# RWM102 Study Guide

 Site: Saylor Academy Course: RWM102: Algebra Book: RWM102 Study Guide
 Printed by: Guest user Date: Friday, January 27, 2023, 7:15 PM

## Navigating this Study Guide

#### Study Guide Structure

In this study guide, the sections in each unit (1a., 1b., etc.) are the learning outcomes of that unit.

Beneath each learning outcome are:

• questions for you to answer independently;
• a brief summary of the learning outcome topic; and
• and resources related to the learning outcome.

At the end of each unit, there is also a list of suggested vocabulary words.

#### How to Use this Study Guide

1. Review the entire course by reading the learning outcome summaries and suggested resources.
2. Test your understanding of the course information by answering questions related to each unit learning outcome and defining and memorizing the vocabulary words at the end of each unit.

By clicking on the gear button on the top right of the screen, you can print the study guide. Then you can make notes, highlight, and underline as you work.

Through reviewing and completing the study guide, you should gain a deeper understanding of each learning outcome in the course and be better prepared for the final exam!

## Unit 1: Variables and Variable Expressions

### 1a. Identify parts of algebraic expressions, including terms, factors, and coefficients

• How do you refer to the different parts of an algebraic expression?

Algebraic expressions are often made up of several parts. For example, the algebraic expression $3x^2+5x-3$, is made up of three parts, which are called terms. The terms are $3x^2$, $5x$, and $-3$. Each term is made up of its own parts, for example $3x^2$ is made up of the coefficient, which is $3$, and the variable, which is $x^2$. The term $-3$ has no variable part, so it is called a constant, since the value of $-3$ does not change.

To review, see Identifying Variable Parts and Coefficients of Terms.

### 1b. Evaluate algebraic expressions for the given values of the variables

• How do you evaluate an algebraic expression for a specific value?

Sometimes, an algebraic expression is given to represent a situation, and a specific value is assigned to the variable. For example, $4x+7$ is an algebraic expression, and you could be asked to evaluate the expression at $x=2$. Since you are being told that $x=2$, you will replace, or substitute, the $2$ in place of the $x$. After substituting, you will find $4(2)+7$. By following the order of operations, you can multiply $4$ and $2$, then add $7$ to find the answer, which is $15$.

Similarly, you could have an expression with multiple variables where you need to substitute for all the variables. For example: Evaluate $5x+2y-3z$ when $x=1$, $y=3$, $z=5$. In this case, you would substitute for all three variables at the same time, then follow the order of operations to find the answer:

$5(1)+2(3)-3(5)=5+6-15=-4$

### 1c. Apply commutative, associative, and distributive properties of real numbers to simplify algebraic expressions

• What properties can you use to simplify algebraic expressions?

When working with algebraic expressions, it is often helpful to first simplify the expression. When simplifying expressions, the following properties are helpful to know and apply.

Commutative Property: Addition and multiplication can be performed in any order without changing the value. For example, $5+3=3+5$ and $2 \times 3=3 \times 2$.

Associative Property: Addition and multiplication can be regrouped without changing the value. For example, $(2+3)+5=2+(3+5)$ and $(3 \times 5) \times 7 = 3 \times (5 \times 7)$.

Distributive Property: Multiplication can be distributed across either addition or subtraction within parentheses. For example, $3(x+5)=3 \times x+3 \times 5=3x+15$.

To review, see:

### Unit 1 Vocabulary

This vocabulary list includes terms listed above that students need to know to successfully complete the final exam for the course.

• algebraic expressions
• term
• coefficient
• variable
• constant
• evaluate
• substitute
• order of operations
• simplify
• commutative property
• associative property
• distributive property

## Unit 2: Linear Equations

### 2a. Determine whether a given real number is a solution of an equation

• How do you know if a value is a solution to an equation?

A value is considered a solution to an equation if you can replace the variable with that value and the outcome is true. To check if a value is a solution, first, substitute the value in place of the variable. Then, simplify the expressions on both sides of the equation. Finally, determine if the final statement is true.

For example, determine if $x = 3$ is a solution to the equation $2x+3=5x-2$. First, substitute 3 in place of $x$ on both sides of the equation: $2(3)+3=5(3)-2$. Next, simplify both sides of the equation: $6+3=15-2$. Finally, determine if the final statement is true: $9=13$ is not true, therefore 3 is not a solution to this equation.

To review, see Solving Linear Equations with One Variable.

### 2b. Simplify equations using addition and multiplication properties

• How do you simplify equations using addition properties?
• How do you simplify equations using multiplication properties?

When solving equations, we can use some basic properties to find the value of the variable. When a constant is added or subtracted with your variable, you can isolate the variable by doing the opposite operation with the same value on both sides of the equation. For example, to solve the equation $x+4=7$, notice that 4 is currently being added to our variable, $x$. To solve this equation, simply subtract 4 from both sides. $x+4-4=7-4$. This will give the answer $x=3$.

Similarly, if the variable is being multiplied or divided by a number you can again perform the opposite operation with the same number on both sides of the equation to solve for the variable. For example to solve the equation $\dfrac{x}{4}=3$, notice the variable $x$ is being divided by 4. Multiply both sides of the equation by 4 to solve for $x$. $\dfrac{x}{4} \times 4=3 \times 4$. Therefore $x = 12$.

### 2c. Find the solution of a given linear equation with one variable

• How do you solve equations using addition and subtraction properties?
• How do you solve equations using multiplication and division properties?
• How do you solve 2-step equations?
• How do you solve equations with variables on both sides?
• How do you solve equations with parentheses?

