## MAT 116 Course Prep

This collection of problems is meant to provide practice for a range of skills needed for MAT 116. Read and attempt each problem first; if you aren’t sure how to start a problem, explore the resources on the right to refresh your memory and try again. A drop-down button is found beneath each problem for you to compare both your logic and final answers. Keep track of the skills you aren’t comfortable with, and reacquaint yourself with them so you’re fully prepared for the topics you’ll grapple with soon. Get help in the Math Lab.  Return to the Course Prep page.

Problem 1: Identify whether the given statement is an expression or an equation. If it is an equation, determine if it is also an identity. Determine whether each term in the statement is constant or variable.

 Part a $x+3=5$ Click here to show solution. Solution: A mathematical expression is just that: a “phrase” which represents a value. An equation is created when two expressions are set equal to each other – therefore, the easiest way to differentiate an expression from an equation is simply seeing whether or not an equal sign is present. An equation is an identity if the equality holds for all possible values of the variables involved. A constant term has a value which cannot change (e.g., “5”), while a variable term can represent different values (e.g., $7n$ with different choices of $n$). This first statement is an equation since there is an equal sign, but it is not an identity since we can choose a value of $x$ that breaks the equality. For example, if $x = 1$, then the statement claims that $1+3=5$, which is obviously not true. The term $x$ is variable, while the terms 3 and 5 are constant. Video: Variables, Expressions, and Equations Part b $3k+2m$ Click here to show solution. Solution: This is an expression since no equal sign is present. By extension, it is not an identity, since identities must first be equations. Both terms are variable here. Part c $45$ Click here to show solution. Solution: This is an expression, even though no variables are present. The only term, 45, is a constant. Part d $n^{10}+1.5$ Click here to show solution. Solution: This is an expression. The term $n^{10}$ is variable, while $1.5$ is constant. Part e $q^3-5q^2$ Click here to show solution. Solution: This is also an expression since there is still no equal sign. Both terms are variable. Part f $\frac{36}{79}y = 11$ Click here to show solution. Solution: This is an equation, but not an identity since the equality does not hold for all values of $y$. Choose $y=1$ for example, which implies that $\frac{36}{79}=11$. This is not true, and therefore the equation is not an identity. The term on the left hand side of this equation is variable, while the one on the right is constant. Part g $xy-2b$ Click here to show solution. Solution: This is an expression. Both terms are variable. Part h $2x+1=4x+2$ Click here to show solution. Solution: This is an equation, though again not an identity. Choose $x=1$, which implies that $2(1)+1=4(1)+2 \Rightarrow 3 = 7$. This is not true, and therefore the equation is not an identity. The terms $2x$ and $4x$ are variable, while 1 and 2 are constant.

Problem 2: Rewrite the following expressions in the form $\boldsymbol{kx^{n}}$, where $\boldsymbol{k}$ and $\boldsymbol{n}$ are real numbers.

 Part a $\sqrt{x}$ Click here to show solution. Solution: In general, the $\textit{n}$th root of a number $x$ can be expressed as $x^{\frac{1}{n}}$. In this case, the simplified form is $x^{\frac{1}{2}}$, where $k=1,n=\frac{1}{2}$. Video: Intro to Rational Exponents Part b $\sqrt[4]{x^3}$ Click here to show solution. Solution: Following the same rule as the previous question, rewrite this first as $(x^3)^{\frac{1}{4}}$. Exponent rules dictate that in general, $(x^a)^b = x^{a\cdot b}$, and so this simplifies to $x^{\frac{3}{4}}$, where $k=1,n=\frac{3}{4}$. Video: Rewriting Roots as Rational Exponents Part c $\sqrt[6]{x^{24}}$ Click here to show solution. Solution: Following the same rules as the previous question, the expression first reduces to $x^{\frac{24}{6}}$, and simplification of the exponent then gives the answer as $x^{4}$, where $k=1,n=4$. Part d $\frac{1}{x^4}$ Click here to show solution. Solution: In general, $\frac{1}{x^a}=x^{-a}$. The answer is therefore $x^{-4}$, where $k=1,n=-4$. Video: Exponents in denominators Part e $\frac{3}{x^7}$ Click here to show solution. Solution: Rewrite the expression as $3\frac{1}{x^7}$. Using the same rule as the previous question, the expression reduces to $3(x^{-7})$, where $k=3,n=-7$. Part f $-\frac{5}{x^2}$ Click here to show solution. Solution: Rewrite the expression as $-5(\frac{1}{x^2})$. This then simplifies to $-5(x^{-2})$, where $k=-5,n=-2$. Part g $\frac{x^3}{x}$ Click here to show solution. Solution: Assuming $x \neq 0$, this simply reduces to $x^2$ as the numerator and denominator share common factor $x$. In this case, $k=1,n=2$. Video: Intro to Rational Expression Simplification Part h $\frac{x^3}{x^{10}}$ Click here to show solution. Solution: Here, the numerator and denominator share common factor $x^3$. Assuming $x \neq 0$, cancellation gives $\frac{1}{x^7}$, which then simplifies to $x^{-7}$, where $k=1,n=-7$.

