frac{1}{2}}}=\frac{x^{-1}}{x^{\frac{1}{2}}} =x^{-\frac{3}{2}}=\frac{1}{x^{\frac{3}{2}}}\)} ,\\ ~ \amp = \amp \frac{3y}{\sqrt{x^3}}, \amp \mbox{since \(\ds x^{\frac{3}{2}}=\sqrt{x^3}\)} . \end{array} \end{equation*}

An answer of \(3yx^{-3/2}\) is equally acceptable, and such an expression may prove to be computationally simpler, although a positive exponent may be preferred.

Subsection 1.1.3 The Quadratic Formula and Completing the Square

The technique of completing the square allows us to solve quadratic equations and also to determine the center of a circle/ellipse or the vertex of a parabola.

The main idea behind completing the square is to turn:

\begin{equation*} ax^2 + bx + c \end{equation*}


\begin{equation*} a(x - h)^2 + k\text{.} \end{equation*}

One way to complete the square is to use the following formula:

\begin{equation*} ax^2+bx+c=a\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}+c\text{.} \end{equation*}

But this formula is a bit complicated, so some students prefer following the steps outlined in the next example.

Example 1.8. Completing the Square.

Solve \(2x^2+12x-32=0\) by completing the square.


In this instance, we will not divide by \(2\) first (usually you would) in order to demonstrate what you should do when the ‘\(a\)’ value is not \(1\text{.}\)

\(2x^2+12x-32=0\) Start with original equation.
\(2x^2+12x=32\) Move the number over to the other side.
\(2(x^2+6x)=32\) Factor out the \(a\) from the \(ax^2+bx\) expression.
\(6~~\to~~\frac{6}{2}=3~~\to~~3^2=9\) Take the number in front of \(x\text{,}\)
divide by \(2\),
then square it.
\(2(x^2+6x+9)=32+2\cdot9\) Add the result to both sides,
taking \(a=2\) into account.
\(2(x+3)^2=50\) Factor the resulting perfect square trinomial.
  You have now completed the square!
\((x+3)^2=25~~\to~~x=2 \mbox{ or } x=-8\) To solve for \(x\text{,}\) simply divide by \(a=2\)
and take square roots.

Suppose we want to solve for \(x\) in the quadratic equation \(ax^2+bx+c=0\text{,}\) where \(a\neq 0\text{.}\) The solution(s) to this equation are given by the quadratic formula.

The Quadratic Formula.

The solutions to \(ax^2+bx+c=0\) (with \(a\neq 0\)) are \(\ds{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}\text{.}\)

To prove the quadratic formula we use the technique of completing the square. The general technique involves taking an expression of the form \(x^2+rx\) and trying to find a number we can add so that we end up with a perfect square (that is, \((x+n)^2\)). It turns out if you add \((r/2)^2\) then you can factor it as a perfect square.

For example, suppose we want to solve for \(x\) in the equation \(ax^2+bx+c=0\text{,}\) where \(a\neq 0\text{.}\) Then we can move \(c\) to the other side and divide by \(a\) (remember, \(a\neq 0\) so we can divide by it) to get

\begin{equation*} x^2+\frac{b}{a}x=-\frac{c}{a}\text{.} \end{equation*}

To write the left side as a perfect square we use what was mentioned previously. We have \(r=(b/a)\) in this case, so we must add \((r/2)^2=(b/2a)^2\) to both sides

\begin{equation*} x^2+\frac{b}{a}x+\left(\frac{b}{2a}\right)^2=-\frac{c}{a}+\left(\frac{b}{2a}\right)^2\text{.} \end{equation*}

We know that the left side can be factored as a perfect square

\begin{equation*} \left(x+\frac{b}{2a}\right)^2=-\frac{c}{a}+\left(\frac{b}{2a}\right)^2\text{.} \end{equation*}

The right side simplifies by using the exponent rules and finding a common denominator

\begin{equation*} \left(x+\frac{b}{2a}\right)^2=\frac{-4ac+b^2}{4a^2}\text{.} \end{equation*}

Taking the square root we get

\begin{equation*} x+\frac{b}{2a}=\pm\sqrt{\frac{-4ac+b^2}{4a^2}}\text{,} \end{equation*}

which can be rearranged as

\begin{equation*} x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\text{.} \end{equation*}

In essence, the quadratic formula is just completing the square.