When solving an equation with multiple steps, always begin by eliminating any addition or subtraction first, before addressing any multiplication or division. For example, we can use addition/subtraction properties in the equation $3x - 2 = 7$, begin by adding 2 to both sides, since there is a -2 already there. This yields, $3x - 2 + 2 = 7 + 2$, which simplifies to $3x = 9$. Now we can use multiplication/division properties to divide both sides by 3, giving us a final answer of $x = 3$.

If there is an equation with numbers and variables on both sides, first you must collect all terms with the variable on one side, and all terms without variables (the constants) on the other side of the equal sign. Then solve the equation as described above. For example, $3x+1=2x+5$ would be solved by subtracting $2x$ from both sides, $3x-2x+1=2x-2x+5$, which gives us $x+1=5$. Then subtract 1 from both sides, $x+1-1=5-1$, which gives the final answer, $x=4$.

If the equation has parentheses, first use the order of operations and/or the distributive property to remove the parentheses, then solve as described above.

To review, see:

### 2d. Determine the number of solutions of a given linear equation in one variable

• Can a linear equation in one variable have no solutions?
• Can a linear equation in one variable have more than one solution?

In certain instances, equations have solutions that do not end with the variable equal to a single specific value, as we saw in 2b and 2c. When solving an equation, if all the variables cancel, we end up with a statement with constants on both sides. For example, $3x+5=3x-1$. When we solve this equation, we end up with $5=-1$. Since this is false, this equation has no solutions.

On the other hand, if the final solution is a true statement, such as $3=3$, then we say the equation has "infinitely many solutions", or "all real numbers", because any value for the variable will still make $3=3$ true.

To review, see Solving Linear Equations with One Variable

### 2e. Solve a literal equation for the given variable

• How do you solve an equation for a specific variable?

Sometimes equations have more than one variable, and instead of finding a value for a specific variable, we need to simply solve the equation for a variable. For example, if we solve the equation $3v+s=t-2$ for the variable $s$, we simply use the same skills we have learned for solving equations by subtracting $3v$ from both sides, giving us an answer of $s=t-2-3v$.

To review, see Solving for One Variable

### 2f. Rearrange formulas to isolate a quantity of interest

• How can you use a formula set up for a specific variable to instead find a different variable?

Sometimes you may find a formula given to you that isn't set up to give you the quantity you desire. For example, in the formula $F+V=E+2$, the formula is not currently set up to give you the value of $V$. You can simply solve the formula for $V$, as we have done before, to get the new formula, $V=E+2-F$. Now you can use the values you are given for $E$ and $F$ to find $V$.

To review, see Solving for One Variable

### Unit 2 Vocabulary

This vocabulary list includes terms listed above that students need to know to successfully complete the final exam for the course.

• solution
• statement
• opposite operation
• parentheses
• no solutions
• infinitely many solutions
• all real numbers

## Unit 3: Word Problems

### 3a. Translate a verbal expression into a variable expression

• How do you create an expression from a word problem?

When solving word problems, one of the challenges is correctly converting the written words to a mathematical expression. When reading a word problem, try to look for these keywords that often represent mathematical operations:

MATHEMATICS DICTIONARY
Word or Phrase
Mathematical Operation
Sum, sum of, added to, increased by, more than, and, plus
$+$
Difference, minus, subtracted from, decreased by, less, less than
$-$
Product, the product of, of, multiplied by, times, per
$\cdot$
Quotient, divided by, ratio, per
$\div$
Equals, is equal to, is, the result is, becomes
$=$
A number, an unknown quantity, an unknown, a quantity
$x$ (or any symbol)

Sometimes the name of the variable is explicitly given, such as "The product of $x$ and 3 is equal to 7", which means $3x=7$.

Other times, the variable is not named, and you can use any variable you choose, such as "Three more than a number is equal to 10", which means $x+3=10$.

To review, see Translating Words into Mathematical Symbols

### 3b. Use the basic percent equation to solve problems involving percents

• How do you solve percent problems?
• How do you convert a percent to a decimal?
• How do you convert a decimal to a percent?

When working with percents, it is important to remember that the word percent means "out of 100". Therefore, we often multiply or divide by 100.

When converting a decimal to a percent, we must multiply by 100. For example, .345 is multiplied by 100 to give 34.5%.

On the other hand, sometimes we have a percent and we want to convert it to a decimal. This is done by dividing by 100. For example, 45% is divided by 100, giving a decimal value of .45.

Similarly, sometimes a fraction needs to be converted to a percent. To do this, simply divide the fraction to get a decimal, then convert the decimal to a percent as we just did.

When solving percent problems, it is helpful to use a standard percent equation, such as the equation:

$part = \% rate \times base$

For example, if you are asked to find 45% of 50, use the percent equation to find $part=45 \% \times 50$. Convert the percent to a decimal, and then multiply to find $part= .45 \times 50=22.5$.

To review, see Solving Percent Word Problems and Practice Solving Percent Word Problems.

### 3c. Apply the basic percent equation to problems involving mixtures, markups, and discounts

• How do you solve percent problems with mixtures?
• How do you solve percent problems with markups and discounts?

A common percent problem is the mixture problem. In these problems, you are trying to combine two different liquids of different strengths to make a mixture of a new strength. For example, if you have 20L of 25% sugar solution, how much of a 10% sugar solution would you have to mix to end up with a 15% sugar solution?

This problem is just the percent equation used 3 times together. Set up the problem with the percent strength times the volume for each of the two mixtures you have, and remember that will equal the total amount of 15% solution in the end. The amount of the 10% solution you need to add should be x.

$.25 \times 20+.10x=.15(x+20)$

By simplifying, using the distributive property, and solving, we will find that $x = 40L$.

When calculating markups or discounts, we can use the percent change equation, which is:

$\% change=\dfrac{final \; amount - original \; amount}{original \; amount} \times 100$

For example, if a dress has an original price of $80, and is on sale for$50, you can calculate the discount by substituting the values into the percent change equation.

