Now that you know how to set up an equation, the next thing you need to do is solve for the value that the question asks for. Above all, the most important thing to remember when manipulating equations is that each side of the equation must be manipulated in the same way. If you divide one side of an equation by 3, you must divide the other side by 3. If you take the square root of one side of an equation, take the square root of the other.

By treating the two sides of the equation in the same way, you won’t violate the equality of the equation. You will, of course, change the form of the equation—that’s the point of manipulating it. But the equation will always remain true as long as you do the same thing to both sides.

For example, let’s look at what happens when you manipulate the equation 3x + 2 = 5, with x = 1.

These examples show that you can tamper with the equation as long as you tamper the same way on both sides. If you follow this rule, you can manipulate the question without affecting the value of the variables.

To solve an equation with one variable, you must manipulate the equation to isolate that variable on one side of the equation. Then, by definition, that variable is equal to whatever is on the other side, and you have successfully “solved for the variable.”

For the quickest results, take the equation apart in the opposite order of the standard order of operations. That is, first add and subtract any extra terms on the same side as the variable. Then, multiply and divide anything on the same side of the variable. Next, raise both sides of the equation to a power or take their roots. And finally, do anything inside parentheses. This process is PEMDAS in reverse (SADMEP!). The idea is to “undo” everything that is being done to the variable so that it will be isolated in the end. Let’s look at an example:

In this equation, the variable x is being squared, multiplied by 3, added to 5, etc. We need to do the opposite of all these operations in order to isolate x and thus solve the equation.

First, subtract 1 from both sides of the equation:

Then, multiply both sides of the equation by 4:

Next, divide both sides of the equation by 3:

Now, subtract 5 from both sides of the equation:

Again, divide both sides of the equation by 3:

Finally, take the square root of each side of the equation:

We have isolated x to show that x = ±5.

Sometimes the variable that needs to be isolated is not located conveniently. For example, it might be in a denominator, or an exponent. Equations like these are solved the same way as any other equation, except that you may need different techniques to isolate the variable. Let’s look at a couple of examples:

Solve for x in the equation .

The key step is to multiply both sides by x to extract the variable from the denominator. It is not at all uncommon to have to move the variable from one side to the other in order to isolate it. Here’s another, slightly more complicated, example:

Solve for x in the equation .

This question is a good example of how it’s not always simple to isolate a variable. (Don’t worry about the logarithm in this problem—we’ll review these later on in the chapter.) However, as you can see, even the thorniest problems can be solved systematically as long as you have the right tools. In the next section we’ll discuss factoring and distributing, two techniques that were used in this example.

Having just given you a very basic introduction to solving equations, we’ll reemphasize two things:

Now we’ll work with some more interesting tools you will need to solve certain equations.

Distributing and factoring are two of the most important techniques in algebra. They give you ways of manipulating expressions without changing the expression’s value. In other words, distributing and factoring are tools of reorganization. Since they don’t affect the value of the expression, you can factor or distribute one side of the equation without doing the same for the other side of the equation.

The basis for both techniques is the following property, called the distributive property:

Similarly,

a can be any kind of term, from a variable to a constant to a combination of the two.

When you “distribute” a factor into an expression within parentheses, you simply multiply each term inside the parentheses by the factor outside the parentheses. For example, consider the expression 3y(y² – 6):

If we set the original, undistributed expression equal to another expression, you can see why distributing facilitates the solving of some equations. Solving 3y(y² – 6) = 3y³ + 36 looks quite difficult. But when you distribute the 3y, you get:

Subtracting 3y³ from both sides gives us:

Factoring an expression is essentially the opposite of distributing. Consider the expression 4x³ – 8x² + 4x, for example. You can factor out the greatest common factor of the terms, which is 4x:

The expression simplifies further:

See how useful these techniques are? You can group or ungroup quantities in an equation to make your calculations easier. In the last example from the previous section on manipulating equations, we distributed and factored to solve an equation. First, we distributed the quantity log 3 into the sum of x and 2 (on the right side of the equation). We later factored the term x out of the expression x log 2 – x log 3 (on the left side of the equation).

Distributing eliminates parentheses, and factoring creates them. It’s your job as a Math IIC mathematician to decide which technique will best help you solve a problem.