Subsection 1.1.4 Inequality Notation

Recall that we use the symbols \(\lt ,>,\leq,\geq\) when writing an inequality. In particular,

  • \(a\lt b\) means \(a\) is to the left of \(b\) (that is, \(a\) is strictly less than \(b\)),

  • \(a\leq b\) means \(a\) is to the left of or the same as \(b\) (that is, \(a\) is less than or equal to \(b\)),

  • \(a>b\) means \(a\) is to the right of \(b\) (that is, \(a\) is strictly greater than \(b\)),

  • \(a\geq b\) means \(a\) is to the right of or the same as \(b\) (that is, \(a\) is greater than or equal to \(b\)).

To keep track of the difference between the symbols, some students use the following mnemonic.


The \(\lt\) symbol looks like a slanted L which stands for “Less than”.

Example 1.9. Inequalities.

The following expressions are true:

\begin{equation*} 1\lt 2, -5\lt -2, 1\leq 2, 1\leq 1, 4\geq\pi>3, 7.23\geq -7.23\text{.} \end{equation*}

The real numbers are ordered and are often illustrated using the real number line:

Subsection 1.1.5 Intervals

Assume \(a,b\) are real numbers with \(a\lt b\) (i.e., \(a\) is strictly less than \(b\)). An interval is a set of every real number between two indicated numbers and may or may not contain the two numbers themselves. When describing intervals we use both round brackets and square brackets.

(1) Use of round brackets in intervals: \((~,~)\text{.}\) The notation (a,b) is what we call the open interval from a to b and consists of all the numbers between \(a\) and \(b\text{,}\) but does not include \(a\) or \(b\text{.}\) Using set-builder notation we write this as:

\begin{equation*} (a,b)=\{x\in\mathbb{R}\,:\,a\lt x\lt b\}\text{.} \end{equation*}

We read \(\{x\in\mathbb{R}\,:\,a\lt x\lt b\}\) as “the set of real numbers \(x\) such that \(x\) is greater than \(a\) and less than \(b\)” On the real number line we represent this with the following diagram:

Note that the circles on \(a\) and \(b\) are not shaded in, we call these open circles and use them to denote that \(a\text{,}\) \(b\) are omitted from the set.

(2) Use of square brackets in intervals: \([~,~]\text{.}\) The notation [a,b] is what we call the closed interval from a to b and consists of all the numbers between \(a\) and \(b\) and including \(a\) and \(b\text{.}\) Using set-builder notation we write this as

\begin{equation*} [a,b]=\{x\in\mathbb{R}\,|\,a\leq x\leq b\}\text{.} \end{equation*}

On the real number line we represent this with the following diagram:

Note that the circles on \(a\) and \(b\) are shaded in, we call these closed circles and use them to denote that \(a\) and \(b\) are included in the set.

To keep track of when to shade a circle in, you may find the following mnemonic useful:


The round brackets \((,)\) and non-shaded circle both form an “O” shape which stands for “Open and Omit”.

Note: Any set which is bound at positive and/or negative infinity is an open interval.

Taking combinations of round and square brackets, we can write different possible types of intervals (we assume \(a\lt b\)):

\((a,b)=\{x\in\mathbb{R}\,:\,a\lt x\lt b\} \) \([a,b]=\{x\in\mathbb{R}\,:\,a\leq x\leq b\} \) \([a,b)=\{x\in\mathbb{R}\,:\,a\,\leq x\lt b\}\)
\((a,b]=\{x\in\mathbb{R}\,:\,a\lt x\leq b\} \) \((a,\infty)=\{x\in\mathbb{R}\,:\,x>a\} \) \([a,\infty)=\{x\in\mathbb{R}\,:\,x\geq a\} \)
\((-\infty,b)=\{x\in\mathbb{R}\,:\,x\lt b\} \) \((-\infty,b]=\{x\in\mathbb{R}\,:\,x\leq b\} \) \((-\infty,\infty)=\mathbb{R}=\mbox{all real numbers} \)

Interactive Demonstration. Investigate interval notation by sliding the points below:

Subsection 1.1.6 Inequality Rules

Before solving inequalities, we start with the properties and rules of inequalities.

Inequality Rules.

Add/subtract a number to both sides:

  • If \(a\lt b\text{,}\) then \(a+c\lt b+c\) and \(a-c\lt b-c\text{.}\)

Adding two inequalities of the same type:

  • If \(a\lt b\) and \(c\lt d\text{,}\) then \(a+c\lt b+d\text{.}\) Add the left sides together, add the right sides together.

Multiplying by a positive number:

  • Let \(c>0\text{.}\) If \(a\lt b\text{,}\) then \(c\cdot a\lt c\cdot b\text{.}\)

Multiplying by a negative number:

  • Let \(c\lt 0\text{.}\) If \(a\lt b\text{,}\) then \(c\cdot a>c\cdot b\text{.}\) Note that we reversed the inequality symbol!

Similar rules hold for each of \(\leq\text{,}\) \(>\) and \(\geq\text{.}\)

Subsection 1.1.7 Solving Basic Inequalities

We can use the inequality rules to solve some simple inequalities.

Example 1.10. Basic Inequality.

Find all values of \(x\) satisfying

\begin{equation*} 3x+1>2x-3\text{.} \end{equation*}

Write your answer in both interval and set-builder notation. Finally, draw a number line indicating your solution set.


Subtracting \(2x\) from both sides gives \(x+1>-3\text{.}\) Subtracting \(1\) from both sides gives \(x>-4\text{.}\) Therefore, the solution is the interval \((-4,\infty)\text{.}\) In set-builder notation the solution may be written as \(\{x\in\mathbb{R}\,:\,x>-4\}\text{.}\) We illustrate the solution on the number line as follows:

Sometimes we need to split our inequality into two cases as the next example demonstrates.

Example 1.11. Double Inequalities.

Solve the inequality

\begin{equation*} 4>3x-2\geq 2x-1\text{.} \end{equation*}

We need both \(4>3x-2\) and \(3x-2\geq 2x-1\) to be true:

\begin{equation*} \begin{array}{ccc} 4>3x-2 \amp \text{and} \amp 3x-2\geq 2x-1,\\ 6>3x \amp \text{and} \amp x-2\geq -1,\\ 2>x \amp \text{and} \amp x\geq 1,\\ x\lt 2 \amp \text{and} \amp x\geq 1. \end{array} \end{equation*}

Thus, we require \(x\geq 1\) but also \(x\lt 2\) to be true. This gives all the numbers between \(1\) and \(2\text{,}\) including \(1\) but not including \(2\text{.}\) That is, the solution to the inequality \(4>3x-2\geq 2x-1\) is the interval \([1,2)\text{.}\) In set-builder notation this is the set \(\{x\in\mathbb{R}\,:\,1\leq x\lt 2\}\text{.}\)

Example 1.12. Positive Inequality.

Solve \(4x-x^2>0\text{.}\)


We provide two methods to solve this inequality.

Method 1: Factor \(4x-x^2\) as \(x(4-x)\text{.}\) The product of two numbers is positive when either both are positive or both are negative, i.e., if either \(x>0\) and \(4-x>0\text{,}\) or else \(x\lt 0\) and \(4-x\lt 0\text{.}\) The latter alternative is impossible, since if \(x\) is negative, then \(4-x\) is greater than 4, and so cannot be negative. As for the first alternative, the condition \(4-x>0\) can be rewritten (adding \(x\) to both sides) as \(4>x\text{,}\) so we need: \(x>0\) and \(4>x\) (this is sometimes combined in the form \(4>x>0\text{,}\) or, equivalently, \(0\lt x\lt 4\)). In interval notation, this says that the solution is the interval \((0,4)\text{.}\)

Method 2: Write \(4x-x^2\) as \(-(x^2-4x)\text{,}\) and then complete the square, obtaining

\begin{equation*} -\Bigl((x-2)^2-4\Bigr)=4-(x-2)^2\text{.} \end{equation*}

For this to be positive we need \((x-2)^2\lt 4\text{,}\) which means that \(x-2\) must be less than 2 and greater than \(-2\text{:}\) \(-2\lt x-2\lt 2\text{.}\) Adding 2 to everything gives \(0\lt x\lt 4\text{.}\)

Both of these methods are equally correct; you may use either in a problem of this type.