$\% change=\dfrac{50-80}{80} \times 100= -37.5 \%$

This means we have a 37.5% discount. If the price was increased, instead of decreased, the percent change equation will calculate the markup, instead of the discount.

To review, see:

### 3d. Use the uniform motion equation to solve problems involving uniform motion

• How do you solve problems involving distance, time, and speed?

When solving problems involving the time it takes to travel a certain distance, you should use the uniform motion equation, $d=rt$, where $d$ is the distance, $r$ is the rate (speed), and t is the time. When solving a uniform motion problem, 2 of the 3 variables will be given, you will need to find the third unknown variable.

For example, if Cindy wants to travel to see her grandmother 240 miles away and she drives at 60 miles per hour, how long will it take her to reach her grandmother's house? In this problem, 240 miles is the distance, and 60 miles per hour is the rate, so if we substitute into the uniform motion equation, we get \)240=60t\). Now, we only need to divide both sides by 60 to find $t = 4$. So, it will take her 4 hours to reach her grandmother's house.

To review, see Solving a Formula for a Specific Value

### 3e. Create equations in one variable and use them to solve problems

• How do you create an equation from a word problem?
• How do you solve one-variable word problems?

When solving word problems, it is often necessary to create a mathematical equation from the information given to solve the problem. When creating an equation for a word problem, you can follow these steps:

1. Let $x$ (or some other letter) represent the unknown quantity.
2. Translate the words to mathematical symbols and form an equation. Draw a picture if possible.
3. Solve the equation.
4. Check the solution by substituting the result into the original statement, not equation, of the problem.
5. Write a conclusion.

For example, consider the question "when 3 times a number is increased by 4, the result is eight more than 5 times the number". In this case, we can translate the words into the equation $3x+4=5x+8$. Using our skills for solving equations, you will find that $x=-2$.

To review, see Solving Algebraic Expressions and Equations Part I.

### Unit 3 Vocabulary

This vocabulary list includes terms listed above that students need to know to successfully complete the final exam for the course.

• converting
• mathematical expression
• percent
• percent equation
• mixture problem
• markups
• discounts
• percent change equation
• uniform motion equation
• word problems

## Unit 4: Inequalities

### 4a. Simplify inequalities using addition and multiplication properties

• How do you solve inequalities using addition and multiplication properties?

Solving inequalities is almost the same to solving equations, with just one exception, which we will discuss shortly. For inequalities involving addition or subtraction, the same techniques of doing the "opposite" that we learned for equations will still work to solve inequalities. For example, $x-3 < 5$ can be solved by adding 3 to both sides, giving a final answer of $x < 8$.

Similarly, inequalities with the multiplication or division of positive numbers can be solved just as they were with equations. For example, $3x > 12$ can be solved by dividing both sides by 3, giving a final answer of $x > 4$.

The only notable difference is when you are multiplying or dividing an inequality by a negative number. Any time you multiply or divide an inequality by a negative number, you must "flip" the inequality symbol. For example, to solve $-3x >12$, you must divide by -3, as normal, but when you do, the greater than symbol will "flip" to become a less than symbol, giving a final answer of $x < -4$.

To review, see One Step Inequalities.

### 4b. Find and graph on a number line the solution of a given linear inequality with one variable

• How do you solve linear inequalities?
• How do you represent the solution to an inequality?

As we previously saw, solving inequalities is mostly the same as equations. Once you have finished solving an inequality, you can leave your answer with the inequality symbol, but sometimes a more visual approach to the solution is required. We can graph the solution to inequalities on a number line. For example, if the final solution to an inequality is $x < 3$, then we can graph that solution on the number line as such: In this case, since x is less than 3, we start our solution at 3, with an open circle, since 3 itself is not a solution. Then we shade all the numbers that are less than 3, as those are the solution to the problem. If the final solution includes a or symbol, then the circle should be filled in, not open.

To review, see Plotting Inequalities on a Number Line

### 4c. Create inequalities in one variable and use them to solve problems

• How do you create an inequality to solve a problem?

When creating inequalities, all of the same rules continue from creating equations. The only difference is that instead of saying something like, "is equal to", inequalities will say something like "is less than", or "is more than or equal to".

Remember the inequality symbols and their definitions, and use them appropriately to write an inequality:

< "is less than"

> "is greater than"

$\leq$ "is less than or equal to"

$\geq$ "is greater than or equal to"

For example, three more than twice a number is less than or equal to 13. Written as an inequality, we get $2x+3 \leq 13$. By solving, we find that $x \leq 5$.

To review, see Multi-step Inequalities with Variables on Both Sides.

### Unit 4 Vocabulary

This vocabulary list includes terms listed above that students need to know to successfully complete the final exam for the course.

• inequalities
• number line
• graph
• inequality symbols

## Unit 5: Graphs of Linear Equations and Inequalities

### 5a. Graph points with given coordinates on the rectangular coordinate plane

• How do you graph $(x,y)$ points on the coordinate plane?

The coordinate plane is made up of a horizontal axis, the x-axis, and a vertical axis, the y-axis. When plotting an $(x,y)$ point on the coordinate plane, you begin at the origin, $(0,0)$ and move left or right based on the value of $x$. Then, from there, move vertically based on the value of $y$. That is the location of your point.

For example, let's plot the point $(3, 7)$ As you can see, we went 3 to the right, because the $x$ value is positive three, and then up 7, since the $y$ value is positive 7. P is located on the point $(3,7)$.

To review, see Points in the Coordinate Plane.

### 5b. Determine coordinates of a point on the rectangular coordinate system

• How do you determine the coordinates of a point on the coordinate plane?

