Tackling basic linear equations

To solve a linear equation, you isolate the variable on one side of the equation. You do so by adding the same number to both sides — or you can subtract, multiply, or divide the same number on both sides.

For example, you solve the equation by adding 7 to each side of the equation, to isolate the variable and the multiplier, and then dividing each side by 4, to leave the variable on its own:

• becomes
• giving you .

When a linear equation has grouping symbols such as parentheses, brackets, or braces, you deal with any distributing across and simplifying within the grouping symbols before you isolate the variable. For instance, to solve the equation , you first distribute the 3 and –4 inside the brackets:

You then combine the terms that combine and distribute the negative sign in front of the bracket; it’s like multiplying through by :

Simplify again, and you can solve for x:

When distributing a number or negative sign over terms within a grouping symbol, make sure you multiply every term by that value or sign. If you don’t multiply each and every term, the new expression won’t be equivalent to the original.

To check your answer from the previous example problem, replace every x in the original equation with . If you do so, you get a true statement. In this case, you get . The solution is the only answer that works — focusing your work on just one answer is what’s nice about linear equations.

Clearing out fractions

The problem with fractions, like cats, is that they aren’t particularly easy to deal with. They always insist on having their own way — in the form of common denominators before you can add or subtract. (Or, with cats, they get hissy.) And division? Don’t get me started!

Seriously, though, the best way to deal with linear equations that involve variables tangled up with fractions is to get rid of the fractions. Your game plan is to multiply both sides of the equation by the least common denominator of all the fractions in the equation.

To solve , for example, you multiply each term in the equation by 70, which is the least common denominator (also known as the least common multiple) for fractions with the denominators 5, 7, and 2:

Now you distribute the reduced numbers over each parenthesis, combine the like terms, and solve for x:

Extraneous (false) solutions can occur when you alter the original format of an equation. When working with fractions and changing the form of an equation to a more easily solved form, always check your answer in the original equation. For the previous example problem, you insert into and get or . This one checks.

Isolating different unknowns

When you see only one variable in an equation, you have a pretty clear idea what you’re solving for. When you have an equation like or , you identify the one variable and start solving for it.

Life isn’t always as easy as one-variable equations, however. Being able to solve an equation for some variable when it contains more than one unknown can be helpful in many situations. If you’re repeating a task over and over — such as trying different widths of gardens or diameters of pools to find the best size — you can solve for one of the variables in the equation in terms of the others.

The equation for example, is the formula you use to find the area of a trapezoid. The letter A represents area, h stands for height (the distance between the two parallel bases), and the two b’s are the two parallel sides called the bases of the trapezoid.

If you want to construct a trapezoid that has a set area, you need to figure out what dimensions give you that area. You’ll find it easier to do the many computations if you solve for one of the components of the formula first — for h, , or .

To solve for h in terms of the rest of the unknowns or letters, you multiply each side by two, which clears out the fraction, and then divide by the entire expression in the parentheses:

You can also solve for , the measure of the second base of the trapezoid. To do so, you multiply each side of the equation by two, and then divide each side by h.

Next, subtract from each side of the equation.

When you rewrite a formula aimed at solving for a particular unknown, you can put the formula into a graphing calculator or spreadsheet to do some investigating into how changes in the individual values change the variable that you solve for (see the “Paying off your mortgage with algebra” sidebar).

Solving linear inequalities

To solve a basic linear inequality, you first move all the variable terms to one side of the inequality and the numbers to the other. After you simplify the inequality down to a variable and a number, you can find out what values of the variable will make the inequality into a true statement. For example, to solve , you add 4x to each side and subtract 4 from each side. The inequality sign stays the same because no multiplication or division by negative numbers is involved. Now you have . Dividing each side by 7 also leaves the sense (direction of the inequality) untouched because 7 is a positive number. Your final solution is . The answer says that any number larger than one can replace the x’s in the original inequality and make the inequality into a true statement.

The rules for solving linear equations (see the section “Linear Equations: Handling the First Degree”) also work with inequalities — somewhat. Everything goes smoothly until you try to multiply or divide each side of an inequality by a negative number.

When you multiply or divide each side of an inequality by a negative number, you have to reverse the sense (change to , or vice versa) to keep the inequality true.

The inequality , for example, has grouping symbols that you have to deal with. Distribute the 4 and 3 through their respective multipliers to make the inequality into . Simplify the terms on each side to get . Now you put your inequality skills to work. Subtract 6x from each side and add 14 to each side; the inequality becomes . When you divide each side by , you have to reverse the sense; you get the answer . Only numbers smaller than or exactly equal to work in the original inequality.

When solving the previous example, you have two choices when you get to the step , based on the fact that the inequality is equivalent to . If you subtract 6x from both sides, you end up dividing by a negative number. If you move the variables to the right and the numbers to the left, you don’t have to divide by a negative number, but the answer looks a bit different. If you subtract 4x from each side and subtract 10 from each side, you get . When you divide each side by 2, you don’t change the sense, and you get . You read the answer as “ is greater than or equal to x.” This inequality has the same solutions as , but stating the inequality with the number coming first is a bit more awkward.