We next present another method to solve more complicated looking inequalities. In the next example we will solve a rational inequality by using a number line and test points. We follow the guideline below.

Guideline for Solving Rational Inequalities.
  1. Move everything to one side to get a \(0\) on the other side.

  2. If needed, combine terms using a common denominator.

  3. Factor the numerator and denominator.

  4. Identify points where either the numerator or denominator is \(0\text{.}\) Such points are called split points.

  5. Draw a number line and indicate your split points on the number line. Draw closed/open circles for each split point depending on if that split point satisfies the inequality (division by zero is not allowed).

  6. The split points will split the number line into subintervals. For each subinterval pick a test point and see if the expression in Step 3 is positive or negative. Indicate this with a \(+\) or \(-\) symbol on the number line for that subinterval.

  7. Now write your answer in set-builder notation. Use the union symbol \(\cup\) if you have multiple intervals in your solution.

Example 1.13. Rational Inequality.

Write the solution to the following inequality using interval notation:

\begin{equation*} \frac{2-x}{2+x}\geq 1\text{.} \end{equation*}

One method to solve this inequality is to multiply both sides by \(2+x\text{,}\) but because we do not know if \(2+x\) is positive or negative we must split it into two cases (Case 1: \(2+x>0\) and Case 2: \(2+x\lt 0\)).

Instead we follow the guideline for solving rational inequalities:

\begin{equation*} \begin{array}{rcl} \mbox{Start with original problem:} \amp ~ \amp \displaystyle{\frac{2-x}{2+x}\geq 1}\\ \\ \mbox{Move everything to one side:} \amp ~ \amp \displaystyle{\frac{2-x}{2+x}-1 \geq 0}\\ \\ \mbox{Find a common denominator:} \amp ~ \amp \displaystyle{\frac{2-x}{2+x}-\frac{2+x}{2+x} \geq 0}\\ \\ \mbox{Combine fractions:} \amp ~ \amp \displaystyle{\frac{(2-x)-(2+x)}{2+x} \geq 0}\\ \\ \mbox{Expand numerator:} \amp ~ \amp \displaystyle{\frac{2-x-2-x}{2+x} \geq 0}\\ \\ \mbox{Simplify numerator:} \amp ~ \amp \displaystyle{\frac{-2x}{2+x} \geq 0(*)} \end{array} \end{equation*}

Now we have the numerator and denominator in fully factored form. The split points are \(x=0\) (makes the numerator \(0\)) and \(x=-2\) (makes the denominator \(0\)). Let us draw a number line with the split points indicated on it:

The point \(x=0\) is included since if we sub \(x=0\) into (*) we get \(0\geq 0\) which is true. The point \(x=-2\) is not included since we cannot divide by zero. We indicate this with open/closed circles on the number line (remember that open means omit):

Now choosing a test point from each of the three subintervals we can determine if the expression \(\frac{-2x}{2+x}\) is positive or negative. When \(x=-3\text{,}\) it is negative. When \(x=-1\text{,}\) it is positive. When \(x=1\text{,}\) it is negative. Indicating this on the number line gives:

Since we wish to solve \(\frac{-2x}{2+x}\geq 0\text{,}\) we look at where the \(+\) signs are and shade that area on the number line:

Since there is a closed circle at \(0\text{,}\) we include it. Therefore, the solution is \((-2,0]\text{.}\)

Example 1.14. Rational Inequality.

