ORCCA (Draft)Open Resources for Community College Algebra

Evaluating Expressions

When we evaluate an expression's value, we substitute each variable with its given value.

Example

Evaluate the value of \(\frac{5}{9}(F - 32)\) if \(F=212\text{.}\)

Substituting a Negative Number

When we substitute a variable with a negative number, it's important to use parentheses around the number.

Example

Evaluate the following expressions if \(x=-3\text{.}\)

1. \begin{align*} x^2\amp=(-3)^2\\ \amp=9 \end{align*}

2. \begin{align*} x^3\amp=(-3)^3\\ \amp=(-3)(-3)(-3)\\ \amp=-27 \end{align*}

3. \begin{align*} -x^2\amp=-(-3)^2\\ \amp=-9 \end{align*}

4. \begin{align*} -x^3\amp=-(-3)^3\\ \amp=-(-27)\\ \amp=27 \end{align*}

Checking Possible Solutions

Given an equation or an inequality (with one variable), checking if some particular number is a solution is just a matter of replacing the value of the variable with the specified number and determining if the resulting equation/inequality is true or false. This may involve some amount of arithmetic simplification.

Example

Is \(-5\) a solution to \(2(x+3)-2=4-x\text{?}\)

To find out, substitute in \(-5\) for \(x\) and see what happens.

So no, \(-5\) is not a solution to \(2(x+3)-2=4-x\text{.}\)

Solving One-Step Equations

When we solve linear equations, we use Properties of Equivalent Equations and follow Steps to Solving Simple (One-Step) Linear Equations.

Example

Solve for \(g\) in \(\frac{1}{2}=\frac{2}{3}+g\text{.}\)

We will subtract \(\frac{2}{3}\) on both sides of the equation:

We will check the solution by substituting \(g\) in the original equation with \(-\frac{1}{6}\text{:}\)

The solution \(-\frac{1}{6}\) is checked and the solution set is \(\left\{-\frac{1}{6}\right\}\text{.}\)

Solving One-Step Inequalities

When we solve linear inequalities, we also use Properties of Equivalent Equations with one small complication: When we multiply or divide by the same negative number on both sides of an inequality, the direction reverses!

Example

Solve the inequality \(-2x\geq12\text{.}\) State the solution set with both interval notation and set-builder notation.

To solve this inequality, we will divide each side by \(-2\text{:}\)

• The inequality's solution set in interval notation is \((-\infty,-6]\text{.}\)

• The inequality's solution set in set-builder notation is \(\{x\mid x\leq-6\}\text{.}\)

Solving One-Step Equations Involving Percentages

An important skill for solving percent-related problems is to boil down a complicated word problem into a simple form like “\(2\) is \(50\%\) of \(4\text{.}\)”

Example

What percent of \(2346.19\) is \(1995.98\text{?}\)

Using \(P\) to represent the unknown quantity, we write and solve the equation:

In summary, \(2346.19\) is approximately \(85.07\%\) of \(1995.98\text{.}\)

Modeling with Equations and Inequalities

To set up an equation modeling a real world scenario, the first thing we need to do is identifying what variable we will use. The variable we use will be determined by whatever is unknown in our problem statement. Once we've identified and defined our variable, we'll use the numerical information provided in the equation to set up our equation.

Examples

A bathtub contains 2.5 ft³ of water. More water is being poured in at a rate of 1.75 ft³ per minute. When will the amount of water in the bathtub reach 6.25 ft³?

Since the question being asked in this problem starts with “when,” we immediately know that the unknown is time. As the volume of water in the tub is measured in ft³ per minute, we know that time needs to be measured in minutes. We'll defined \(t\) to be the number of minutes that water is poured into the tub. Since each minute there are 1.75 ft³ of water added, we will add the expression \(1.75t\) to \(2.5\) to obtain the total amount of water. Thus the equation we set up is:

Rules of Exponents

Let \(x,\) and \(y\) represent real numbers, variables, or algebraic expressions, and let \(m\) and \(n\) represent positive integers. Then the following properties hold:

Product of Powers

\(x^m\cdot x^n=x^{m+n}\)

Power to Power

\((x^m)^n=x^{m\cdot n}\)

Product to Power

\((xy)^n = x^n\cdot y^n\)

Examples

Simplify the following expressions using the rules of exponents.

Algebraic Properties

Let \(a,\text{,}\) \(b\text{,}\) and \(c\) represent real numbers, variables, or algebraic expressions. Then the following properties hold:

Commutative Property of Multiplication:

\begin{equation} a\cdot b=b\cdot a\tag{2.9.1} \end{equation}

Associative Property of Multiplication:

\(a\cdot(b\cdot c)=(a\cdot b)\cdot c\)

Commutative Property of Addition:

\(a+b=b+a\)

Associative Property of Addition:

\(a+(b+c)=(a+b)+c\)

Distributive Property:

\begin{equation} a(b+c)=ab+ac\tag{2.9.2} \end{equation}

Examples

Use the associative, commutative, and distributive properties to simplify the following expression as much as possible:

Solution: We will remove parentheses by the distributive property, and then combine like terms: