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Section 1.4 Algebra Part 1 Activity

Subsection 1.4.1 Order Of Operations

Activity 1.4.1.

Evaluate this expression, noting the order of operations.

\begin{equation*} \sqrt{\frac{(2)(73) + 254}{4}} - \sqrt{9} \end{equation*}
Solution
\begin{equation*} \sqrt{\frac{(2)(73) + 254}{4}} - \sqrt{9} \end{equation*}

The second square root just has a number in it, so it can be evaluted. The first square root has several operations inside it, so those operations must be done before the square root. We'll start with the multiplication in the numerator.

\begin{equation*} = \sqrt{\frac{146 + 254}{4}} - 3 \end{equation*}

There is a division (or fraction) inside the square root. However, the numerator of the fraction has a sum in it; we must do this sum before we can complete the fraction, as it is an inside piece of the fraction.

\begin{equation*} = \sqrt{\frac{400}{4}} - 3 \end{equation*}

How we can do the division.

\begin{equation*} = \sqrt{100} - 3 \end{equation*}

Now all the pieces inside the square root have been finished and we can evaluate the square root.

\begin{equation*} = 10 - 3 = 7 \end{equation*}
Activity 1.4.2.

Evaluate this expression, noting the order of operations.

\begin{equation*} 3 - \frac{(6)(10)}{12} + (5-3)^2 \end{equation*}
Solution
\begin{equation*} 3 - \frac{(6)(10)}{12} + (5-3)^2 \end{equation*}

We'll evalute the term in brackets first.

\begin{equation*} = 3 - \frac{(6)(10)}{12} + 2^2 \end{equation*}

Next we'll evaluate the exponent.

\begin{equation*} = 3 - \frac{(6)(10)}{12} + 4 \end{equation*}

Then we can do the multiplication and divisiosn in any order. I'll do the multiplication first.

\begin{equation*} = 3 - \frac{60}{12} + 4 \end{equation*}

Then I'll do the division.

\begin{equation*} = 3 - 5 + 4 \end{equation*}

Finally, we do the addition and subtraction.

\begin{equation*} 3 - 5 + 4 = -2 + 4 = 2 \end{equation*}

Subsection 1.4.2 Solving

Activity 1.4.3.

Solve for the variable \(c\text{.}\)

\begin{equation*} 4c + 3(2-5c) + 9 = 0 \end{equation*}
Solution
\begin{equation*} 4c + 3(2-5c) + 9 = 0 \end{equation*}

First we'll distribute the 3.

\begin{equation*} 4c + 6 - 10c + 9 = 0 \end{equation*}

Then we can add the constants and the variable terms.

\begin{equation*} 4c + 6 - 10c + 9 = 0 \end{equation*}

Then we can add the constants and the variable terms.

Activity 1.4.4.

Solve for the variable \(z\text{.}\)

\begin{equation*} \end{equation*}
Solution
\begin{equation*} 4z - 8z + 5z = 9z \end{equation*}

All the terms are mutliples of the variable. Let's get them all on one side of the equation. We can do this by subtracting \(9z\) from both sides of the equation.

\begin{equation*} 4z - 8z + 5z -9z = 0 \end{equation*}

Then we can add/subtract all the termx on the right.

\begin{align*} - 4z + 5z -9z \amp = 0 \\ z -9z \amp = 0 \\ -8z \amp = 0 \end{align*}

Finally, we can divide by \(-8\)

\begin{equation*} z = \frac{0}{-8} = 0 \end{equation*}

\(z=0\) is the only solution.

Activity 1.4.5.

Solve for the variable \(r\text{.}\)

\begin{equation*} \sqrt{5r + 9} = 7 \end{equation*}
Solution
\begin{equation*} \sqrt{5r + 9} = 7 \end{equation*}

We need to get the variable out of the square root. We can do this by squaring both sides of the equation. (Again, we always perform the same operation to both sides to preserve the inequality.)

\begin{gather*} (\sqrt{5r + 9})^2 = (7)^2 \\ 5r + 9 = 49 \end{gather*}

Then we can subtract 9 from both sides of the equation.}

\begin{equation*} 5r = 40 \end{equation*}

Finally, we divide both sides of the equation by 5.

\begin{equation*} r = 8 \end{equation*}

Subsection 1.4.3 Conceptual Questions

Activity 1.4.6.
  • What is a variable? What does it mean to solve an equation?
  • How can (some) real world problems be translated into algebraic language? What becomes a constant and what becomes a variable?
  • Fundamentally, what does equality mean in mathematics? What does it mean to perform operations and preserve equality?