Let’s look at a few examples:

There are other steps you can take to simplify expressions or equations. Combining like terms is one of the simpler techniques you can use, and it involves adding or subtracting the coefficients of variables that are raised to the same power. For example, by combining like terms, the expression

can be simplified to

by adding the coefficients of the variable x³ together and the coefficients of x² together.

The point is, you’d rather have one term, 7x², instead of x², 3x², –3x², 2x², and 4x² all floating around in your expression. A general formula for combining like pairs looks like this:

When the product of any number of terms is zero, you know that at least one of the terms is equal to zero. For example, if xy = 0, you know that either:

This is useful in a situation like the following:

By the zero product rule, you know that (x + 4) = 0 or (x – 3) = 0. In this equation, either x = –4 or x = 3, since one of the expressions in parentheses must be equal to 0. Consider this equation:

Again, since 3x² or (x + 2) must equal 0, we know that either x = 0 or x = –2.

Keep your eye out for a zero product—it’s a big time-saver, especially when you have multiple answers to choose from.

To solve an equation in which the variable is within absolute value brackets, you must divide the equation into two equations. The two equations are necessary because an absolute value really defines two equal values, one positive and one negative. The most basic example of this is an equation of the form |x| = c. In this case, either x = c or x = –c.

A slightly more complicated example is this:

In this problem, you must solve two equations: First, solve for x in the equation x + 3 = 5. In this case, x = 2. Then, solve for x in the equation x + 3 = –5. In this case, x = –8. So the solutions to the equation |x + 3| = 5 are x = {–8, 2}.

Generally speaking, to solve an equation in which the variable is within absolute value brackets, first isolate the expression within the absolute value brackets, and then create two equations. Keep one of these two equations the same, while in the other equation, negate one side of the equation. In either case, the absolute value of the expression within brackets will be the same. This is why there are always two solutions to absolute value problems (unless the variable is equal to 0, which is neither positive nor negative).

Here is one more example:

First, isolate the expression within the absolute value brackets:

Then solve for the variable as if the expression within absolute value brackets were positive:

Next, solve for the variable as if the expression within absolute value brackets were negative:

The solution set for x is {y² – 3, –y² –1}.

Before you get too comfortable with expressions and equations, we should introduce inequalities. An inequality is like an equation, but instead of relating equal quantities, it specifies exactly how two expressions are not equal.

Solving inequalities is exactly like solving equations except for one very important difference: when both sides of an inequality are multiplied or divided by a negative number, the relationship between the two sides changes and so the direction of the inequality must be switched.

Here are a few examples:

Notice that in the last example, the inequality had to be reversed. Another way to express the solution is x ≥ –2. To help you remember that multiplication or division by a negative number reverses the direction of the inequality, recall that if x > y, then –x > –y, just as 5 > 4 and –5 < –4. Intuitively, this makes sense, and it might help you remember this special rule of inequalities.

There is a critical difference between the solutions to equalities and solutions to inequalities. The number of solutions to an equation is usually equal to the highest power of the equation. A linear equation (highest term of x) has one solution, a quadratic equation (highest term of x²) has two solutions, and a cubic equation (highest term of x³) has three solutions. In an inequality, this rule does not hold true. As you can see from the above examples, there are often infinite solutions to inequalities: the solutions are often graphed as planes rather than points.

An equation without any absolute values generally results in, at most, only a few different solutions. Solutions to inequalities are often large regions of the x-y plane, such as x < 5. The introduction of the absolute value, as we’ve seen before, usually introduces two sets of solutions. The same is true when absolute values are introduced to inequalities: the solutions often come in the form of two regions of the x-y plane.

If the absolute value is less than a given quantity, then the solution is a single range, with a lower and an upper bound. For example,

First, solve for the upper bound:

Second, solve for the lower bound:

Now combine the two bounds into a range of values for x. –1 ≤ x ≤ 5 is the solution.

The other solution for an absolute value inequality involves disjoint ranges: one whose lower bound is negative infinity and whose upper bound is a real number, and one whose lower bound is a real number and whose upper bound is infinity. This occurs when the absolute value is greater than a given quantity. For example,

First, solve for the upper range:

Then, solve for the lower range:

Now combine the two ranges to form the solution, which is two disjoint ranges: –∞ < x < – ²⁰/₃ or 4 < x < ∞.