Problem 3: Given that $\boldsymbol{f(x)=x^2}$, $\boldsymbol{g(x) = -x^2}$, $\boldsymbol{h(x)=1-2x}$, simplify the following expressions. If possible, find exact values.

 Part a $f(-1)$ Click here to show solution. Solution: $f(-1) = (-1)^2 = 1.$ Video: Worked Example: Evaluating Functions from Equation Part b $g(-1)$ Click here to show solution. Solution: $g(-1) = -(-1)^2 = -(1) = -1.$ Part c $h(-1)$ Click here to show solution. Solution: $h(-1) = 1-2(-1) = 1+2 = 3.$ Part d $f(x)-g(x)$ Click here to show solution. Solution: $f(x)-g(x) = (x^2) - (-x^2) = x^2+x^2=2x^2.$ Since no value of $x$ is given, an exact value is not possible as an answer. Part e $g(x)+f(x)$ Click here to show solution. Solution: $g(x)+f(x) = (-x^2)+x^2 = 0.$ Part f $\frac{f(x)}{g(x)}$ Click here to show solution. Solution: $\frac{f(x)}{g(x)} = \frac{x^2}{-x^2} = -1,$ since the top and bottom share a common factor of $x^2$ which cancels (assuming $x \neq 0$). Part g $f(x)g(x)$ Click here to show solution. Solution: $f(x)g(x)=(x^2)(-x^2)=-x^4$. Part h $g(x)h(x)$ Click here to show solution. Solution: $g(x)h(x) = (-x^2)(1-2x) = -x^2+2x^3,$ by the distributive property.

Problem 4: Solve the following equations for x, and verify your solution.

