Math Study Guide for the NLN NEX

Algebra

Algebra is a branch of mathematics that deals with finding unknown values. These unknown values are represented with symbols, and there are rules for manipulating those symbols that will help you solve equations and understand mathematical relationships.

Algebraic Terms

There are several fundamental terms that form the building blocks of algebra. Here are five common terms:

variable—This is a symbol that is used to represent an unknown or an arbitrary number. It is usually represented by $x$, but $y$, $a$, $b$, or any other letter is also possible.
constant—This is a number that is fixed and is not dependent on the equation in question. It can be thought of as the opposite of a variable. Examples of constants include $2$, $\pi$, and $\sqrt{5}$.
expression—This is a combination of variables, constants, and operators. There is no equal sign in an algebraic expression. For example, $2x$ and $4y + 10$ are algebraic expressions.
equation—This is a mathematical statement that shows two expressions are equal. If you put an equal sign ($=$) in between two expressions, that makes it an equation. For example, $2x + 10 = 20$ and $x^2 =64$ are equations.
term—This is a single mathematical expression within an equation. In the equation $2x + 16 = 4x$, the individual terms are $2x$, $16$, and $4x$.
like terms—These are terms in an expression that have the same variables raised to the same power. In the expression $3x - 5y^2+6x-2y$, the like terms are $3x$ and $6x$.

The Order of Operations

In algebra, adhering to the order of operations is paramount to avoiding mistakes and ensuring consistent calculations. You must perform mathematical operations in a particular sequence or you will get the wrong answer. PEMDAS is an acronym that can help you remember the sequence:

P—parentheses
E—exponents
M/D—multiplication and division
A/S—addition and subtraction

Note: All operations are performed from left to right within their respective steps in the sequence. For example, if there are multiple addition operations, perform the left-most one first.

Let’s try a problem that requires following the order of operations.

What is the value of the expression $3 + 5 \times ( 2- 4)$?

Using PEMDAS, we know that we must first evaluate the parentheses:

\[3 + 5 \times ( 2- 4)\] \[= 3 + 5 \times -2\]

Now, we need to multiply $5$ and $-2$:

\[= 3 + -10\]

Finally, we do addition and get our answer:

\[= -7\]

If you didn’t follow PEMDAS and, for instance, performed addition before multiplication, your result would have been different and incorrect. Following the order of operations is essential in math. Let’s look at another example.

What is the value of the expression $10 \times 2 - (2^2 + 3) - 10$?

Using PEMDAS, we will first deal with what’s inside the parentheses:

\[10 \times 2 - (2^2 + 3) - 10\] \[= 10 \times 2 - (4 + 3) - 10\] \[= 10 \times 2 -7 - 10\]

Note: We followed PEMDAS even inside the parentheses.

Now, let’s multiply $10$ and $2$:

\[= 20 - 7 - 10\]

Finally, we will subtract, starting from the left and going to the right:

\[= 3\]

Helpful Algebraic Properties

Algebraic properties provide a systematic way to manipulate mathematical expressions, making complex problems more manageable. There are many algebraic properties that help us manipulate equations. Here, we will learn about two fundamental properties that play a pivotal role in algebraic operations: the distributive property and the property of zero.

The Distributive Property

At the heart of algebraic simplification is the distributive property, a foundational principle for the manipulation of expressions. It involves multiplication and addition or subtraction. The distributive property is:

\[a \times (b+c) = (a \times b) + (a \times c) \text{ and } a \times (b-c) = (a \times b) - (a \times c)\]

As you can see, multiplying a term by the sum or difference of two others is equivalent to multiplying the term by each individually and then combining the results.

Consider the expression $2y(b+c)$. Using the distributive property, we can expand it as follows:

\[2y(b + c) = 2y \times (b+c)\] \[= (2y \times b) + (2y \times c)\] \[= 2yb + 2yc\]

This simplification demonstrates the value of the distributive property in handling algebraic terms, enabling a more concise representation of complex expressions.

The Property of Zero

Another fundamental property in algebra is the property of zero. It is expressed as follows:

\[x \times 0 = 0\]

It tells us that any value (except zero) multiplied by zero will result in zero. While seemingly straightforward, the implications of the property of zero reverberate throughout algebra.

Computation

Equations and expressions are powerful tools that enable you to model real-world problems and systematically find solutions. Here, we will look at simplifying expressions, solving equations, and clearing fractions

Expressions

Simplifying expressions involves streamlining mathematical statements by combining like terms and applying various algebraic rules. This process enables you to represent complex expressions in a more straightforward and manageable form.

For example, consider the expression $-10x + 2y - 7y + 2x$. By combining the like terms, you can simplify it to $-8x - 5y$, making it more concise and easier to work with.

Example 1

Simplify the expression $3a + 5b - 2a - 7b + 4a$.

Solution

To simplify this expression, start by grouping the like terms.

The terms involving $a$ are $3a$, $-2a$, and $4a$.
The terms involving $b$ are $5b$ and $-7b$.

Now, combining the like terms, we have:

\[(3a - 2a + 4a) + (5b - 7b)\] \[= (3-2+4)a + (5-7)b\] \[= 5a -2b\]

Example 2

Simplify the expression $2x^2 + 3x - 4 + 5x^2 - x + 6$.

Solution

Start by identifying and grouping the like terms:

The terms involving $x^2$ are $2x^2$ and $5x^2$.
The terms involving $x$ are $3x$ and $-x$.
The constant terms are $-4$ and $6$.

Now, combining the like terms, we have:

\[(2x^2 + 5x^2) + (3x - x) + (-4 + 6)\] \[= (2+5)x^2 + (3-1)x +2\] \[= 7x^2 + 2x + 2\]

Equations

Equations are mathematical sentences expressing the equality of two expressions. Solving equations involves determining the values of the variables that make the equation true.

For instance, let’s take the equation $5p - 10 = 15$. To solve for the value of $p$, you need to first add $10$ to both sides to get $5p = 25$. Then, you will divide both sides by $5$ to get $p = 5$. This process allows you to find the specific values that satisfy the given equation.

Example 1

Solve for $y$:

\[3y + 7 = 22\]

Solution

The goal is to isolate the variable $y$. Start by subtracting $7$ from both sides of the equation:

\[3y + 7 - 7 = 22 - 7\]

This simplifies to:

\[3y = 15\]

Next, divide both sides by $3$ to solve for $y$:

\[\frac{3y}{3} = \frac{15}{3}\] \[y = 5\]

Example 2

Solve for $x$:

\[4(x - 2) = 3x + 6\]

Solution

Start by expanding the left side:

\[4x - 8 = 3x + 6\]

Next, move all terms involving $x$ to one side of the equation by subtracting $3x$ from both sides:

\[4x - 3x - 8 = 3x + 6 -3x\]

This simplifies to:

\[x - 8 = 6\]

Now, add $8$ to both sides to solve for $x$:

\[x - 8 + 8 = 6 + 8\] \[x = 14\]

Clearing Fractions from Equations

Clearing fractions from equations is a fundamental skill in algebraic manipulation. This process involves eliminating denominators to make solving equations more straightforward. Consider the equation $\frac{3}{4}x = 6$. To get rid of the fraction, you would multiply both sides of the equation by $4$:

\[4 \times \left(\frac{3}{4}x\right) = 4 \times 6\] \[3x = 24\]

Now, you can divide both sides by $3$ to get $x = 8$. This manipulation simplified the equation, making it easier to find the value of $x$.

Inequalities

Inequalities in mathematics express relationships between two quantities that are not equal. These relationships are represented by inequality symbols, allowing us to compare values without specifying an exact equality. The following are examples of inequalities:

\[2x + 10 > 9\] \[3x - 5 \leq 4x + 11\]

Inequality Symbols

In mathematics, there are four basic inequality symbols to denote different relationships between numbers:

greater than—This indicates that the quantity on the left is greater than the quantity on the right. The symbol is $>$. For example, $10 > 2$ tells us that $10$ is greater than $2$.
less than—This indicates that the quantity on the left is less than the quantity on the right. The symbol is $<$. For example, $2 < 8$ tells us that $2$ is less than $8$.
greater than or equal to—This indicates that the quantity on the left is greater than or equal to the quantity on the right. The symbol is $\geq$. For example, $2x \geq 10$ tells us that $2x$ is greater than or equal to $10$.
less than or equal to—This indicates that the quantity on the left is less than or equal to the quantity on the right. The symbol is $\leq$. For example, $8x \leq 64$ tells us that $8x$ is less than or equal to $64$.

