Algebra

As we get started in CHEM 110, it's important to quickly review algebra. Many of the problems you'll encounter in CHEM 110 will require you to manipulate and solve algebraic expressions. Instead of providing a comprehensive review of all topics in algebra, this page will only cover a few targeted topics where students often have difficulty.

Order of Operations

When calculating an algebraic expression, it is important to carry out operations in the order listed below:

1. Parentheses (carry out operations enclosed by parentheses)
2. Exponents
3. Multiplication and Division
4. Addition and Subtraction

Let's take a look a couple of examples:

Example 1: calculate (4+3)/5

First carry out the operation in the parentheses:

(4+3)/5 $\Longrightarrow$ 7/5

There are no exponents, so next we proceed to multiplication/division. Let's divide to get our final answer:

7/5 $\Longrightarrow$ 1.4

Example 2: calculate: 4+3/5 =

There are no parentheses or exponents, so we begin with multiplication/division. Let's divide:

4+3/5 $\Longrightarrow$ 4 + 0.6

Next we proceed to addition/subtraction. Let's add to get our final answer:

4+0.6 $\Longrightarrow$ 4.6

Example 3: Order of operations often presents problems when students enter calculations into calculators. Be sure to use parentheses as needed. The most common mistakes we see are when students are dividing by a product, for example:

$\frac{10}{5\cdot4}=$

If you type "10/5*4" into your calculator, you will get an answer of 8, which is not correct.

If you include parentheses around the product on bottom of the fraction, and type "10/(5*4)" into your calculator, you will get the correct answer, 0.5.

Similarly, typing "10/5/4" will also give you the correct answer.

Multiplying and Dividing Exponents

When multiplying two exponents together, you can add the exponents together provided both numbers have the same base. Generically, this looks like this (where "n" is any number or variable):

$n^x\cdot n^y=n^{x+y}$

Similarly, when you divide two of the same base, you can subtract the exponents. Generically, this looks like this (where "n" is any number or variable):

$\frac{n^x}{n^y}=n^{x-y}$

Example 1:

$10^2\cdot10^{-6}=$

Since we're multiplying two exponent numbers with the same base, the answer will have the same base (10), and the exponent will simply be the sum of the two exponents:

$10^2\cdot10^{-6}=10^{-4}$

If you're asked to express this as a decimal, keep in mind that the negative exponent is equivalent to saying 1/10⁴, or 1/10,000, or simply 0.0001.

Example 2:

$10^2\div10^{-6}=$

Since we're dividing two exponent numbers with the same base, the answer will have the same base (10), and the exponent will simply be the difference between the two exponents:

$10^2\div10^{-6}=10^{8}$

If you're asked to express this as a decimal number, it would equal 100,000,000

Solving Systems of Equations

Whenever you have a single equation that contains two or variables, it's impossible to solve for the value of a single variable. To do that, you must have a second equation that contains additional information about one or more of the variables found in the original equation. We call this collection of equations a system of equations. One way of solving systems of equations is by using a substitution strategy in which you solve one equation for a particular variable and substitute your result in place of that same variable in the second equation.

Example 1: Solve the following system of equations for x and y.

x - 2 = y + 2

2x = y + 1

Here we are given two equations and asked to solve the system of equations for both x and y. We can do this using the substitution strategy. I need to first pick one of the equations and solve it for one of its variables (i.e., rearrange the equation such that one side of the equation contains x or y by itself). Let's pick the first equation and solve it for x. To do that, we will add 2 to both sides of the equation, this will leave us with x by itself on the left side of the equation:

x = y + 4

[Note: We could have chosen to solve for y, or we could have chosen to solve the second equation for x or y; we would have ultimately obtained the same values of x and y in the end.]

Now that we have solved the first equation for x, let's substitute this equation in place of x in the second equation:

2(y + 4) = y + 1

Now we're left with an equation that contains a single variable, which we can solve! To do so, let's first distribute the 2 across the parentheses on the left side of the equation:

2y + 8 = y + 1

Now let's subtract y from both sides of the equation:

y + 8 = 1

Now let's subtract 8 from both sides of the equation:

y = -7

Now that we know the value of y, let's use it to obtain the value of x. Since I know that y = -7, I can simply substitute -7 in place of y in either of the two original equations. Let's substitute -7 in place of y in the first original equation:

x - 2 = -7 + 2

Now I can solve for x by adding 2 to both sides of the equation and adding the constants together:

x = -7 + 2 + 2 = -3

Now I know the values of both x and y; x = -3 and y = -7

Video Examples

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Click here to try out the Part A quiz for this section.

When you're confident in the skill, take the Part B quiz!