 Part a $100 = 0.2x$ Click here to show solution. Solution: Divide both sides by 0.2: $500=x$. To verify, plug this value of $x$ back into the original question and make sure that the statement is true. In this case, the choice of $x = 500$ gives us $100 = 0.2(500) \Rightarrow 100 = 100$. Since this is true, we know our solution is correct. To give you an idea of what it would look like if we somehow got the incorrect answer, pretend we solved the original equation and got $x = 80$. Then, in the verification step, we would have $100 = 0.2(80) \Rightarrow 100 = 16$. This is obviously false, and so $x= 80$ is not a valid solution to the equation. Video: Solving Linear Equations I Write-up: One Step Equations Part b $-80=\frac{4}{3}x$ Click here to show solution. Solution: Divide both sides by $\frac{4}{3}$. This has the effect of multiplying each side by $\frac{3}{4}$: $\frac{-240}{4} = x$, which reduces to show that $x=-60$. We verify by plugging this value back into the original question: $-80 = \frac{4}{3}(-60) \Rightarrow -80 = \frac{-240}{3} \Rightarrow -80 = -80$. This is true, and so we’ve verified our answer. Part c $44 = 8-6x$ Click here to show solution. Solution: Begin by subtracting 8 from each side, yielding $36 = -6x$. Division by -6 gives $x=-6$. Verify the solution by plugging this value of $x$ back into the original equation: $44 = 8-6(-6) \Rightarrow 44 = 8-(-36) \Rightarrow 44 = 8+36 \Rightarrow 44=44$. This is true, and so we’ve verified our answer. Video: Solving Linear Equations II Write-up: Two Step Equations Part d $\frac{x^2-6x}{x}=3$ Click here to show solution. Solution: On the left-hand side, a factor of $x$ is common to each term in the numerator, and cancels with the $x$ found in the denominator assuming $x \neq 0$: $\frac{x^2-6x}{x} = \frac{x(x-6)}{x} = x-6$. Simplification gives $x-6=3$, and adding 6 to both sides gives $x=9$. Verify the solution by substitution: $\frac{(9)^2-6(9)}{9} = 3 \Rightarrow \frac{81-54}{9} = 3 \Rightarrow \frac{27}{9} = 3 \Rightarrow 3 = 3$. This is true, and so we’ve verified our answer. Part e $x^5=243$ Click here to show solution. Solution: To isolate the $x$, raise both sides to the $\frac{1}{5}$ power. This gives $x=243^{\frac{1}{5}} = \sqrt[5]{243} = 3$. We plug in this value of $x$ to verify the answer: $(3)^5 = 243 \Rightarrow 3\cdot3\cdot3\cdot3\cdot3=243 \Rightarrow (9)\cdot3\cdot3\cdot3 = 243 \Rightarrow (27)\cdot3\cdot3 = 243 \Rightarrow \dots \Rightarrow 243 = 243$. This is true, and so we’ve verified our answer. Video: Solving Equations with Rational Exponents Part f $x^9=27.8$ Click here to show solution. Solution: In a similar fashion as the previous problem, raise both sides to the $\frac{1}{9}$ power, giving $x=27.8^{\frac{1}{9}} = \sqrt[9]{27.8}$. This is approximately 1.447. A calculator is suggested to verify this one: $(1.447)^9 = 27.8 \Rightarrow 27.811 = 27.8$. Our solution was rounded to begin with, which creates the small discrepancy in the verification, but this is approximately correct. Part g $xy=k$ Click here to show solution. Solution: In earlier questions, $x$ was the only variable present, and “solving the equation” ended with a numerical solution. In cases like these, we must recall that “solving for $x$” is synonymous with isolating $x$ on one side of the equation. In this equation, divide both sides by $y$: $\frac{xy}{y} = \frac{k}{y} \Rightarrow x = \frac{k}{y}$ due to the cancellation on the left-hand side of the equation. We can still verify this solution via substitution for $x$ in the original equation: $\frac{k}{y} y = k \Rightarrow \frac{ky}{y} = k \Rightarrow k = k$ where the final step is possible due to the cancellation of $y$‘s on the left hand side. This statement is true, and so we’ve verified our answer. Part h $y=xk$ Click here to show solution. Solution: In a similar fashion as the previous problem, we simply aim to isolate $x$ on one side of the equation. Divide by $k$ on both sides: $\frac{y}{k} = \frac{xk}{k} \Rightarrow \frac{y}{k} = x$. Verify via substitution: $y = \frac{y}{k}k \Rightarrow y = \frac{yk}{k} \Rightarrow y = y$, where the final step is possible due to the cancellation of $k$‘s on the right hand side. This statement is true, and so we’ve verified our answer.

Problem 5: Given $\boldsymbol{y=5}$, either find the value of the given expression or solve the given equation for x.