Write the solution to the following inequality using interval notation:

\begin{equation*} \frac{2}{x+2}>{3}{x+3}\text{.} \end{equation*}

We provide a brief outline of the solution. By subtracting \((3x+3)\) from both sides and using a common denominator of \(x+2\text{,}\) we can collect like terms and simplify to get:

\begin{equation*} \frac{-(3x^2+9x+4)}{x+2}>0\text{.} \end{equation*}

The denominator is zero when \(x=-2\text{.}\) Using the quadratic formula, the numerator is zero when \(x=\frac{-9\pm\sqrt{33}}{6}\) (these two numbers are approximately \(-2.46\) and \(-0.54\)). Since the inequality uses “\(>\)” and \(0>0\) is false, we do not include any of the split points in our solution. After choosing suitable test points and determining the sign of \(\frac{-(3x^2+9x+4)}{x+2}\) we have

Looking where the \(+\) symbols are located gives the solution:

\begin{equation*} \left(-\infty,\frac{-9-\sqrt{33}}{6}\right)\bigcup\left(-2,\frac{-9+\sqrt{33}}{6}\right)\text{.} \end{equation*}

When writing the final answer we use exact expressions for numbers in mathematics, not approximations (unless stated otherwise).

Subsection 1.1.8 The Absolute Value

The absolute value of a number \(x\) is written as \(|x|\) and represents the distance \(x\) is from zero. Mathematically, we define it as follows:

\begin{equation*} |x|=\left\{\begin{array}{cl} x, \amp \mbox{if \(x\geq 0\),} \\ -x, \amp \mbox{if \(x\lt 0\).} \end{array} \right. \end{equation*}

Thus, if \(x\) is a negative real number, then \(-x\) is a positive real number. The absolute value does not just turn minuses into pluses. That is, \(|2x-1|\neq 2x+1\text{.}\) You should be familiar with the following properties.

Absolute Value Properties.
  1. \(|x|\geq 0\text{.}\)

  2. \(|xy|=|x||y|\text{.}\)

  3. \(|1/x|=1/|x|\) when \(x\neq 0\text{.}\)

  4. \(|-x|=|x|\text{.}\)

  5. \(|x+y|\leq |x|+|y|\text{.}\) This is called the triangle inequality.

  6. \(\sqrt{x^2}=|x|\text{.}\)

Example 1.15. \(\sqrt{x^2}=|x|\).

Observe that \(\sqrt{(-3)^2}\) gives an answer of \(3\text{,}\) not \(-3\text{.}\)

When solving inequalities with absolute values, the following are helpful.

  • Case 1: \(a>0\)
    • \(|x|=a\) has solutions \(x=\pm a\text{.}\)

    • \(|x|\leq a\) means \(x\geq -a\) and \(x\leq a\) (that is, \(-a\leq x\leq a\)).

    • \(|x|\lt a\) means \(x\lt -a\) and \(x\lt a\) (that is, \(-a\lt x\lt a\)).

    • \(|x|\geq a\) means \(x\leq -a\) or \(x\geq a\text{.}\)

    • \(|x|> a\) means \(x\lt -a\) or \(x> a\text{.}\)

  • Case 2: \(a\lt 0\)
    • \(|x|=a\) has no solutions.

    • Both \(|x|\leq a\) and \(|x|\lt a\) have no solutions.

    • Both \(|x|\geq a\) and \(|x|>a\) have solution set \(\{x|x\in\R\}\text{.}\)

  • Case 3: \(a=0\)
    • \(|x|=0\) has solution \(x=0\text{.}\)

    • \(|x|\lt 0\) has no solutions.

    • \(|x|\leq 0\) has solution \(x=0\text{.}\)

    • \(|x|>0\) has solution set \(\{x\in\R|x\neq 0\}\text{.}\)

    • \(|x|\geq 0\) has solution set \(\{x|x\in\R\}\text{.}\)

Subsection 1.1.9 Solving Inequalities that Contain Absolute Values

We start by solving an equality that contains an absolute value. To do so, we recall that if \(a\geq 0\) then the solution to \(|x|=a\) is \(x=\pm a\text{.}\) In cases where we are not sure if the right side is positive or negative, we must perform a check at the end.

Example 1.16. Absolute Value Equality.

Solve for \(x\) in \(|2x+3|=2-x\text{.}\)


This means that either:

\begin{equation*} \begin{array}{ccc} 2x+3=+(2-x) \amp \text{or} \amp 2x+3=-(2-x)\\ 2x+3=2-x \amp \text{or} \amp 2x+3=-2+x\\ 3x=-1 \amp \text{or} \amp x=-5\\ x=-1/3 \amp \text{or} \amp x=-5 \end{array} \end{equation*}

Since we do not know if the right side \(a=2-x\) is positive or negative, we must perform a check of our answers and omit any that are incorrect.

If \(x=-1/3\text{,}\) then we have \(LS=|2(-1/3)+3|=|-2/3+3|=|7/3|=7/3\) and \(RS=2-(-1/3)=7/3\text{.}\) In this case \(LS=RS\text{,}\) so \(x=-1/3\) is a solution.

If \(x=-5\text{,}\) then we have \(LS=|2(-5)+3|=|-10+3|=|-7|=7\) and \(RS=2-(-5)=2+5=7\text{.}\) In this case \(LS=RS\text{,}\) so \(x=-5\) is a solution.

We next look at absolute values and inequalities.

Example 1.17. Absolute Value Inequality.

Solve \(|x-5|\lt 7\text{.}\)


This simply means \(-7\lt x-5\lt 7\text{.}\) Adding \(5\) to each gives \(-2\lt x\lt 12\text{.}\) Therefore the solution is the interval \((-2,12)\text{.}\)

In some questions you must be careful when multiplying by a negative number as in the next problem.

Example 1.18. Absolute Value Inequality.

Solve \(|2-z|\lt 7\text{.}\)


This simply means \(-7\lt 2-z\lt 7\text{.}\) Subtracting \(2\) gives: \(-9\lt -z\lt 5\text{.}\) Now multiplying by \(-1\) gives: \(9>z>-5\text{.}\) Remember to reverse the inequality signs! We can rearrange this as \(-5\lt z\lt 9\text{.}\) Therefore the solution is the interval \((-5,9)\text{.}\)

Example 1.19. Absolute Value Inequality.

Solve \(|2-z|\geq 7\text{.}\)


Recall that for \(a>0\text{,}\) \(|x|\geq a\) means \(x\leq -a\) or \(x\geq a\text{.}\) Thus, either \(2-z\leq -7\) or \(2-z\geq 7\text{.}\) Either \(9\leq z\) or \(-5 \geq z\text{.}\) Either \(z\geq 9\) or \(z \leq -5\text{.}\) In interval notation, either \(z\) is in \([9,\infty)\) or \(z\) is in \((-\infty,-5]\text{.}\) All together, we get our solution to be: \((-\infty,-5]\cup [9,\infty)\text{.}\)

In the previous two examples the only difference is that one had \(\lt\) in the question and the other had \(\geq\text{.}\) Combining the two solutions gives the entire real number line!

Example 1.20. Absolute Value Inequality.

Solve \(0\lt |x-5|\leq 7\text{.}\)


We split this into two cases.

(1) For \(0\lt |x-5|\) note that we always have that an absolute value is positive or zero (i.e., \(0\leq |x-5|\) is always true). So, for this part, we need to avoid \(0=|x-5|\) from occurring. Thus, \(x\) cannot be \(5\text{,}\) that is, \(x\neq 5\text{.}\)

(2) For \(|x-5|\leq 7\text{,}\) we have \(-7\leq x-5\leq 7\text{.}\) Adding \(5\) to each gives \(-2\leq x\leq 12\text{.}\) Therefore the solution to \(|x-5|\leq 7\) is the interval \([-2,12]\text{.}\)

To combine (1) and (2) we need combine \(x\neq 5\) with \(x\in[-2,12]\text{.}\) Omitting \(5\) from the interval \([-2,12]\) gives our solution to be: \([-2,5)\cup(5,12]\text{.}\)

Exercises for Section 1.1.