Solving Inequalities

Solving inequalities is similar to solving linear equations, though there is a key difference. Consider the following example, which shows how an inequality is solved in almost the same way as its corresponding equation. The only difference is the use of an inequality symbol instead of an equal sign:

OLD10 Solving Inequalities.png

The key difference between solving inequalities and equations comes in when you multiply or divide by a negative number. In those situations, the inequality symbol is reversed.

Example 1

\[10x - 60 \leq 30x\] \[10x - 30x \leq 60\] \[-20x \leq 60\] \[\frac{-20x}{-20} \geq \frac{60}{-20}\] \[x \geq - 3\]

Example 2

Solve for the range of values of $p$:

\[2(p-2) > 4p - 6\]

Solution

Using the distributive property, we simplify the left side:

\[2p - 4 > 4p - 6\]

Next, we will subtract $4p$ from both sides to get:

\[2p - 4 - 4p > 4p - 6 - 4p\] \[-2p - 4 > - 6\]

Now, we add $4$ to both sides to isolate the term with $p$:

\[-2p - 4 + 4 > -6 + 4\] \[-2p > -2\]

Dividing both sides by $-2$ will give us our answer. Remember to reverse the inequality symbol since we are dividing by a negative number:

\[\frac{-2p}{-2} < \frac{-2}{-2}\] \[p < 1\]

Problem Solving

Solving word problems often involves translating verbal descriptions into algebraic expressions or equations to find a mathematical solution. This process requires identifying key information, determining the unknowns, and establishing relationships between them. Below, we outline the steps to achieve this:

Determine what needs to be found and identify the unknowns. Assign variables to unknown quantities, such as $x$ or $y$.
Express the given information as equations/expressions. For instance, “twice the [unknown variable]” can be translated as $2x$. Identifying keywords is the key here.
Relate information using symbols. For instance, “the sum of two numbers” can be conveyed as $x + y$.
Form equations that represent relationships, such as $2x = x + y$.

To translate words to algebraic equations and expressions, you need to be familiar with certain common keywords. Some are listed below:

sum—indicates an addition operation (e.g., $x + y$)
difference—indicates a subtraction operation (e.g., $x - y$)
product—indicates a multiplication operation (e.g., $xy$, $4x$)
quotient—indicates a division operation (e.g., $\frac{x}{y}$)
is—denotes equality (e.g., $3x = y$)
more than—indicates greater than (e.g., $x > y$)
fewer than—indicates less than (e.g., $x < y$)
twice—represents multiplication by $2$ (e.g., $2y$)

By recognizing these keywords and employing the outlined steps, you can effectively set up and solve algebraic equations to find solutions to word problems.

Word Problems Using Expressions

Expressions are useful in word problems when you need to simplify or represent a situation without necessarily solving for an exact value. These problems often involve translating a verbal statement into a mathematical expression representing relationships or changes. We’ll see how this works with a sample problem.

A movie theater sells two types of tickets, adult tickets and child tickets. The cost of an adult ticket is $\$12$, and the cost of a child ticket is $\$7$. Write an expression that represents the total cost for buying $a$ adult tickets and $c$ child tickets.

Solution

To solve this problem, we need to express the total cost in terms of the number of adult and child tickets. For $a$ adult tickets priced at $\$12$ each, we have:

\[12a\]

For $c$ child tickets priced at $\$7$ each, we have:

\[7c\]

Thus, the total cost, $T$, is $T = 12a + 7c$.

This expression can now be used to calculate the total cost for any number of adult and child tickets. For example, if someone buys three adult tickets and two child tickets, the total cost would be:

\[T = 12(3) + 7(2) = 36 + 14 = \$50\]

Word Problems Using Equations

Equations come into play when a word problem requires finding the exact value of an unknown quantity. This typically involves setting up an equation based on the relationships described in the problem and then solving for the unknown variable. Consider the following problem.

A chemist needs to prepare $200$ milliliters of a $50\%$ alcohol solution. She has a $70\%$ alcohol solution and a $30\%$ alcohol solution. How many milliliters of each solution should she mix to create the $50\%$ solution?

