Sample Algebra I Final Exam Questions and Solutions

Rename $\displaystyle\,\frac{1}{\sqrt 2}\,$ as a fraction with no radical in the denominator:

$$ \begin{align} \frac1{\sqrt 2} &= \frac1{\sqrt 2}\cdot \frac{\sqrt 2}{\sqrt 2}\cr\cr &= \frac{\sqrt 2}2 \end{align} $$

Key concept: multiplying by $\,1\,$ changes the name of a number, but not where it ‘lives’ on a number line! Also, by definition of square roots, $\,\sqrt 2\cdot\sqrt 2 = 2\,.$

Lesson: Practice with Radicals

Rename $\,7\,$ as a quotient of integers, where the numerator is greater than $\,10\,$:

$$ \begin{align} 7 &= 7\cdot \frac 22\cr\cr &= \frac 71\cdot\frac 22\cr\cr &= \frac{14}2 \end{align} $$

Many other answers are possible.

Rename $\,23{,}000{,}000\,$ in scientific notation:

$$ \begin{align} 23{,}000{,}000 &= 23{,}000{,}000\times \overbrace{10^{-7}\times 10^7}^{=\, 10^0\, =\, 1}\cr\cr &= (23{,}000{,}000 \times 10^{-7})\times 10^7\cr\cr &= 2.3 \times 10^7 \end{align} $$

Rename $\,7\,$ in the form $\,x - y\,,$ where $\,x\,$ and $\,y\,$ are not integers:

$$ \begin{align} 7 &= 7 + \overbrace{(\frac 12 - \frac 12)}^{\text{add zero}}\cr\cr &= (7 + \frac 12) - \frac 12\cr\cr &= \frac{15}2 - \frac 12 \end{align} $$

Many other answers are possible.

Key concept: adding $\,0\,$ changes the name of a number, but not where it ‘lives’ on a number line!

Rename $\,7\,$ in the form $\,\frac 12 x\,$:

$$ \begin{align} 7 &= \overbrace{\frac12 \cdot 2}^{=\,1}\,\cdot 7 \cr &\qquad \text{(bring $\,\frac 12\,$ into the picture)}\cr\cr &= \frac 12(14) \end{align} $$

Rename $\,(x-2)(x+3)\,$ as a sum (i.e., multiply out):

$$ \begin{align} (x-2)(x+3) &= x\cdot x + 3x - 2x - 6\cr &\qquad \text{(using FOIL)}\cr\cr &= x^2 + x - 6 \end{align} $$

FOIL stands for Firsts, Outers, Inners, Lasts.

Rename $\,x^2 - y^2\,$ as a product (i.e., factor):

$$ x^2 - y^2 = (x-y)(x+y) $$

Lesson: Factoring a Difference of Squares

Rename $\displaystyle\,\frac 12\,$ in the form $\displaystyle\,\frac 3x\,$:

Approach #1: Multiply by $\,1\,$ in a form that brings $\,3\,$ into the picture:

$$ \frac 12 = \frac 12\cdot\frac 33 = \frac 36 $$

Approach #2: Solve an equation:

$$ \begin{gather} \frac 12 = \frac 3x\cr\cr (2x)(\frac 12) = (2x)(\frac 3x)\cr\cr x = 6 \end{gather} $$

So, $\,\frac 12 = \frac 36\,.$

Rename $\displaystyle\,\frac 1x - \frac 2{x+3}\,$ as a single fraction:

$$ \begin{align} &\frac 1x - \frac 2{x+3}\cr\cr &\quad = \frac 1x\cdot\frac{x+3}{x+3} - \frac 2{x+3}\cdot\frac{x}x\cr\cr &\quad = \frac{x+3}{x(x+3)} - \frac{2x}{x(x+3)}\cr\cr &\quad = \frac{x+3 - 2x}{x(x+3)}\cr\cr &\quad = \frac{-x + 3}{x(x+3)} = \frac{3-x}{x(x+3)} \end{align} $$

Either of these final names is equally good!

Lesson: Adding and Subtracting Fractions With Variables

Rename $\,0.25\,$ as a percent:

$$ \begin{align} 0.25 &= \frac{25}{100}\cr &= 25\cdot \frac 1{100}\cr &= 25\% \end{align} $$

Lesson: Changing Decimals to Percent

Rename $\displaystyle\,\frac{x^4x^{-1}}{(x^2)^3 x}\,$ in the form $\,x^k\,$:

$$ \begin{align} \frac{x^4x^{-1}}{(x^2)^3 x} &= \frac{x^{4+(-1)}}{x^6x^1}\cr\cr &= \frac{x^3}{x^7}\cr\cr &= x^{3-7}\cr\cr &= x^{-4} \end{align} $$

There are many other ways that this could be done.

Lesson: Multi-Step Exponent Law Practice

Rename $\,300\text{ ft/sec}\,$ in the form $\,x\,\text{ mph}\,.$ Here, ‘mph’ is ‘miles per hour’, and there are $\,5{,}280\,$ feet in one mile.

$$ \begin{align} 300\text{ ft/sec} &= \frac{300\text{ ft}}{\text{sec}}\cr\cr &= \frac{300 \text{ ft}}{\text{sec}}\cdot \frac{1\text{ mile}}{5280\text{ ft}}\cdot \frac{60\text{ sec}}{1\text{ min}}\cdot \frac{60\text{ min}}{1\text{ hr}}\cr\cr &= \frac{300(60)(60)\text{ miles}}{5280 \text{ hr}}\cr\cr &\approx 204.5 \text{ mph} \end{align} $$

Lesson: Rate Problems

Solve: $\,3x(1-5x)(x^2-16) = 0$

$$ \begin{gather} 3x(1-5x)(x^2-16) = 0\cr\cr 3x = 0\ \ \text{or}\ \ 1-5x = 0\ \ \text{or}\ \ x^2-16 = 0\cr\cr x = 0\ \ \text{or}\ \ 1 = 5x\ \ \text{or}\ \ x^2 = 16\cr\cr x = 0\ \ \ \text{or}\ \ \ x = \frac15\ \ \ \text{or}\ \ \ x = \pm 4 \end{gather} $$

Key idea: recognize that a product equals zero, and use the Zero Factor Law. Recall that ‘$\,x=\pm 4\,$’ means ‘$\,x = 4\ \ \text{or}\ \ x = -4\,$’.

Solve: $\displaystyle\,\frac 12x - 7 = 3x + \frac x5$

$$ \begin{gather} \frac 12x - 7 = 3x + \frac x5\cr\cr 10(\frac 12x - 7) = 10(3x + \frac x5)\cr \text{(clear fractions)}\cr\cr 5x - 70 = 30x + 2x\cr\cr -27x = 70\cr\cr x = -\frac{70}{27} \approx -2.6 \end{gather} $$

Lesson: Solving Linear Equations Involving Fractions

Solve: $\,|x-3|\gt 1$

$$ \begin{gather} |x-3|\gt 1\cr\cr x-3\gt 1\ \ \ \text{or}\ \ \ x-3 \lt -1\cr\cr x\gt 4\ \ \ \text{or}\ \ \ x\lt 2 \end{gather} $$

Lesson: Solving Absolute Value Inequalities Involving ‘Greater Than’

Advanced solution: if you've already learned that $\,|x-y|\,$ gives the distance between $\,x\,$ and $\,y\,,$ then this can be solved by inspection: we want all numbers $\,x\,$ whose distance from $\,3\,$ is greater than $\,1\,.$

Solve: $\,2\lt |x|\lt 3$

We want all numbers $\,x\,$ whose distance from $\,0\,$ is between $\,2\,$ and $\,3\,.$

$$ \begin{gather} 2\lt |x|\lt 3\cr\cr 2 \lt |x|\ \ \ \text{and}\ \ \ |x|\lt 3\cr\cr |x| \gt 2\ \ \ \text{and}\ \ \ |x|\lt 3 \end{gather} $$

In other words, we want all numbers $\,x\,$ whose distance from $\,0\,$ is greater than $\,2\,$ and less than $\,3\,.$

Lessons:
Simplifying Basic Absolute Value Expressions
Solving Absolute Value Inequalities Involving ‘Greater Than’
Solving Absolute Value Inequalities Involving ‘Less Than’

Solve: $\,1-2x\lt 3$

$$ \begin{gather} 1-2x\lt 3\cr\cr -2x\lt 2\cr\cr x\gt -1 \end{gather} $$

Remember: when you multiply/divide both sides of an inequality by a negative number, you must change the direction of the inequality symbol!

Lesson: Solving Simple Linear Inequalities with Integer Coefficients

Solve: $\,x^2 = x + 2$

$$ \begin{gather} x^2 = x + 2\cr\cr x^2 - x - 2 = 0\cr\cr (x-2)(x+1) = 0\cr\cr x = 2\ \ \ \text{or}\ \ \ x=-1 \end{gather} $$

Lesson: Solving Simple Quadratic Equations by Factoring

Solve: $\,1\lt x\ \ \ \text{or}\ \ \ x\le -1\,$

$$ \begin{gather} 1\lt x\ \ \ \text{or}\ \ \ x\le -1\cr\cr x\gt 1\ \ \ \text{or}\ \ \ x\le -1 \end{gather} $$

Lesson: Practice With the Mathematical Words ‘and’, ‘or’, ‘is equivalent to’

Solve: $\,0.005(x - 0.01) = 0.003 + 0.4x$

$$ \begin{gather} 0.005(x - 0.01) = 0.003 + 0.4x\cr\cr (1000)\bigl(0.005(x - 0.01)\bigr) = (1000)\bigl(0.003 + 0.4x\bigr)\cr\cr 5(x-0.01) = 3 + 400x\cr\cr 5x - 0.05 = 3 + 400x\cr\cr -395x = 3.05\cr\cr x = \frac{3.05}{-395} = -\frac{305}{39500} = -\frac{61}{7900} \approx -0.0077 \end{gather} $$

Lesson: Multiplying and Dividing Decimals by Powers of Ten

Graph $\,x\gt 3\,,$ viewed as an inequality in two variables.

We are viewing this as: $\,x + 0y \gt 3$

We want the set of all points $\,(x,y)\,,$ where $\,x\gt 3\,$ and $\,y\,$ is any real number. To show this set:

• Dash the vertical line $\,x = 3\,$ (to show that it is not included)
• Shade everything to the right

Lesson: Linear Inequalities in Two Variables (Part 2)

Graph $\,2y - 3 = 0\,,$ viewed as an equation in two variables.

We are viewing this as: $\,0x + 2y - 3 = 0$

$$ \begin{gather} 2y - 3 = 0\cr\cr 2y = 3\cr\cr y = \frac 32 \end{gather} $$

We want the set of all points $\,(x,y)\,$ where $\,y = \frac 32\,$ and $\,x\,$ is any real number. This is the horizontal line that crosses the $y$-axis at $\,\frac 32\,$:

Lesson: Horizontal and Vertical Lines

Graph the compound sentence: $\,x = 3\ \ \text{and}\ \ y = 2$

This is the mathematical word ‘and’.

It is the intersection of the vertical line $\,x = 3\,$ and the horizontal line $\,y = 2\,,$ which is the single point $\,(3,2)\,.$

Lessons:
Horizontal and Vertical Lines
Practice With the Mathematical Words ‘and’, ‘or’, ‘is equivalent to’

Graph the compound sentence: $\,x = 3\ \ \text{or}\ \ y = 2$

This is the mathematical word ‘or’.

It is the union of the vertical line $\,x = 3\,$ and the horizontal line $\,y = 2\,.$

Lessons:
Horizontal and Vertical Lines
Practice With the Mathematical Words ‘and’, ‘or’, ‘is equivalent to’

Graph the equation in two variables: $\,y = 2x - 1$

This is the line with slope $\,2\,$ that crosses the $y$-axis at $\,-1\,.$

Lesson: Graphing Lines

Graph the equation in two variables: $\,y = \sqrt{x}$

Eventually, this will be one of the ‘basic models’ that you should have on your fingertips! For a first time, though, plot a few easy points, and see the shape that emerges:

$x$	$y$
$0$	$\sqrt{0} = 0$
$1$	$\sqrt{1} = 1$
$4$	$\sqrt{4} = 2$
$9$	$\sqrt{9} = 3$

Lesson: Introduction to Equations and Inequalities in Two Variables

Graph the equation in two variables: $\,|x| = 2$

We are viewing this as: $\,|x| + 0y = 2$

Note that:

$$ \begin{gather} |x| = 2\cr \text{is equivalent to}\cr x = 2\ \ \text{or}\ \ x = -2 \end{gather} $$

We need the set of all points $\,(x,y)\,$ for which $\,x = 2\,$ or $\,x = -2\,.$ Here, $\,y\,$ can be any real number. This is the union of two vertical lines:

Graph the inequality in two variables: $\,y \le 2$

We are viewing this as: $\,0x + y \le 2$

We need the set of all points $\,(x,y)\,$ for which $\,y \le 2\,.$ Here, $\,x\,$ can be any real number. First, graph the horizontal line $\,y = 2\,,$ solid (showing that it is included in the solution set). Then, shade everything beneath this line:

Graph the equation in two variables: $\displaystyle\,\frac{y-2}3 = 2x - 1$

$$ \begin{gather} \frac{y-2}3 = 2x - 1\cr\cr y - 2 = 3(2x-1)\cr\cr y - 2 = 6x - 3\cr\cr y = 6x - 1 \end{gather} $$

This is the line with slope $\,6\,$ that crosses the $y$-axis at $\,-1\,.$

$$ \begin{align} \frac{1+\sqrt 2}{\root3\of 5 - 7} &= \frac{(1+\sqrt 2)}{(\root3\of 5 - 7)}\cr\cr &\approx -0.45637 \end{align} $$

Evaluate $\,3x^2 - 5x + 1\,,$ where $\,x = -1.8\,$:

$$ 3(-1.8)^2 - 5(-1.8) + 1 = 19.72 $$

Evaluate $|1 - 2x|\,,$ where $\,x = \sqrt 3\,$:

$$ |1-2(\sqrt 3)| \approx 2.46410 $$

$$ \begin{align} &(2.03\times 10^{-9})(-4.1\times 10^7)\cr\cr &\quad = (2.03)(-4.1)(10^{-2})\cr\cr &\quad \approx -0.08323 \end{align} $$

(a) What is the domain of the function $\displaystyle\,f(x) = \frac{1-3x}{x-2}\,$?

The domain of $\,f\,$ is the set of all real numbers $\,x\,$ for which the denominator is not zero. Therefore, only the number $\,2\,$ must be excluded from the domain.

Letting $\,\text{dom}(f)\,$ denote the domain of $\,f\,,$ using interval notation, and recalling that ‘$\,\cup\,$’ is the union operation, we have:

$$ \text{dom}(f) = (-\infty,2)\cup (2,\infty) $$

(b) Use your graphing calculator to graph the function $\,f\,$ in the window $\,-1\lt x\lt 3\,$ and $\,-15\lt y\lt 10\,.$

(c) Find the $x$-intercept of the graph.

For this problem, you could use your calculator (and get an estimate) or find the intercept algebraically, which is done here:

• An $x$-intercept is a point with $y$-value equal to $\,0\,.$ So, you need to set $\,\frac{1-3x}{x-2}\,$ equal to $\,0\,.$
• The only way a fraction can equal zero is when its numerator is zero: $$ \begin{gather} 1-3x = 0\cr 1 = 3x\cr x = \frac 13 \approx 0.33333 \end{gather} $$
• The only $x$-intercept is: $$ (\frac 13,0) \approx (0.33333,0) $$ (Note that this agrees with the graph.)

(d) Use your calculator to estimate a value of $\,x\,$ for which $\,f(x) = 5\,$. (Zoom, as necessary, to get $\,f(x)\,$ within $\,0.01\,$ of $\,5\,.$)

When you zoom in this close, the graph looks nearly vertical! From this graph, it looks like the $x$-value of the intersection of the line $\,y = 5\,$ and $\,y = \frac{1-3x}{x-2}\,$ is a bit less than $\,1.5\,$ (here, $\,x\,$ is still from $\,-1\,$ to $\,3\,.$) Your calculator can get much closer!

Finding it algebraically (which you don't need to do for this problem): $$ \begin{gather} \frac{1-3x}{x-2} = 5\cr\cr 1-3x = 5(x-2)\cr\cr 1-3x = 5x - 10\cr\cr 11 = 8x\cr\cr x = \frac{11}8 = 1.375 \end{gather} $$

(a) Take a number, multiply by $\,2\,,$ then subtract $\,3\,.$

$2x - 3$

This expression is written in the most conventional way.

(b) Take a number, subtract $\,3\,,$ then multiply by $\,2\,.$

$2(x-3)$

This expression is written in the most conventional way.

(c) Take a number, multiply it by $\,2\,,$ cube the result, add $\,1\,,$ then divide by the original number.

$\displaystyle\frac{(2x)^3 + 1}x$

This expression is written in a conventional way.

$3x-1$

Take a number, multiply by $\,3\,,$ then subtract $\,1\,.$

$2(x+1)^3 - 5$

Take a number, add $\,1\,,$ cube the result, multiply by $\,2\,,$ then subtract $\,5\,.$

$\displaystyle\frac{x-3}7 - 1$

Take a number, subtract $\,3\,,$ divide by $\,7\,,$ then subtract $\,1\,.$

(a) $f(0)$

$$ \begin{align} f(0) &= 0^2 - 2\cdot 0 + 1\cr &= 1 \end{align} $$

(b) $f(1) - 2$

$$ \begin{align} f(1) - 2 &= (1^2 - 2\cdot 1 + 1) - 2\cr &= 0 - 2\cr &= -2 \end{align} $$

(c) $f(f(-1))$

$$ \begin{align} f(f(-1)) &=f\bigl( \overbrace{(-1)^2 - 2(-1) + 1}^{f(-1)}\bigr)\cr &= f(4)\cr &= 4^2 - 2\cdot 4 + 1\cr &= 16 - 8 + 1\cr &= 9 \end{align} $$

Notice that we worked from the inside to the outside!

(a) slope $\,3\,,$ passing through the point $\,(2,-1)\,$

This is a perfect candidate for point-slope form, which is then put into $\,y = mx+b\,$ form:

$$ \begin{gather} y - y_1 = m(x - x_1)\cr\cr y - (-1) = 3(x - 2)\cr\cr y + 1 = 3x - 6\cr\cr y = 3x - 7 \end{gather} $$

(b) the horizontal line that crosses the $y$-axis at $\,2$

$y = 2$

(c) the line that is perpendicular to $\,x - 3y = 5\,$ and passes through the point $\,(0,3)$

• Find the slope of $\,x-3y = 5\,$: $$ \begin{gather} x - 3y = 5\cr\cr -3y = -x + 5\cr\cr y = \frac x3 - \frac 53 \end{gather} $$ The slope is $\,\frac 13\,.$
• The slope of a perpendicular line is the opposite reciprocal (‘flip’ and take the opposite). This gives $\,-3\,.$
• The line with slope $\,-3\,$ that passes through $\,(0,3)\,$ is: $ y = -3x + 3$

A certain telephone company charges a $\,\$4.95\,$ monthly fee, for which the customer gets $\,100\,$ minutes of calls anywhere in the United States. Each additional minute costs $\,7\,$ cents. Carol always talks more than $\,100\,$ minutes. Let $\,x\gt 100\,$ denote the number of minutes that Carol talks in a month, and let $\,C(x)\,$ denote the dollar amount owed to the telephone company. Find a formula for $\,C(x)\,.$

• Fixed charge, with units of dollars: $\,4.95$
• Carol gets $\,100\,$ minutes ‘free’. The number of additional minutes is: $\,x - 100$
• Each additional minute costs $\,7\,$ cents, which is $\,0.07\,$ dollars.
• The dollar cost of the additional minutes is $\,0.07(x - 100)\,.$
• Thus: $\,C(x) = 4.95 + 0.07(x - 100)$

Of course, there are always ‘hidden’ charges (like taxes), but we're ignoring those!