If we see a point on the coordinate plane, we can identify its coordinates in the reverse way from how we plotted the point. Let's find the coordinates of the point $D$. First, consider the $x$-coordinate of the point. Since $D$ is 3 to the left, it has an $x$-coordinate of -3. Similarly, $D$ has a $y$-coordinate of -3. Therefore, the coordinates of $D$ are (-3,-3).

To review, see Points in the Coordinate Plane.

### 5c. Determine whether a given ordered pair is a solution of the equation with two variable

• How can you check if a certain point is the solution to an equation?

When we graph an equation, every point on the graph is a solution to the equation that was graphed. Because of that, we can check if a certain point is a solution to the equation by simply checking if that point is on the graph.

For example, consider the equation $y=x^2+4x+4$. Is the point $(3,1)$ a solution to the equation?

The graph is: Since we have been given the graph, all we need to do is check if the point $(3,1)$ is on the graph. If we move to 3 on the positive \(x)-axis, and then up 1, we find a point that is on the graph. Therefore, $(3,1)$ is a solution to the equation $y=x^2+4x+4$.

To review, see Ordered Pair Solutions to Equations.

### 5d. Find and graph solutions of the equation in two variables

• How do you find and graph the solution to an equation?

When you have an equation you want to graph the solution of, you should start by finding some specific solutions using an x-y table. Then plot those points on the coordinate plane, and finally connect the points to draw the graph.

For example, to graph the solutions to the equation $y=3x-1$, we will make an $x-y$ table, and select some $x$-values which we will substitute into the equation to find the corresponding $y$-values. First, let's set up the $x-y$ table. You can select any $x$-values you want, but values near the middle of your graph are generally good.

 $X$ $Y$ -1 0 1 2

Now we will substitute those $x$-values in for $x$ in the equation to find the $y$-values.

 $X$ $Y$ -1 3(-1)-1= -4 0 3(0)-1= -1 1 3(1)-1= 2 2 3(2)-1= 5

Now we have 4 points on our graph. Plot those points, then connect them to graph the equation. We now have the graph of the solutions to the equation.

To review, see Graphing Linear Equations with Two Variables.

### 5e. Graph a straight line given either its equation, or a slope and y-intercept

• What is slope?
• How do you find and use slope when graphing?

When graphing a linear equation, a key point to focus on is the slope. The slope is the change in $y$ divided by the change in $x$. We often use the letter "$m$" to represent slope. The slope formula is:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

When graphing, the slope of a line can be seen and calculated visually as well.

To calculate the slope visually, simply identify two points on the line, then count the change in y and change in x between those points, sometimes called "rise over run". Be sure to be careful to consider if the points are changing positively (up/right) or negatively (down/left) to accurately calculate the slope. For example, we will calculate the slope of the following line: If we focus on the points (-5,1) and (0,3), we can see that between these points, the y went up 2, and the $x$ went to the right 5. Therefore our slope is $\dfrac{2}{5}$.

When graphing a line, you can use any point along with the slope to make your graph.

For example, let's graph a line passing through the point (-3, 1) with a slope of ⅔.

First, we will plot a point at (-3,1). Then from that point, we will move according to the slope, ⅔. We will move up 2 and to the right 3, and arrive at another point on the line, the point (0,3). Finally, connect these points and you will have the graph of your line. To review, see Understanding the Slope of a Line

### 5f. Find slope and intercepts of a straight line given its equation or its graph

• How do you find the $y$-intercept of a line?
• How do you find the $x$-intercept of a line?
• How do you graph a line in slope-intercept form?

When graphing a line, one easy way to find some important points is to find the x-intercept and y-intercept. When viewing a graph, the intercepts can be found by simply looking where the line crosses the $x$-axis and $y$-axis. For example, the linear function above has a $y$-intercept of (0,-3) and an $x$-intercept of (2,0).

If you have the equation of a line, finding the intercepts is quite simple. To find the $y$-intercept, you simply let the $x=0$, and solve for $y$. Similarly, to find the $x$-intercept, let $y=0$ and solve for $x$.

For example, to find the intercepts of $3x-2y=6$, first we will let $x=0$ to find the $y$-intercept. $3(0)-2y=6$, $-2y=6$, $y=-3$. Therefore the $y$-intercept is (0,-3). Similarly, we can find the $x$-intercept by letting $y=0$. $3x-2(0)=6$, $3x=6$, $x=2$. Therefore the $x$-intercept is (2,0).

One way the equation of a line can be written is called slope-intercept form. Slope-intercept form is $y=mx+b$, where the $m$ is the slope, and $b$ is the $y$-intercept. In slope-intercept form, we can start by plotting the $y$-intercept, then use the slope to find another point and graph the line.

For example, the line $y=2x-3$, has a $y$-intercept of (0,-3) and a slope of 2. To graph, we begin by plotting the y-intercept, then from that point, graphing a slope of 2 to find another point and draw the graph. ### 5g. Write the equation of a line passing through two given points

• How do you write the equation of a line passing through two points?

Another way to write the equation of a line is called point-slope form. Point-slope form is $y-y_1=m(x-x_1)$. In point-slope form, $(x_1,y_1)$ is a point on the line, and $m$ is still the slope. This form is ideal for problems where you are asked to write the equation of a line, as you only need any point on the line and the slope to write the equation.

For example, to find the equation of the line passing through (-2,3) and (-1,-2), first we must find the slope. Using the slope equation, the slope is $m=\dfrac{-2-3}{-1-(-2)}=\dfrac{-5}{1}=-5$. Now that we have the slope, we can use either point to write the equation of the line. If we use the point (-2,3), then the equation of the line is $y-3=-5(x+2)$.

To review, see Linear Equations in Point-Slope Form

### 5h. Write the equation of a line with a given slope passing through a given point

• How do you write the equation of a line given a slope and a point?

