How to Simplify Math Expressions

Using the Order of Operations

Know the order of operations. When simplifying math expressions, you can't simply proceed from left to right, multiplying, adding, subtracting, and so on as you go. Some math operations take precedence over others and must be done first. In fact, doing operations out of order can give you the wrong answer. The order of operations is: terms in parentheses, exponents, multiplication, division, addition, and, finally, subtraction. A handy acronym you can use to remember this is "Please excuse my dear Aunt Sally," or "PEMDAS". Note that, while the basic knowledge of the order of operations makes it possible to simplify most basic expressions, specialized techniques are needed to simplify many variable expressions, including nearly all polynomials. See Method Two below for more information.

Start by solving all of the terms in parentheses. In math, parentheses indicate that the terms inside should be calculated separately from the surrounding expression. Regardless of the operations being performed within them, be sure to tackle the terms in parentheses as your first act when you attempt to simplify an expression. Note that, however, within each pair of parentheses, the order of operations still applies. For instance, within parentheses, you should multiply before you add, subtract, etc. As an example, let's try to simplify the expression 2x + 4(5 + 2) + 3 - (3 + 4/2). In this expression, we would solve the terms in parentheses, 5 + 2 and 3 + 4/2, first. 5 + 2 = 7. 3 + 4/2 = 3 + 2 = 5. The second parenthetical term simplifies to 5 because, owing to the order of operations, we divide 4/2 as our first act inside the parentheses. If we simply went from left to right, we might instead add 3 and 4 first, then divide by 2, giving the incorrect answer of 7/2. Note - if there are multiple parentheses nested inside one another, solve the innermost terms first, than the second-innermost, and so on.

Solve the exponents. After tackling parentheses, next, solve your expression's exponents. This is easy to remember because, in exponents, the base number and the power are positioned right next to each other. Find the answer to each exponent problem, then substitute the answers back into your equation in place of the exponents themselves. After dealing with the parentheses, our example expression is now 2x + 4(7) + 3 - 5. The only exponent in our example is 3, which equals 9. Add this back into the equation in the place of 3 to get 2x + 4(7) + 9 - 5.

Solve the multiplication problems in your expression. Next, perform any necessary multiplication in your expression. Remember that multiplication can be written several ways. A × symbol, a dot, or an asterisk are all ways to show multiplication. However, a number hugging parentheses or a variable (like 4(x)) also denotes multiplication. There are two instances of multiplication in our problem: 2x (2x is 2 × x) and 4(7). We don't know the value of x, so let's leave 2x as it is.. 4(7) = 4 × 7 = 28. We can rewrite our equation as 2x + 28 + 9 - 5.

Move on to division. As you search for division problems in your expression, keep in mind that, like multiplication, division can be written multiple ways. The simple ÷ symbol is one, but also remember that slashes and bars in a fraction (like 3/4, for instance) signify division. Because we already solved a division problem (4/2) when we tackled the terms in parentheses, our example no longer has any division in it, so we will skip this step. This brings up an important point - you don't have to perform every operation in the PEMDAS acronym when simplifying an expression, just the ones that are present in your problem.

Add. Next, perform any addition problems in your expression. You can simply proceed from left to right through your expression, but you may find it easiest to add numbers that combine in simple, manageable ways first. For instance, in the expression 49 + 29 + 51 +71, it's easier to add 49 + 51 = 100, 29 + 71 = 100, and 100 + 100 = 200, rather than 49 + 29 = 78, 78 + 51 = 129, and 129 + 71 = 200. Our example expression has been partially simplified to "2x + 28 + 9 - 5". Now, we must add what we can - let's look at each addition problem from left to right. We can't add 2x and 28 because we don't know the value of x, so let's skip it. 28 + 9 = 37, so let's rewrite or expression as "2x + 37 - 5".

Subtract. The very last step in PEMDAS is subtraction. Proceed through your problem, solving any remaining subtraction problems. You may address the addition of negative numbers in this step, or in the same step as the normal addition problems - it won't affect your answer. In our expression, "2x + 37 - 5", there is only one subtraction problem. 37 - 5 = 32

Review your expression. After proceeding through the order of operations, you should be left with your expression in the simplest terms. However, if your expression contains one or more variables, understand that the variable terms will remain largely untouched. Simplifying variable expressions requires you to find the values of your variables or to use specialized techniques to simplify the expression (see below). Our final answer is "2x + 32". We can't address this final addition problem until we know the value of x, but when we do, this expression will be much easier to solve than our initial lengthy-expression.