 Part a $y-y^2$ Click here to show solution. Solution: This is an expression, so we are instructed to find the value. Plugging in $y=5$, we have $5-5^2 = 5-25 = -20$. Part b $-\frac{4y}{y^3+1}$ Click here to show solution. Solution: This is also an expression, so we plug in $y=5$ and simplify: $-\frac{4(5)}{5^3+1} = -\frac{20}{126}$. This can simplify down to $-\frac{10}{63}$ since both the numerator and denominator are divisible by 2. Write-up: Simplifying Fractions Part c $5x+4y=40$ Click here to show solution. Solution: This is an equation, so we are instructed to solve for $x$. The first step is still to plug in $y=5$, giving \begin{aligned} &5x+4(5) = 40\\ \Rightarrow &5x+20 = 40\\ \Rightarrow &5x=20\\ \Rightarrow &x = 4. \end{aligned} Part d $(x+y)^2=9$ Click here to show solution. Solution: This is also an equation, though there are two common trains of thought when it comes to solving it. Both start by plugging in the known value of $y$, giving $(x+5)^2 = 9$. In the first train of thought, consider expanding the left hand side: \begin{aligned} &(x+5)(x+5) = 9\\ \Rightarrow &x^2+5x+5x+25=9\\ \Rightarrow &x^2+10x+25=9\\ \Rightarrow &x^2+10x+16=0. \end{aligned} From here, we can use the quadratic formula to find two solutions: $x=-8,-2$. Try them both and notice that plugging in both of these values into the original equation makes the statement true. In the second train of thought, note that you can take the square root of both sides of the equation, giving $x+5 = \pm 3 \Rightarrow x=(-3-5),(3-5) = -8,-2$. This is trickier! Keep in mind that the $\pm$ symbol, meaning “plus or minus”, means that the 3 here could be positive or negative. This is because $(-3)^2 = 9 = (3)^2$. Video: Worked Example: Using the Quadratic Formula Write-up: The Quadratic Formula Write-up: Solving Quadratics with the Square Root Method

Problem 6: Expand the following expressions.

 Part a $(a+b)^2$ Click here to show solution. Solution: This can be rewritten as $(a+b)(a+b)$, which can be FOILed (recall this tells you the order in which to multiply terms – First, Outer, Inner, Last). This gives $(a)(a)+(a)(b)+(b)(a)+(b)(b) = a^2+2ab+b^2$. Video: Multiplying Binomials Write-up: FOIL Method Part b $(a-b)^2$ Click here to show solution. Solution: Following the same logic as the last problem, the multiplication gives $(a)(a)+(a)(-b)+(-b)(a)+(-b)(-b) = a^2-2ab+b^2$.

Problem 7: Find the slope of the line passing through the following points.

 Part a (0,5) and (4,19) Click here to show solution. Solution: Slope (commonly denoted by the symbol $m$) can be found via the familiar phrase “rise over run”. More accurately, we are concerned with the change in rise $(y_2-y_1)$ over the change in run $(x_2-x_1)$. In this case, this comes out to $m = \frac{19-5}{4-0} = \frac{14}{4} = \frac{7}{2}$. Worked Example: Slope from Two Points Write-up: Slope Formula Part b (2,11) and (3,-3) Click here to show solution. Solution: Following the same logic as the previous problem, we have $m = \frac{-3-11}{3-2} = \frac{-14}{1} = -14$. Part c $(x_1,y_1)$ and $(x_2,y_2)$ Click here to show solution. Solution: This represents the general case: $m = \frac{y_2-y_1}{x_2-x_1}$.

Problem 8: Find the equation of the line passing through the following points.