When a slope and a point are given, rather than two points, writing the equation of a line is even simpler with point-slope form. Since a point and the slope are all that are needed to write the equation, you simply need to plug in the information given.

For example, to find the equation of the line passing through (-2,5) with a slope of ⅓, simply substitute into the point-slope equation,$y-5=\dfrac{1}{3}(x+2)$.

To review, see Linear Equations in Point-Slope Form

### 5i. Locate on a coordinate plane all solutions of a given inequality in two variables

• How do you graph the solutions to a linear inequality?

Linear inequalities are very similar to linear equations, except instead of just finding solutions on the line, we will be finding an entire area of the graph that has solutions to our inequality.

To graph a linear inequality, such as $y < 3x-1$, start by graphing the equivalent equation, $y=3x-1$. The y-intercept is (0,-1) and the slope is 3. When graphing, draw a dashed line, instead of a solid line. For inequalities with the < or > symbols, always use a dashed line. For inequalities with the $\leq$ or $\geq$ symbols, you can use a solid line. Now, pick any point on one side of the line. We will test that point in our inequality to see if it satisfies the inequality. If it does, then we will shade that side. If it doesn't, then we will shade the other side. For example, we will test the point (0,0), which is on the left/upper side of the line. $0 < 3(0)-1$ simplifies to $0 < -1$. This is not true. Therefore we must shade the other side. To review, see Graphs of Linear Inequalities

### 5j. Represent relationships between quantities as an equation or inequality in two variables

• How do you represent the relationship between quantities in an inequality?

Inequalities are used every day in our lives. For example, if you want to buy gas and snacks, but only have $20, you have solved an inequality. For example, if gas is$3 per gallon, and snacks are $4 each, you can create an inequality such as $3g+4s \leq 20$. You can test values for g and s to find possible solutions, such as $g=2$, $s=2$. Since $3(2)+4(2)=14$, and $14 \leq 20$, then $g=2$, $s=2$ is a possible solution. To review, see Graphs of Linear Inequalities. ### 5k. Interpret the meaning of slope and intercepts of the graph of a relationship between quantities • How can you identify parallel lines from their slopes? • How can you identify perpendicular lines from their slopes? Parallel lines are two lines that never intersect. Think of parallel lines like the lines on a highway, they never intersect. When looking at the equations of two lines, the key to determining if the lines are parallel is to examine their slopes. Parallel lines must have the same slope. For example, the lines $y = 2x + 1$ and $y = 2x - 3$ are parallel because they both have a slope of 2. Perpendicular lines are two lines that intersect at a 90 degree angle. The product of the slopes of two perpendicular lines is always equal to -1. Another way to identify perpendicular lines is that the slope of one line is the opposite reciprocal of the other line. The opposite means change the sign, and reciprocal means to flip the number, making the numerator the denominator, and vice versa. For example, the lines $y=2x-1$ and $y=-\dfrac{1}{2}x+5$ are perpendicular since the opposite reciprocal of 2 is $-\dfrac{1}{2}$. To review, see Parallel and Perpendicular Lines ### Unit 5 Vocabulary This vocabulary list includes terms listed above that students need to know to successfully complete the final exam for the course. • coordinate plane • x-axis • y-axis • origin • point • coordinates • x-y table • slope • rise over run • x-intercept • y-intercept • slope-intercept form • point-slope form • linear inequalities • equivalent equation • parallel lines • perpendicular lines • product • opposite reciprocal ## Unit 6: Systems of Linear Equations and Inequalities ### 6a. Determine the number of solutions of a given system of linear equations • How do you verify if a point is a solution to a system of equations? A solution to a system of equations is just like the solution to a single linear equation, except that the point must satisfy both equations in order to be considered the solution to the system of equations. For example, the system of equations: $x+2y=13$ $3x-y=-11$ Let's check if the point (-1,7) is a solution. To check, first we will substitute the point into the first equation. $-1+2(7)=-1+14=13$ So far, the point works, but we must make sure it works in the other equation as well: $3(-1)-7=-3-7=-10$ Since this does not satisfy both equations, (-1,7) is not a solution to this system. To review, see Checking Solutions for Systems of Linear Equations ### 6b. Classify systems of linear equations according to the number of solutions • How do you know the number of solutions of a system of linear equations? Systems of linear equations can have 0, 1, or infinite solutions. Most commonly, two lines intersect at only one point, meaning the system has 1 solution. If the two lines are parallel, then they never intersect, and therefore the system has no solution. Finally, if the system has two equations that are actually representative of the same line, then all the points on each line are also a solution to the other equation, meaning there are infinitely many solutions. To review, see Using Graphs to Solve Linear Equations. ### 6c. Solve systems of linear equations using graphing, substitution, or elimination • How do you solve a system of linear equations with graphing? • How do you solve a system of linear equations with substitution? • How do you solve a system of linear equations with elimination? Systems of linear equations can be solved through 3 methods, each with advantages and disadvantages. First, systems of linear equations can be solved by graphing. To use graphing, you only need to graph each line on the same coordinate plane, and then find the point where the lines cross. That point is the solution to the system. For example, consider the following system of equations: $y=3x-1$ $y=-x+3$ We can graph both lines and look for the point where they intersect. Since the lines intersect at (1,2), that is the solution to the system. The graphing method works well when the solution is a lattice point, with whole number values, but is not as effective if the answers are fractions or decimals. Another method is substitution. Substitution is an algebraic method, rather than the geometric method of graphing. In substitution, we solve one equation for either $x$ or $y$, and then substitute that value into the other equation to find the value of one variable. Once we have found that value, we can substitute it to find the value of the other variable. For example, let us once again consider our example: $y=3x-1$ $y=-x+3$ Since $y=-x+3$ in the second equation, we can replace the $y$ in the first equation with that value: $-x+3=3x-1$ Now we can solve for $x$. $4=4x$, therefore $x=1$. We can now use that value to find the value of $y$: $y=3(1)-1=2$. Therefore the solution is (1,2) Finally, we can solve a system of equations by elimination. This method is best for systems where one variable can't be isolated that easily. In this method, we multiply one or both equations so that when we add them together, one of the variables cancels. This will allow us to solve for one variable, and then as we did with substitution, we can use that value to find the other remaining value. For example consider the following system of equations: $3x+2y=7$ $-x-4y=6$ First, we will multiply the top equation by 2, so that when we add the equations, the $y$ terms will cancel: $6x+4y=14$ $-x-4y=6$ Now we add the two equations together and solve for $x$: $5x=20$, $x=4$ Now that we know $x=4$, we can substitute into one of the original equations to find $y$: $3(4)+2y=7$ Now we can solve for $y$: $2y=-5$, $y=-52$ Therefore the solution to this system of linear equations is (4, -52) ### 6d. Locate on a coordinate plane all solutions of a given system of inequalities • How do you graph the solutions to a system of linear inequalities? When solving a system of inequalities, graph the solution to each inequality, and shade the side with the solutions. When you have done both, look for the area where the shading overlaps. That is the area with the solutions that work for both inequalities, and are therefore the solutions to the system of inequalities. For example:    To review, see Using Graphs to Solve Linear Equations ### 6e. Create systems of equations and use them to solve world problems • How do you use systems of equations to solve real-world problems? As we have seen, systems of equations are helpful in solving real-world problems. When given a real-world problem, we can create a system of equations to find the solution. For example, consider the following problem: Juan is considering two cell phone plans. The first company charges$120 for the phone and $30 per month for the calling plan that Juan wants. The second company charges$40 for the same phone but charges $45 per month for the calling plan that Juan wants. After how many months would the total cost of the two plans be the same? First, we need to create two linear equations to represent the problem: First company: $120+30x=y$ Second company: $40+45x=y$ Since 120 and 40 are the fixed costs, they are the constants, and the monthly cost is the coefficient of $x$, since each month you have to pay that amount. Now we can solve this system of equations in any of the ways we have already learned, such as elimination, substitution, or graphing. In this case, we can use substitution to get $x$: $120+30x=40+45x$ We can solve this to find $80=15x$, $x=\dfrac{80}{15}=5\dfrac{1}{3}$ months. #### 6f. Use systems of inequalities to model word problems and interpret their solutions in the context of the problem • How can you use systems of inequalities to solve word problems? Sometimes, a system of equations isn't appropriate for our problem. Instead, a system of inequalities should be used. For example, consider the following problem: Jake does not want to spend more than$50 on bags of fertilizer and peat moss for his garden. Fertilizer costs $2 a bag and peat moss costs$5 a bag. Jake's van can hold at most 20 bags.