Solution

Let $x$ represent the number of milliliters of the $70\%$ alcohol solution, and $200 - x$ represent the number of milliliters of the $30\%$ alcohol solution.

We can set up an equation based on the amount of pure alcohol in each solution:

\[0.70x + 0.30(200 - x) = 0.50(200)\]

Now, we solve for $x$:

\[0.70x + 60 - 0.30x = 100\] \[0.40x + 60 = 100\] \[0.40x = 40\] \[x = \frac{40}{0.40} = 100\]

So, the chemist should mix $100$ milliliters of the $70\%$ solution with $100$ milliliters of the $30\%$ solution to create $200$ milliliters of a $50\%$ solution.

Word Problems Using Inequalities

Inequalities are used in word problems when involving a range of possible values rather than a single solution. These problems require setting up an inequality to express a condition that must be met. Consider the following example problem.

A company wants to buy computers for its office. Each computer costs $\$800$, and the company has a budget of $\$10\text{,}000$. They also need to reserve $\$2\text{,}000$ from the budget for other expenses. What is the maximum number of computers the company can buy?

Solution

Let $n$ represent the number of computers the company can buy.

The cost for $n$ computers is $\$800n$. The company needs to reserve $\$2\text{,}000$, so the budget available for computers is $\$10\text{,}000 - \$2\text{,}000 = \$8\text{,}000$.

Now, set up the inequality:

\[800n \leq 8\text{,}000\]

To find the maximum number of computers, solve for $n$:

\[n \leq \frac{8\text{,}000}{800} = 10\]

So, the company can buy a maximum of $10$ computers to stay within their budget.

Multi-Step Problem Solving

Multi-step problems require performing several operations or processes to reach a solution. These problems are more complex than single-step problems because they involve combining different mathematical techniques, such as using multiple operations or working with various units. In a medical context, multi-step problems often arise when calculating medication dosages, converting units, or determining the correct proportions of substances.

Let’s look at a couple multi-step problems.

A patient is prescribed a medication that needs to be taken in a dosage of $250$ milligrams every eight hours for five days. The medication comes in $500$-milligram tablets, which need to be split in half for the correct dosage. Calculate the total number of tablets the patient will need to take over the five days.

Solution

Let’s go over the thinking process and solve the problem.

Determine the dosage per day.—Since the medication is taken every $8$ hours and there are $24$ hours per day, we divide by $8$ to get $3$ doses a day.

Calculate the total dosage over five days.—We multiply the number of doses per day ($3$) by the number of days ($5$), which gives us $15$ doses in total.

Determine the number of tablets required.—Each tablet is $500$ milligrams, but the required dosage is $250$ milligrams, so each tablet provides two doses. Therefore, the total number of tablets needed is:

\[\frac{15}{2} = 7.5\]

The total number of tablets the patient will take is $7.5$. However, since you can’t have half a tablet in this context, the patient will need 8 tablets.

Here’s another problem like this.

A nurse needs to administer $1.5$ liters of IV fluids to a patient over six hours. The IV setup delivers fluids at a rate of $20$ drops per milliliter, and the fluid bag has a capacity of $500$ milliliters. How many drops per minute should the nurse set the IV to deliver, and how many full fluid bags will be needed?

Solution

Let’s go over the thinking process and solve the problem.

First, convert liters to milliliters:

\[1.5 \text{ L} = 1\text{,}500 \text{ mL}\]

Now, calculate the total number of drops needed:

\[1\text{,}500 \text{ mL} \times 20 \text{ drops/mL} = 30\text{,}000 \text{ drops}\]

Then, determine the drops per minute:

\[6 \text{ hr} = 360 \text{ min}\] \[\frac{30\text{,}000 \text{ drops}}{360 \text{ min}} \approx 83.33 \text{ drops/min}\]

Last, calculate the number of fluid bags needed:

\[\frac{1\text{,}500 \text{ mL}}{500 \text{ mL/bag}} = 3 \text{ bags}\]

Therefore, the nurse should set the IV to deliver approximately $83$ drops per minute, and three full fluid bags will be required to administer the $1.5$ liters over six hours.