MTHS 105 - Applied College Algebra

Question 1

One number is seven more than four times another number. If the sum of the numbers is 22, find the other numbers.

\[\begin{align*} x+(4x+7) = 22 & \hspace{1.3cm}\mbox{Set Up Equation}\\ x+4x+7 = 22 & \hspace{1.3cm}\mbox{Drop Parenthesis}\\ 5x+7= 22 & \hspace{1.3cm}\mbox{Combine Like Terms}\\ 5x =15 & \hspace{1.3cm}\mbox{Subtract 7 on Both Sides}\\ x= 3 & \hspace{1.3cm}\mbox{Divide by 3}\\ \end{align*}\]

Therefore, the other number is \[22 - 3 = 19\]

Question 2

Solve the following equation for \(x\). Answer as a fraction simplified in lowest terms

\[\begin{align*} -\frac{2x}{3}-\frac{7}{3}=-2 & \hspace{1.3cm}\mbox{Original Problem}\\[1ex] -\frac{2x}{3}=-2+\frac{7}{3} & \hspace{1.3cm}\mbox{Add $\frac{7}{3}$ to Both Sides}\\[1ex] -\frac{2x}{3}=-\frac{6}{3}+\frac{7}{3} & \hspace{1.3cm}\mbox{Find LCD}\\[1ex] -\frac{2x}{3}=\frac{1}{3} & \hspace{1.3cm}\mbox{Combine Fractions}\\[1ex] -6x = 3 & \hspace{1.3cm}\mbox{Cross Multiplied}\\[1ex] x = -\frac{1}{2} & \hspace{1.3cm}\mbox{Divide by $-6$, Simplify, Final Answer} \end{align*}\]

Question 3

Solve the following equation for \(x\). Leave your answer as a fraction, simplified in lowest terms.

\[\begin{align*} -4(x-3)-1=4-7(x-5) & \hspace{1.3cm}\mbox{Original Problem}\\[1ex] -4x+12-1=4-7x+35 & \hspace{1.3cm}\mbox{Distributed}\\[1ex] -4x+11=-7x+39 & \hspace{1.3cm}\mbox{Combine Like Terms}\\[1ex] 3x+11=39 & \hspace{1.3cm}\mbox{Added $7x$}\\[1ex] 3x=28 & \hspace{1.3cm}\mbox{Subtracted $11$}\\[1ex] x=\frac{28}{3} & \hspace{1.3cm}\mbox{Divide by 3, Final Answer}\\[1ex] \end{align*}\]

Question 4

Find the average rate of change of the given function on the interval \([0,5]\). Simplify your answer.

The function is \[h(x)=x^3+2x^2-x-1\]

The Formula for Average Rate of Change is:

\[\frac{f(x_2)-f(x_1)}{x_2-x_1}\]

Substitute the interval values into the function, you’ll get:

\[\begin{align*} \frac{(5)^3+2(5)^2-(5)-1 - [(0)^3+2(0)^2-0-1]}{5-0} & \hspace{1.3cm}\mbox{Substituted}\\[1ex] \frac{125+2(25)-(5)-1 -[-1]}{5}& \hspace{1.3cm}\mbox{Simplified}\\[1ex] \frac{125+50-5-1+1}{5} & \hspace{1.3cm}\mbox{Simplified Further}\\[1ex] \frac{170}{5} & \hspace{1.3cm}\mbox{Simplified Again}\\[1ex] \frac{f(x_2)-f(x_1)}{x_2-x_1} = 34 & \hspace{1.3cm}\mbox{Final Answer}\\ \end{align*}\]

Question 5

Given the graph of the function \(f(x)\) as shown below. Find the average rate of change of the function \(f(x)\) on the interval \([2,6]\). Simplify your answer.

Caption: Graph of a Function f(x)

The ordered pairs for each ends of the interval on \([2,6]\) is \((2, 10)\) and \((6,30)\)

The Formula for Average Rate of Change is:

\[\frac{f(x_2)-f(x_1)}{x_2-x_1}\]

Substitute the interval values into the function, you’ll get:

\[\begin{align*} \frac{30-10}{6-2}& \hspace{1.3cm}\mbox{Substituted}\\[1ex] \frac{20}{4} & \hspace{1.3cm}\mbox{Simplified}\\[1ex] 5 & \hspace{1.3cm}\mbox{Final Answer}\\ \end{align*}\]

Question 6

Given the table of values, find the average rate of change of the function \(f(x)\) on the interval \([2,7]\).

\(x\)	1	2	3	4	5	6	7	8	9
\(f(x)\)	10	16	19	21	26	29	36	36	40

The ordered pairs for each ends of the interval on \([2,7]\) is \((2, 16)\) and \((7,36)\)

The Formula for Average Rate of Change is:

\[\frac{f(x_2)-f(x_1)}{x_2-x_1}\]

Substitute the interval values into the function, you’ll get:

\[\begin{align*} \frac{36-16}{7-2} & \hspace{1.3cm}\mbox{Substituted}\\[1ex] \frac{20}{5}& \hspace{1.3cm}\mbox{Simplified}\\[1ex] 4& \hspace{1.3cm}\mbox{Final Answer}\\[1ex] \end{align*}\]

Question 7

A college student takes the same number of credits each semester. She had 12 credits when she started, and after 3 semesters, she had 54 credits. Express the rate at which she is earning credits? Be sure to include units in your answer.

The ordered pairs for this problem are \((0, 12)\) and \((3,54)\)

The Formula for Rate of Change is:

\[\frac{y_2-y_1}{x_2-x_1}\]

Substitute the interval values into the formula, you’ll get:

\[\begin{align*} \frac{54-12}{3-0} & \hspace{1.3cm}\mbox{Substituted}\\[1ex] \frac{42}{3}& \hspace{1.3cm}\mbox{Simplified}\\[1ex] 14& \hspace{1.3cm}\mbox{Final Answer}\\[1ex] \end{align*}\]

She is earning 14 credits per semester.

Question 8

John is saving money at a constant rate. Suppose he initially has $110 saved, and after 2 months, he has $160 saved. Expresses the rate at which John is saving? Be sure to include units in your answer.

The ordered pairs for this problem are \((0, 110)\) and \((2,160)\)

The Formula for Rate of Change is:

\[\frac{y_2-y_1}{x_2-x_1}\]

Substitute the interval values into the formula, you’ll get:

\[\begin{align*} \frac{160-110}{2-0} & \hspace{1.3cm}\mbox{Substituted}\\[1ex] \frac{50}{2}& \hspace{1.3cm}\mbox{Simplified}\\[1ex] 25& \hspace{1.3cm}\mbox{Final Answer}\\[1ex] \end{align*}\]

Therefore, he is saving at a rate of $25 per month