When working with absolute values, it is important to first isolate the expression within absolute value brackets. Then, and only then, should you solve separately for the cases in which the quantity is positive and negative.

Inequalities are also used to express the range of values that a variable can take on. a < x < b means that the value of x is greater than a and less than b. Consider the following word problem example:

Let a be the age of people for whom the board game is appropriate. The lower bound of a is 40, and the upper bound is 65. The range of a does not include its lower bound (it is appropriate for people “older than 40”), but it does include its upper bound (“no older than 65,” i.e., 65 is appropriate, but 66 is not). Therefore, the range of the age of people for which the board game is appropriate can be expressed by the inequality:

Here is another example:

The boundary weights of this car part are 0.98 21.5 = 21.07 and 1.02 21.5 = 21.93 grams. The problem states that the piece cannot weigh less than the minimum weight or more than the maximum weight in order for it to work. This means that the part will function at boundary weights themselves, and the lower and upper bounds are included. The answer to the problem is 21.07 ≤ x ≤ 21.93, where x is the weight of the part in grams.

Finding the range of a particular variable is essentially an exercise in close reading. Every time you come across a question involving ranges, you should carefully scrutinize the problem to pick out whether or not a particular variable’s range includes its bounds. This inclusion is the difference between “less than or equal to” and simply “less than.”

Operations like addition, subtraction, and multiplication can be performed on ranges just as they are performed on variables or inequalities. For example:

To solve this problem, simply manipulate the range like an inequality until you have a solution. Begin with the original range:

Then multiply the inequality by 2:

Add 3 to the inequality, and you have the answer.

There is one crucial rule that you need to know about multiplying ranges: if you multiply a range by a negative number, you must flip the greater than or less than signs. For instance, if you multiply the range 2 < x < 8 by –1, the new range will be –2 > x > –8. Math IIC questions that ask you to perform operations on ranges of one variable will often test your alertness by making you multiply the range by a negative number.

Some range problems on the Math IIC will be made slightly more difficult by the inclusion of more than one variable. In general, the same basic procedures for dealing with one-variable ranges apply to adding, subtracting, and multiplying two-variable ranges.

Simply add the ranges. The lower bound is –2 + 0 = –2. The upper bound is 8 + 5 = 13. Therefore, –2 < x + y < 13.

In this case, you have to find the range of –t. By multiplying the range of t by –1 and reversing the direction of the inequalities, we find that 1 < –t < 3. Now we can simply add the ranges again to find the range of s – t. 4 + 1 = 5, and 7 + 3 = 10. Therefore, 5< s – t < 10.

In general, to subtract ranges, find the range of the opposite of the variable being subtracted, and then add the ranges like usual.

First, multiply the lower bound of one variable by the lower and upper bounds of the other variable:

Then, multiply the upper bound of one variable with both bounds of the other variable:

The least of these four products becomes the lower bound, and the greatest is the upper bound. Therefore, –12 < jk < 48.

Let’s try one more example of performing operations on ranges:

The first step is to find the range of x + y:

We have our bounds for the range of x + y, but are they included in the range? In other words, is the range 0 < x + y < 11, 0 ≤ x + y ≤ 11, or some combination of these two?

The rule to answer this question is the following: if either of the bounds that are being added, subtracted, or multiplied is non-inclusive (< or >), then the resulting bound is non-inclusive. Only when both bounds being added, subtracted, or multiplied are inclusive (≤ or ≥) is the resulting bound also inclusive.

The range of x includes its lower bound, 3, but not its upper bound, 7. The range of y includes both its bounds. Therefore, the range of x + y is 0 ≤ x + y < 11, and the range of 2(x + y) is 0 ≤ 2(x + y) < 22.

An alternate way of expressing the range of a variable might appear on the Math IIC test. A range can be written by enclosing the lower and upper bounds in parentheses or brackets, depending on whether they are included in the range. Parentheses are used when the bound is not included in the range, and brackets are used when the bound is included in the range. For example, the statement a < x < b can be rewritten “the range of x is (a, b).” The statement a ≤ x ≤ b can be rewritten “the range of x is [a, b].” Finally, the statement a < x ≤ b can be rewritten “the range of x is [a, b].”