First, we must create our inequalities. We will let Fertilizer $= x$, and Peat Moss $= y$:

$x+y \leq 20$

$2x+5y \geq 50$

In addition to these two inequalities that we can create from the problem, remember that $x \geq 0$ and $y \geq 0$, since Jake obviously can't have a negative number of bags of something.

We can now graph the solution to this system and then interpret the answers: As you can see in the solution above, the area with the diagonal lines is the solution to our system of equations. Every point in that area is a solution. For example, (5,5) is a solution, meaning Jake could buy 5 bags of fertilizer and 5 bags of peat moss.

### Unit 6 Vocabulary

This vocabulary list includes terms listed above that students need to know to successfully complete the final exam for the course.

• system of equations
• intersect
• parallel
• graphing
• lattice point
• substitution
• elimination
• system of inequalities

## Unit 7: Operations with Monomials

### 7a. Apply the rules of exponents to simplify algebraic exponential expressions

• What happens when you multiply two monomials?
• What does a negative exponent mean, and how can you change a negative exponent to a positive exponent?

Exponents are simply a shorter way to write repeated multiplication. Since $2^4=2 \times 2 \times 2 \times 2$, when we try to simplify $2^4 \times 2^3$, we need to remember this is four 2's multiplied with three 2's, meaning we have seven 2's multiplied together, or $2^7$. We can therefore see that multiplication property states: $a^m \times a^n = a ^ {m+n}$. Similarly, $(3x)^3 \times (3x)^7=(3x)^{10}$.

When dealing with exponents, you may come across a negative exponent. A negative exponent doesn't mean the answer will be negative. Instead, it means to take the reciprocal of the value, what you might call "flipping it". For example, $2^{-3}$ simply means the reciprocal of $2^3$, which is $\dfrac{1}{2^3}$. Similarly, if there is a negative exponent in the denominator of a fraction, it moves the term to the numerator. For example, $\dfrac{1}{3^{-2}}=3^2$.

To review, see:

### 7b. Multiply, divide, and simplify the powers of monomials

• How do you divide monomials with exponents?
• How do you take an exponent to another exponent?
• How do you multiply and divide different monomials?

When we divide monomials with exponents, we subtract our exponents, rather than adding, like we do when we multiply. For example, $\dfrac{4^5}{4^3}=4^2$. Therefore, the division property states: $\dfrac{a^m}{a^n} = a^{m-n}$. Similarly, $\dfrac{x^3}{x^7}=x^{-4}$.

Sometimes, you might even have an exponent taken to another exponent, such as $(x^2)^3$. When this happens, you need to multiply the exponents, giving us $(x^2)^3=x^{2 \times 3}=x^6$. Therefore, the power property states: $(^am)^n=a^{m \times n}$. Similarly, $(3x^3)^4=3^4 x^{3 \times 4} =81x^{12}$.