Introducing interval notation

You can alleviate the awkwardness of writing answers with inequality notation by using another format called interval notation. You use interval notation extensively in calculus, where you’re constantly looking at different intervals involving the same function. Much of higher mathematics uses interval notation, although I really suspect that book publishers pushed its use because it’s quicker and neater than inequality notation. Interval notation uses parentheses, brackets, commas, and the infinity symbol to bring clarity to the murky inequality waters.

And, surprise surprise, the interval-notation system has some rules:

• You order any numbers used in the notation with the smaller number to the left of the larger number.
• You indicate “or equal to” by using a bracket.
• If the solution doesn’t include the end number, you use a parenthesis.
• When the interval doesn’t end (it goes up to positive infinity or down to negative infinity), use or , whichever is appropriate, and a parenthesis.

Here are some examples of inequality notation and the corresponding interval notation:

Notice that the second example has a bracket by the , because the “greater than or equal to” indicates that you include the also. The same is true of the 4 in the third example. The last example shows you why interval notation can be a problem at times. Taken out of context, how do you know if represents the interval containing all the numbers between and 7 or if it represents the point on the coordinate plane? You can’t tell. You consider the context. A problem containing such notation has to give you some sort of hint as to what it’s trying to tell you.

Compounding inequality issues

A compound inequality is an inequality with more than one comparison or inequality symbol — for instance, . To solve compound inequalities for the value of the variables, you use the same inequality rules (see the intro to this section), and you expand the rules to apply to each section (intervals separated by inequality symbols).

To solve the inequality , for example, you add 5 to each of the three sections and then divide each section by 3:

You write the answer, , in interval notation as .

Here’s a more complicated example. You solve the problem by subtracting 5 from each section and then dividing each section by . Of course, dividing by a negative means that you turn the senses around:

You write the answer, , backward as far as the order of the numbers on the number line; the number is smaller than 3. To flip the inequality in the opposite direction, you reverse the inequalities, too: . In interval notation, you write the answer as .

Solving absolute value equations

A linear absolute value equation is an equation that takes the form . You don’t know, taking the equation at face value, if you should change what’s in between the bars to its opposite, because you don’t know if the expression is positive or negative. The sign of the expression inside the absolute value bars all depends on the size and sign of the variable x. To solve an absolute value equation in this linear form, you have to consider both possibilities: may be positive, or it may be negative.

To solve for the variable x in , you solve both and .

• For example, to solve the absolute value equation , you write the two linear equations and solve each for x by subtracting 5 and dividing by 4: If , then and .
• If , then and .

You have two solutions: 2 and . Both solutions work when you replace the x in the original equation with their values.

One restriction you should be aware of when applying the rule for changing from absolute value to individual linear equations is that the absolute value term has to be alone on one side of the equation.

For instance, to solve , you have to subtract 7 from each side of the equation and then divide each side by 3. Subtracting 7, you have ; then, when you divide by 3, the problem becomes .

Now you can write the two linear equations and solve them for x:

• If , then and .
• If , then and .

Seeing through absolute value inequality

An absolute value inequality contains both an absolute value, , and an inequality: , or . But, then, you knew that was coming.

To solve an absolute value inequality, you have to change from absolute value inequality form to just plain inequality form. The way to handle the change from absolute value notation to inequality notation depends on which direction the inequality points with respect to the absolute value term. The methods, depending on the direction, are quite different:

• To solve for x in you solve .
• To solve for x in |, you solve both and .

The first change sandwiches the between c and its opposite. The second change considers values greater than c (toward positive infinity) and smaller than (toward negative infinity).

Exposing an impossible inequality imposter

The rules for solving absolute value inequalities are relatively straightforward. You change the format of the inequality and solve for the values of the variable that work in the problem. Sometimes, however, amid the flurry of following the rules, an impossible situation works its way in to try to catch you off guard.

For example, say you have to solve the absolute value inequality . It doesn’t look like such a big deal; you just subtract 8 from each side and then divide each side by 2. The dividing value is positive, so you don’t reverse the sense. After performing the initial steps, you use the rule where you change from an absolute value inequality to an inequality with the variable term sandwiched between inequalities. So, what’s wrong with that? Here are the steps:

Subtracting 8, you get , and dividing by 2, you get .

Under the format , the inequality looks curious. Do you sandwich the variable term between and 1 or between 1 and (the first number on the left, and the second number on the right)? It turns out that neither works. First of all, you can throw out the option of writing . Nothing is bigger than 1 and smaller than at the same time. The other version seems, at first, to have possibilities, so you try to solve by adding 7 to each interval, giving you . Dividing each interval by 3, you have .

The solution says that x is a number between 2 and . If you check the solution by trying a number — say, 2.1 — in the original inequality, you get the following:

Because 9.4 isn’t less than 6, you know the number 2.1 doesn’t work. You won’t find any number that works. So, you can’t find an answer to this problem. Did you miss a hint of the situation before you dove into all the work? Yes. (Sorry for the tough love!)

You want to save yourself some time and work? You can do that in this case by picking up on the pesky negative number. When you subtract 8 from each side of the original problem and get , the bells should be ringing and the lights flashing. This statement says that 2 times the absolute value of a number is smaller than , which is impossible. Absolute value is either positive or zero — it can’t be negative — so this expression can’t be smaller than . If you caught the situation before doing all the work, hurrah for you! Good eye. Often, though, you can get caught up in the process and not notice the impossibility until the end — when you check your answer.