 Part a (0,5) and (4,19) Click here to show solution. Solution: There are two lanes of thought here, corresponding to the two different generic forms of a line: slope-intercept form $[y = mx+b]$ and point-slope form $[y-y_i = m(x-x_i)]$. In slope-intercept form, the slope and y-intercept of the line need to be known (unsurprisingly). We calculated the slope between these points above: $m = \frac{7}{2}$. Separately, when $x=0$, we know that $y=5$ since $(0,5)$ is given as a point on this line. Therefore, the y-intercept, commonly denoted with the symbol $b$, is 5. The equation of the line is therefore $y=\frac{7}{2}x+5$. In point-slope form, the slope and any one point on the line must be known (unsurprisingly). Choose $(x_i,y_i) = (0,5)$, for example: then $y-5=\frac{7}{2}(x-0),$ which reduces to $y-5=\frac{7}{2}x$. Adding 5 to both sides gives the same answer as before: $y=\frac{7}{2}x+5$. Point-slope form is generally more helpful, as needing any point on the line is a less strict requirement than knowing the y-intercept. The slope is needed in either case. Video: Point-Slope and Slope-Intercept Equations Write-up: Slope-Intercept Form Write-up: Point-Slope Form Part b (2,11) and (3,-3) Click here to show solution. Solution: We don’t know the y-intercept here, so use point-slope form with either of the points provided. For the purpose of this example, we’ll use $(2,11)$; recall from the earlier question that the slope between these points is -14. The equation of the line is $y-11=-14(x-2) = -14x+28$. Adding 11 to both sides gives $y=-14x+39$. (Note that from this manipulation, we can actually determine the y-intercept to be 39, as the final simplified answer is in slope-intercept form with $m=-14, b=39$. This isn’t obvious from the information given initially.) Part c $(x_1,y_1)$ and $(x_2,y_2)$ Click here to show solution. Solution: Use point-slope form again. For slope, recall that $m = \frac{y_2-y_1}{x_2-x_1}$, and note that either point can be plugged into the point-slope formula. Both $y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$ and $y-y_2=\frac{y_2-y_1}{x_2-x_1}(x-x_2)$ are correct.

Problem 9: From the table below, identify the points and write them in the form (x,y).

Part a
 x 0 2 4 6 8 y -20 -10 0 -10 20

Solution:
The first row, consisting of the $x$-values, can be thought of as function inputs and these values will be the first coordinates of the ordered pairs we’re looking for. The $y$-values in the second row can be thought of as the function outputs corresponding to the respective inputs, and these will be the second coordinates of the ordered pairs. The answers are:
\begin{aligned} (x,y) =&(0,-20),\\ &(2,-10),\\ &(4,0),\\ &(6,-10),\\ &(8,20). \end{aligned}

Problem 10: Identify the points on the following x-y plane and write them in the form (x,y).

 Part a Click here to show solution. Solution: From left-to-right, the points are \begin{aligned} (x,y) &= (-5,0)\\ &= (-3,-3)\\ &= (-2,1) \\ &= (0,0) \text{ and } (0,5)\\ &= (1,2)\\ &= (2,4)\\ &= (5,-2). \end{aligned} Video: Introduction to the Coordinate Plane

Problem 11: Find the area of the following figures (including units).

 Part a A rectangle with a length of 5 meters and width of 9 meters. Click here to show solution. Solution: The area of a rectangle of length $x$ and width $y$ is simply $xy$. In this case, the area is 45 $m^2$. Part b A square with side length 7 inches. Click here to show solution. Solution: The area of a square with side length $x$ is simply $x^2$. In this case, the area is 49 $in^2$. Part c A triangle whose base is 20 cm and whose height is 11 cm. Click here to show solution. Solution: A triangle with base $b$ and height $h$ has area $\frac{1}{2}bh$. In this case, the area is 110 $cm^2$. Part d A circle of radius 4 feet. Click here to show solution. Solution: The area of a circle of radius $r$ is $\pi r^2$. In this case, the area is $16\pi$, or approximately 50.265 $ft^2$.

Problem 12: Your annual salary is $40,000 for the year 2018. Use the information below to calculate your salary in 2019, 2020, and 2021. Part a Assume your salary increases by$1500 per year.

 Year 2018 2019 2020 2021 Salary $40,000 ??? ??? ??? Click here to show solution. Solution:  Year 2018 2019 2020 2021 Salary$40,000 $41,500$43,000 $44,500 Part b Assume your salary increases by 3% per year.  Year 2018 2019 2020 2021 Salary$40,000 ??? ??? ???
Let’s start with the calculation for 2019. An increase of 3% over the salary in 2018 means that $40,000+\frac{3}{100}(40,000)$ is earned in 2019. Factor 40,000 out of the left hand side to find $40,000(1+\frac{3}{100})=40,000(1+0.03) = (1.03)(40,000) = 41,200$. This logic can be applied recursively for the other future years:
 Year 2018 2019 2020 2021 Salary $40,000 (1.03)(40,000) =$41,200 (1.03)(41200) = $42,436 (1.03)(42436) =$43,709.08