We don't have to just multiply and divide the same monomial, we can multiply different monomials as well. To simplify the expression $(3x^2)(5x^3)$, we will multiply the numbers as normal, and then add the exponents on the variable, giving us $15x^5$.

Finally, we can divide different monomials. For example, $\dfrac{8x^4}{4x^3}$ can be simplified by first simplifying the numbers in the fraction, then using the division property to subtract the exponents, giving us an answer of $2x$.

To review, see:

### Unit 7 Vocabulary

This vocabulary list includes terms listed above that students need to know to successfully complete the final exam for the course.

• exponents
• multiplication property
• negative exponents
• reciprocal
• denominator
• numerator
• monomials
• division property
• power property

## Unit 8: Operations with Polynomials

### 8a. Identify the degree of a polynomial

• How do you identify the degree of a polynomial?

Polynomials are often classified by their degree, which is the highest power of any of the terms in the polynomial. For example, in the polynomial $y=5x^3-4x+7$, the highest power on the variable is 3, therefore this is considered a 3rd degree polynomial. The degree is always the highest power on the variable, even if that isn't the first term, in standard form.

To review, see Introduction to Polynomials

### 8b. Classify polynomial according to the number of terms

• What do you call a polynomial?

We can also classify polynomials by their number of terms. For example, a single term, like $2x^3$ is called a monomial. Two monomials added or subtracted together form a binomial, and three monomials together are called a trinomial. Each of those monomials is called a term. If there are more than 3 terms, we generally just call it a polynomial.

For review, see Introduction to Polynomials

### 8c. Add, subtract, and multiply polynomials

• How do you add polynomials?
• How do you subtract polynomials?
• How do you multiply polynomials?

When adding polynomials, we will follow the rules we already know about collecting like terms. Like terms must have the same variable and exponents in order to add them, and the same is true for polynomials. For example, to add the polynomials $3x^4-5x^2+x+5$ and $2x^4+x^3+4x+3$, we identify the like terms, and add their coefficients.

$3x^4+2x^4=5x^4, x+4x=5x$, and $5+3=8$

Note that $-5x^2$ and $x^3$ have no like terms and therefore won't be added to anything. Now we can write our final answer, $5x^4+x3-5x^2+5x+8$.

When subtracting two polynomials, we can simply apply the subtraction to each term in the second polynomial and then add as we just did. For example, $(3x^3+2x^2+1)-(x^3-3x^2+5)$ can be rewritten as $(3x^3+2x^2+1)+(-x^3+3x^2-5)$. Now, this problem is the same as an addition problem, and we can simply collect the like terms.

Multiplying polynomials is a bit more work, and the approach depends somewhat on the types of polynomials being multiplied. First, let's consider a monomial multiplied by any other polynomial, such as this:

$3x(5x^2-3x+1)$. To simplify, we need to apply the distributive property, which states that we can multiply the $3x$ times each term inside the parentheses individually:

$3x(5x^2-3x+1)=(3x)(5x^2)+(3x)(-3x)+(3x)(1)=15x^3-9x^2+3x$

If we multiply anything other than a monomial by our polynomial, we have to use a slightly more complex version of the distributive property, where every term in the first polynomial is individually multiplied by every term in the second polynomial, then all like terms are collected and the final answer is found. For example:

$(x+2)(x^2-4x-1)=(x)(x^2)+(x)(-4x)+(x)(-1)+2(x^2)+2(-4x)+2(-1)$.

Now, as we multiply, we must also take note that there are like terms that will need to be collected:

$x^3-4x^2-x+2x^2-8x-2$

Once we collect our like terms, we get our final answer: $x3-2x^2-9x-2$.

To review, see:

### 8d. Divide a polynomial by a monomial and a binomial

• How do you divide a polynomial by a monomial?
• How do you divide a polynomial by a binomial?

Dividing polynomials requires us to carefully consider the type of polynomial that we are dividing by.

First, let's consider dividing by a monomial. When we divide a polynomial by a monomial, we can simply divide each term of the polynomial in the numerator by the monomial in the denominator. Each of these are then simplified to get our final answer. For example, consider $\dfrac{3x^4+2x^2-6}{3x^2}$. We first rewrite this problem so each term in the numerator is divided individually, then we simplify each of those fractions as such:

$\dfrac{3x^4}{3x^2}+\dfrac{2x^2}{3x^2}-\dfrac{6}{3x^2}=x^2+\dfrac{2}{3}-\dfrac{2}{x^2}$

Dividing a polynomial by a binomial is a bit more challenging. We will use long division, similar to what you learned when you first learned how to divide by hand, to divide these polynomials. Consider the following example: When dividing a polynomial by a binomial, you simply need to follow this division algorithm until you finish the problem.

To review, see:

### 8e. Identify special products of binomials by completing the square and finding the difference of two squares

• How do you square a binomial?
• How do you multiply a difference of squares?

Although by now we are comfortable with exponents, an exponent on a binomial can be computed incorrectly without the correct approach. When squaring a binomial, you cannot simply square each term in the binomial. You must write the binomial twice, and use the distributive property to multiply each term in each binomial with the terms in the other binomial.

For example: $(x+3)^2=(x+3)(x+3)$. Now, we multiply as we have learned, and collect the like terms:

$(x+3)(x+3)=x^2+3x+3+9=x^2+6x+9$

This is the correct way to multiply binomials. Be careful not to forget this step.

Another special kind of binomial multiplication is called a difference of squares. A difference of squares is when the two binomials being multiplied are of the form $(x+a)(x-a)$. When multiplied, the inner and outer terms always cancel, giving an answer of $x^2-a^2$. For example, $(x+5)(x-5)=x^2-25$.

### Unit 8 Vocabulary

This vocabulary list includes terms listed above that students need to know to successfully complete the final exam for the course.

