Basic Algebra Study Guide for the Math Basics

Using Algebra for Word Problems

When solving word problems, the first step is to identify the question, but immediately after that, you should assign a variable to answer the question. For example, if the question is “how many miles did I run?”, you might assign the variable d=distance I ran in miles. Once you’ve assigned a variable, you’ll usually want to write an equation. It’s helpful to know a few basic formulas.

\[\text{Distance} = \text{speed} \cdot \text{time}\] \[\text{Percent of change} = \dfrac{\text{new amount} - \text{original amount}}{\text{original amount}} \cdot 100\] \[\text{Probability} = \dfrac{\text{number of ways to succeed}}{\text{total number of possible outcomes}}\] \[\text{Average} = \dfrac{ \text{sum of amounts}}{\text{total number of objects}}\]

Also helpful is translating words into expressions:

\[\begin{array}{|c|c|} \hline \textbf{Words} & \textbf{Operation}\\ \hline \text{sum, added to, increased by, total} & +\\ \hline \text{difference, subtracted from, decreased by} & - \\ \hline \text{product, times, of} & \times \\ \hline \text{quotient, divided by, ratio} & \div \\ \hline \text{equals, is, are, totals to, equivalent} & = \\ \hline \text{is less than} & \lt \\ \hline \text{is greater than, is more than} & \gt \\ \hline \text{is less than or equal to, at most, no more than} & \le \\ \hline \text{is greater than or equal to, at least, no less than} & \ge \\ \hline \end{array}\]

Manipulating Equations

The essence of algebra is the manipulation of equations to isolate variables to get useful data.

Simplify

An expression is simplified if there are no parentheses, no like terms, and all fractions are reduced. Common types of simplifications you can do might include: distributing, combining like terms, and cancelling parts of fractions.

We’ve already seen some examples that require combining like terms, but let’s see one that requires distribution as well. Simplify:

\[4(x+5) - 2x(3x+3)\]

Start by getting rid of parentheses by using the distributive property. Remember to distribute the negative sign with the \(2x\).

\[4x+20-6x^2-6x\] \[(4x-6x)+20-6x^2\] \[-2x+20-6x^2\]

Usually, when dealing with a polynomial (an expression with multiple terms), it’s common practice to place the terms in descending order, according to the exponents of the variables. So, the final simplified form would usually look like:

\[-6x^2 - 2x +20\]

When dealing with an equation, it’s always best to simplify each side of the equation before actually solving. Like in the case of:

\[5x-7 -3x = 8x + 8 - 8x + 5\] \[(5x-3x)-7 = (8x-8x)+(8+5)\] \[2x-7=0x+13\] \[2x-7=13\]

Now you can move on to solving the equation.

Solve

Remember the most important rule of algebra: “Whatever you do to one side of the equation, you must do to the other side.” Following this rule, you can do whatever you want, but it’s helpful to do only things that isolate the variable.

Here’s a basic example of solving the equation from above that we had already simplified to \(2x-7=13\)

\[\begin{array}{|c|c|} \hline \textbf{Equation} & \textbf{Reason} \\ \hline 2x-7=13 & \text{original}\\ \hline 2x-7+7=13+7 & \text{addition property of equality} \\ \hline 2x =20 & \text{simplify} \\ \hline 2x\div 2= 20\div 2 & \text{division property of equality}\\ \hline x = 10 & \text{simplify} \\ \hline \end{array}\]

How do you know if you’ve solved the equation, or, more specifically, how do you know if \(x=10\) is a solution to the equation \(2x-7=13\)? A value is a solution to an equation if it can replace the variable in the original equation and the result is a true statement. Let’s see if \(x=10\) is a solution:

\[2x-7=13\] \[2(10)-7\; \underline{?}\;13\] \[20 - 7 \;\underline{?}\; 13\] \[13 = 13\]

Yes, \(x=10\) is a solution to the equation \(2x-7=13\).

Now, let’s try an inequality: solve \(-\dfrac{2}{3}x-5 \le 7\).

\[-\dfrac{2}{3}x-5 \le 7\] \[-\dfrac{2}{3}x-5 + 5 \le 7 + 5\] \[-\dfrac{2}{3}x \le 12\] \[-\dfrac{2}{3}x \cdot - \dfrac{3}{2} \le 12 \cdot - \dfrac{3}{2}\] \[x \ge -18\]

Remember, we flipped the inequality symbol because both sides were multiplied by \(-\dfrac{3}{2}\) (a negative number).

Notice that there are many solutions to an inequality. In fact there are infinitely many solutions. In this case, any number greater than or equal to \(-18\) is a solution. That means that \(-18\), \(-16.349\), \(0\), and \(523\) are all solutions, and any of them can be checked by substituting them back in. If \(0\) is a solution, checking it is pretty easy.

\[-\dfrac{2}{3}x-5 \le 7\] \[-\dfrac{2}{3} \cdot 0 - 5 \;?\;7\] \[0 - 5 \;?\; 7\] \[-5 \le 7\]

Which is true. When you check an inequality, you’re basically looking to see if the symbol is facing the right direction. You could have easily forgotten to flip the sign when multiplying by \(-\dfrac{3}{2}\), and the check would have told you so.