Simplify the following expressions as much as possible assuming \(x,y>0\text{:}\)

  1. \(\ds\frac{x^3y^{-1/3}}{\sqrt[3]{y^2}x^2}\)

  2. \(\ds\frac{3x^{-1/3}y^{-2}\sqrt[3]{x^4}}{\sqrt{9x}y^{-3}}\)

  3. \(\ds\left(\frac{16x^2y}{x^4}\right)^{1/2}\frac{\sqrt[3]{x^2}}{2\sqrt{y}}\)


Find the constants \(a,b,c\) if the expression

\begin{equation*} \frac{4x^{-1}y^2\sqrt[3]{x}}{2x\sqrt{y}} \end{equation*}

is written in the form \(ax^by^c\text{.}\)

\(a=2\text{,}\) \(b=-\frac{5}{3}\text{,}\) \(c=\frac{3}{2}\text{.}\)

We want to put

\begin{equation*} \dfrac{4x^{-1}y^{2}\sqrt[3]{x}}{2x\sqrt{y}} \end{equation*}

in the form \(ax^{b}y^{c}\text{.}\) First, note that we may write

\begin{equation*} \frac{1}{x} = x^{-1} \ \ \text{ and } \ \ \frac{1}{\sqrt{y}} = y^{-\frac{1}{2}}\text{.} \end{equation*}

By rearranging and using the above facts, we can simplify the expression:

\begin{equation*} \dfrac{4x^{-1}y^{2}\sqrt[3]{x}}{2x\sqrt{y}} = \left(\dfrac{4x^{-1}\sqrt[3]{x}}{2x}\right) \left(\dfrac{y^{2}}{\sqrt{y}}\right) = \frac{4}{2} \left(x^{-1} x^{-1} x^{\frac{1}{3}} \right) \left(y^{2} y^{-\frac{1}{2}} \right) = 2 x^{-\frac{5}{3}} y^{\frac{3}{2}}\text{.} \end{equation*}

And so it must be that \(a = 2\) , \(b = -\frac{5}{3}\) , and \(c = \frac{3}{2}\text{.}\)

Find the roots of the quadratic equation

\begin{equation*} x^2-2x-24=0\text{.} \end{equation*}
\(x=-4\) and \(x=6\text{.}\)

Solve the equation

\begin{equation*} \frac{x}{4x-16}-2=\frac{1}{x-3}\text{.} \end{equation*}
\(x = \dfrac{7}{2} \pm \dfrac{\sqrt{\frac{23}{7}}}{2} = \dfrac{7}{2} \pm \sqrt{\frac{23}{28}} \)

We wish to find all solutions to

\begin{equation*} \dfrac{x}{4x-16} - 2 = \dfrac{1}{x-3}\text{.} \end{equation*}

First, we rewrite the above equation as

\begin{equation*} \dfrac{x}{4x-16} - \dfrac{1}{x-3} = 2\text{,} \end{equation*}

and group the left-hand side terms using a common denominator:

\begin{equation*} \begin{split} \dfrac{x(x-3) - (4x-16)}{(4x-16)(x-3)} \amp = 2 \\ \dfrac{x^{2} - 7x + 16}{(4x-16)(x-3)} \amp = 2 \end{split} \end{equation*}

Now, for \(x \neq 4\) and \(x \neq 3\text{,}\) we can write this as

\begin{equation*} \begin{split} x^{2} - 7x + 16 \amp = 2 (4x-16)(x-3) \\ x^{2} - x + 16 \amp = 8x^{2} - 56x + 96 \\ -7x^{2}+49x - 80 \amp = 0. \end{split} \end{equation*}

Applying the quadratic formula, we find two solutions:

\begin{equation*} x = \dfrac{7}{2} \pm \dfrac{\sqrt{\frac{23}{7}}}{2} = \dfrac{7}{2} \pm \sqrt{\frac{23}{28}} \end{equation*}

Solve the following inequalities. Write your answer as a union of intervals.

  1. \(3x+1>6\)


    \(3x+1 > 6\) \(\implies x > \frac{5}{3}\text{.}\) All solutions therefore lie in the interval \(\left(\frac{5}{3}, \infty \right)\text{.}\)

  2. \(0\leq 7x-1 \lt 1\)


    \(7x - 1 \lt 1\) \(\implies x \lt \frac{2}{7}\) and \(7x - 1 \geq 0\) \(\implies x \geq \frac{1}{7}\text{.}\) So, \(0 \leq 7x-1 \lt 1\) when \(x \in \left[ \frac{1}{7}, \frac{2}{7} \right)\text{.}\)