• polynomials
• highest power
• degree
• monomial
• binomial
• trinomial
• like terms
• coefficients
• long division
• division algorithm
• squaring
• difference of squares

## Unit 9: Factoring Polynomials

### 9a. Rewrite polynomial as a product of two or more polynomials

• How can you rewrite a polynomial as a product of two polynomials?

Some polynomials have a common factor which all the terms can be divided by evenly. This common factor can be factored out, creating the product of the common factor and what remains of the polynomial. This is commonly referred to as factoring out a common factor.

For example, consider the polynomial $6x^3+12x^2-3x$. First, we must inspect the three terms and find the greatest common factor (GCF) of the coefficients. In this case, the GCF is $3$. Similarly, we look for the GCF of the variable, which is the lowest degree of any of the terms. In this case, it is $x$. We will now factor out the $3x$, by dividing each term and writing what is left inside the parentheses:

$6x^3+12x^2-3x=3x(2x^2+4x-1)$

This is now the factored expression. You can verify that the answer is correct by multiplying the factor outside to the polynomial inside and checking that you get what you started with.

### 9b. Identify polynomials that cannot be factored

• How do you know if a polynomial can be factored?

Not all polynomials can be factored. The first type of factoring to check for is always a common factor. If there is no common factor, there may be no other way to factor the polynomial. For trinomial quadratics, you should attempt one of the methods for factoring into two binomials and exhaust all possibilities before deciding the polynomial doesn't factor. A polynomial that does not factor is called a prime polynomial.

To review, see More on Factoring Polynomials.

### 9c. Choose appropriate factoring strategy for a given polynomial

• What factoring strategies are appropriate for polynomials with 4 terms?
• What factoring strategies are appropriate for polynomials with 3 terms?
• What factoring strategies are appropriate for polynomials with 2 terms?

The strategy with which you attempt to factor a polynomial will depend on the number of terms in the polynomial. Regardless of the type of polynomial, you should always attempt to find and factor out a common factor before attempting any of these methods, as factoring out a common factor will always make the process easier.

When a polynomial has 4 terms, the only method we know how to use is factor by grouping. Group the first two terms together, and the last two terms together, then factor out the common factor for each group. If the polynomial can be factored by grouping, the factors that remain in the parentheses after taking out the common factor will be identical. If they are not, either you didn't factor completely/correctly, or the polynomial can't be factored by grouping.

For example, consider $3x^3+x^2+6x+2$. First we group our terms, and factor out the common factor from each: $(3x^3+x^2)+(6x+2)=x^2(3x+1)+2(3x+1)$. Now, notice the binomial in each set of parentheses is the same. We then factor that common binomial out, leaving us with the factored polynomial:

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

When we have a quadratic trinomial, we can factor either by guessing and checking the factors in two binomials, or using the "AC" method.

When factoring a binomial, the available methods depend on the degree. If the binomial is of the 1st degree, then only a common factor method can be used. If the binomial is of the 2nd degree (a quadratic), then check if it can be factored by difference of squares. If not, it cannot be factored.

If the binomial is of the 3rd degree, check if it can be factored by the sum or difference of cubes. The sum of cubes factors as $(x^3+a3)=(x+a)(x^2-ax+a^2)$ and the difference of cubes factors as $(x^3-a^3)=(x-a)(x^2+ax+a^2)$. For example, the binomial $x^3+27$ can be factored as $x^3+27=x^3+3^3=(x+3)(x^2-3x+6)$.

To review, see:

### 9d. Solve quadratic equations by factoring

• How do you factor quadratics with a leading coefficient of 1?
• How do you factor quadratics with a leading coefficient that is not 1?

When factoring a quadratic expressions of the form $ax^2+bx+c$, the approach you take may depend on the values for $a$, $b$, and $c$. First we will discuss factoring when $a = 1$.

When factoring a quadratic with a leading coefficient of 1, you just need to focus on finding two numbers that add to give the value of $b$, and the same two numbers that multiply to give the value of $c$. For example, when factoring the quadratic expression $x^2-2x-8$, you must find two numbers that add to make -2, and multiply to make -8. Those two numbers are 2 and -4. Therefore, $x^2-2x-8$ can be factored as $(x-4)(x+2)$. You can always multiply your answer using the distributive property to check that it is correct.

When factoring a quadratic in which a does not equal 1, we have to use a different approach. One approach is to use a guess and check method, where the leading coefficients in each binomial multiply to equal the leading coefficient in the quadratic, and the constants in each binomial multiply to make the constant in the quadratic. For example, when we factor $2x^2-13x-45$, we can try different combinations of $2x$ and $x$ for our leading term in each binomial, along with pairs of factors of -45, such as 1 and 45, 3 and 15, and 5 and 9. By trial and error we can see that $(2x-3)(x+15)=2x^2+27x-45$, which means this factoring is incorrect. Eventually you will find the correct factoring, which is $(2x+5)(x-9)$.

The other method for factoring a quadratic in which a does not equal 1, is to use the "ac method". This method uses factor by grouping strategies to make factoring more standardized. For this approach, let's consider the same problem, $2x^2-13x-45$. To begin, we will find ac, which is $(2)(-45)=-90$. Therefore we must now find two numbers that multiply to make -90, and also add to make b, which is -13. The two numbers that satisfy this condition are (-18) and (5). Now we will replace b with these two numbers, making our expression $2x^2-18x+5x-45$. This can now be factored by grouping, giving us $2x(x-9)+5(x-9)$. By factoring out the $(x-9)$, we get the final answer, $(x-9)(2x+5)$.

To review, see:

### Unit 9 Vocabulary

This vocabulary list includes terms listed above that students need to know to successfully complete the final exam for the course.

• common factor
• factoring out
• greatest common factor (GCF)