First–Degree Equations and Inequalities

The simplest equations and inequalities that you will have to solve on the Math 1 test have only one variable and no exponents. They are called first-degree equations and inequalities. You can always use the six-step method described below to solve them.

EXAMPLE 1: The solution of the equation $\frac{1}{2}x+3(x-2)=(x+1)+1$ illustrates each of the six steps.

Step	What You Should Do	$\frac{1}{2}\mathit{x}+3(\mathit{x}-2)=(\mathit{x}+1)+1$
1 2 3 4 5 6	Get rid of fractions (and decimals) by multiplying both sides by a common denominator. Get rid of all parentheses by using the distributive law. Combine like terms on each side. By adding or subtracting, get all the variables on one side. By adding or subtracting, get all the constants onto the other side. Divide both sides by the coefficient of the variable.*	Multiply each term on both sides of the equation by 2: x + 6(x – 2) = 4(x + 1) + 2 x + 6x – 12 = 4x + 4 + 2 7x – 12 = 4x + 6 Subtract 4x from each side: 3x – 12 = 6 Add 12 to each side: 3x = 18 Divide both sides by 3: x = 6

*If you use this method on an inequality and in Step 6 you divide by a negative number, remember to reverse the inequality.

Example 1 is unusual because it required all six steps. When you solve an equation on the Math 1 test, you will probably not need to do every step. Think of the six steps as a list of questions that must be answered in that order. Ask if each step is necessary. If it isn’t, move on to the next one; if it is necessary, do it. Example 2 does not require all six steps.

EXAMPLE 2: For what value of x is 5(x – 10) = x + 10?

Step	Question	Yes/No	5(x − 10) = x + 10
1 2 3 4 5 6	Are there any fractions or decimals? Are there any parentheses? Are there any like terms to combine? Are there variables on both sides? Is there a constant on the same side as the variable? Does the variable have a coefficient?	No Yes  No Yes Yes Yes	Distribute: 5x – 50 = x + 10 Subtract x from each side: 4x – 50 = 10 Add 50 to each side: 4x = 60 Divide both sides by 4: x = 15

TACTIC E1 Memorize these six steps in order, and use this method whenever you have to solve this type of equation or inequality.

When you use the six-step method, do not actually write out a table. Rather, use the method as a guideline and mentally go through each step, doing whichever ones are required. From now on, we will just do each step, using the symbol ⇒ to move from one step to the next.

Sometimes an equation on the Math 1 test contains several variables and you have to solve for one of them in terms of the others.

TACTIC E2  When you have to solve for one variable in terms of others, treat all the others as if they were numbers, and apply the six-step method.

EXAMPLE 3: If $\frac{2}{3}x=3y-4z$, what is the value of y in terms of x and z?

$\begin{array}{crcl}& \frac{2}{3}x& =& 3y-4z\\ \mathrm{Multiply\; both\; sides\; by} 3:& 2x& =& 3(3y-4z)=9y-12z\\ \mathrm{Add} 12z \mathrm{to\; each\; side}:& 2x+12z& =& 9y\\ \mathrm{Divide\; both\; sides\; by} 9:& y& =& \frac{2x + 12z}{9}\end{array}$

EXAMPLE 4: If v is an integer and the average (arithmetic mean) of 1, 2, 3, and v is less than 20, what is the greatest possible value of v?

Write the inequality ${\displaystyle \frac{1+2+3+\nu }{4}}<20$ and use the six-step method:

$\begin{array}{rcl}\frac{1 + 2 + 3 + v}{4}<20& \Rightarrow & \frac{6 + v}{4}<20\Rightarrow \\ 6+v<80& \Rightarrow & v<74\hfill \end{array}$

Since v is an integer, the greatest possible value of v is 73.