  3. \(\ds{\frac{x^2(x-1)}{(x+2)(x+3)^3}\leq 0}\)


    We want to find where

    \begin{equation*} \dfrac{x^{2}(x-1)}{(x+2)(x+3)^{3}} \leq 0\text{.} \end{equation*}

    First, look at the denominator. Notice that it is negative on the interval \(\left(-3,-2\right)\text{,}\) zero when \(x = -3\text{,}\) \(-2\) (there are no solutions for these \(x\)-values) and positive everywhere else. Consider the positive case. Here, the expression reduces to

    \begin{equation*} x^{2}(x-1) \leq 0\text{.} \end{equation*}

    Since \(x^{2}\) is always positive, this inequality is true where \(x \leq 1\text{.}\) This then gives solutions in \(\left(-\infty, -3 \right) \cup \left(-2, 1\right]\text{.}\) Now consider the negative case. Here, the expression reduces to

    \begin{equation*} x^{2}(x-1) \geq 0\text{,} \end{equation*}

    which would require \(x \geq 1\text{,}\) so this gives no more solutions in the interval \(\left(-3,-2\right)\text{.}\) It follows that all solutions lie in \(\left(-\infty, -3 \right) \cup \left(-2, 1\right]\text{.}\)

  4. \(x^2+1>0\)


    \(x^{2} + 1 > 0 \implies x^{2} > -1\text{,}\) which is true for all \(x \in \mathbb{R}\text{.}\) So \(x \in \left(-\infty, \infty \right)\text{.}\)

  5. \(x^2+1\lt 0\)

    No solution

    \(x^{2} + 1 \lt 0 \implies x^{2} \lt -1\text{,}\) which is not true for any \(x \in \mathbb{R}\text{.}\) No solution!

  6. \(x^2+1>2x\)

  7. \(x^3>4x\)

  8. \(x^3\geq4x^2\)

  9. \(\dfrac{1}{x}>2\)

  10. \(\dfrac{x}{x+2}\leq\dfrac{2}{x-1}\)


Solve the equation \(|6x+2|=1\text{.}\)

\(x=-\frac{1}{2}\) and \(x=-\frac{1}{6}\text{.}\)

Find solutions to the following absolute value inequalities. Write your answer as a union of intervals.

  1. \(|x|\geq 2\)


    We want to find where

    \begin{equation*} \left| x + 2 \right| \lt 3x + 6\text{.} \end{equation*}

    Notice that there is one intersection point, \(x = -2\text{.}\) For \(x > -2\text{,}\) the inequality is equivalent to

    \begin{equation*} x + 2 \lt 3x + 6\text{,} \end{equation*}

    which is true for all \(x\) in this interval. For \(x \lt -2\text{,}\) notice that \(3x+6\) is negative, and so cannot be greater than \(\left| \cdot \right|\text{.}\) Thus, the solutions lie in \(\left(-2, \infty \right)\text{.}\)

  2. \(|x-3|\leq 1\)

  3. \(|2x+5|\geq 4\)

  4. \(|x+2|\lt 3x-6\)

  5. \(|2x+5|+4\geq 1\)


    First, we rearrange

    \begin{equation*} \left| 2x + 5 \right| + 4 \geq 1\text{,} \end{equation*}


    \begin{equation*} \left| 2x + 5 \right| \geq -3\text{.} \end{equation*}

    As we had noted above, \(\left| \cdot \right|\) will always give a positive output, and so this inequality will be true for all \(x \in \left(-\infty, \infty \right)\text{.}\)

  6. \(5\lt |x+1|\lt 8\)


    \(\left| x+1 \right| > 5\) when \(x \in \left(-\infty, -6\right) \cup \left(4, \infty \right)\) and \(\left| x+1 \right| \lt 8\) when \(x \in \left(-9, 7 \right)\text{.}\) So, \(5 \lt \left| x+1 \right| \lt 8\) when \(x \in \left(-9, -6\right) \cup \left(4, 7 \right)\text{.}\)

Solve the equation \(\sqrt{1-x}+x=1\text{.}\)

\(x = 0\) and \(x = 1\text{.}\)