Simplifying Complex Expressions

Add like variable terms. When dealing with variable expressions, it's important to remember that terms with the same variable and exponent (or "like terms") can be added and subtracted like normal numbers. The terms must not only have the same variable, but also the same exponent. For example, 7x and 5x can be added to each other, but 7x and 5x can not. This rule also extends to terms with multiple variables. For instance, 2xy can be added to -3xy, but not -3xy or -3y. Let's look at the expression x + 3x + 6 - 8x. In this expression, we can add the 3x and -8x terms because they are like terms. Simplified, our expression is x - 5x + 6.

Simplify numerical fractions by dividing or "canceling out" factors. Fractions that have only numbers (and no variables) in both the numerator and denominator can be simplified in several ways. First, and perhaps easiest, is to simply treat the fraction as a division problem and divide the numerator by the denominator. In addition, any multiplicative factors that appear both in the numerator and denominator can be "canceled" because they divide to give the number 1. In other words, if both the numerator and denominator share a factor, this factor can be removed from the fraction, leaving a simplified answer. For example, let's consider the fraction 36/60. If we have a calculator handy, we can divide to get an answer of .6. If we don't, however, we can still simplify by removing common factors. Another way to think of 36/60 is (6 × 6)/(6 × 10). This can be rewritten as 6/6 × 6/10. 6/6 = 1, so our expression is actually 1 × 6/10 = 6/10. However, we're not done yet - both 6 and 10 share the factor 2. Repeating the above procedure, we are left with 3/5. In other words, you have to divide both the numerator and denominator by their greatest common factor (which in the example above is 12: (12 × 3)/(12 × 5)).

In variable fractions, cancel out variable factors. Variable expressions in the form of fractions offer unique opportunities for simplification. Like normal fractions, variable fractions allow you to remove factors that are shared by both the numerator and denominator. However, in variable fractions, these factors can be both numbers and actual variable expressions. Let's consider the expression (3x + 3x)/(-3x + 15x).This fraction can be rewritten as (x + 1)(3x)/(3x)(5 - x), 3x appears both in the numerator and in the denominator. Removing these factors from the equation leaves (x + 1)/(5 - x). Similarly, in the expression (2x + 4x + 6)/2, since every term is divisible by 2, we can write the expression as (2(x + 2x + 3))/2 and thus simplify to x + 2x + 3. Note that you can't cancel just any term - you can only cancel multiplicative factors that appear both in the numerator and denominator. For instance, in the expression (x(x + 2))/x, the "x" cancels from both the numerator and denominator, leaving (x + 2)/1 = (x + 2). However, (x + 2)/x does not cancel to 2/1 = 2.

Multiply parenthetical terms by their constants. When dealing with variable terms in parentheses with an adjacent constant, sometimes, multiplying each term in the parentheses by the constant can result in a simpler expression. This holds true for purely numeric constants and for constants that include variables. For instance, the expression 3(x + 8) can be simplified to 3x + 24, while 3x(x + 8) can be simplified to 3x + 24x. Note that, in some cases, such as in variable fractions, the constant adjacent to the parentheses gives an opportunity for cancellation and thus shouldn't be multiplied through the parentheses. In the fraction (3(x + 8))/3x, for instance, the factor 3 appears both in the numerator and the denominator, so we can cancel it and simplify the expression to (x + 8)/x. This is simpler and easier to work with than (3x + 24x)/3x, which would be the answer we would get if we had multiplied through.

Simplify by factoring. Factoring is a technique by which some variable expressions, including polynomials, can be simplified. Think of factoring as the opposite of the "multiplying through parentheses" step above - sometimes, an expression can be rendered more simply as two terms multiplied by each other, rather than as one unified expression. This is especially true if factoring an expression allows you to cancel part of it (as you would in a fraction). In special cases (often with quadratic equations), factoring even allows you to find answers to the equation. Let's consider the expression x - 5x + 6 once more. This expression can factor to (x - 3)(x - 2). So, if x - 5x + 6 is the numerator of a certain expression with one of these factor terms in the denominator, like is the case with the expression (x - 5x + 6)/(2(x - 2)), we may want to write it in factored form so that we can cancel it with the denominator. In other words, with (x - 3)(x - 2)/(2(x - 2)), the (x - 2) terms cancel, leaving us with (x - 3)/2. As hinted at above, another reason you may want to factor your expression has to do with the fact that factoring can reveal answers to certain equations, especially when those equations are written as expressions equal to 0. For instance, let's consider the equation x - 5x + 6 = 0. Factoring gets us (x - 3)(x - 2) = 0. Since any number times zero equals zero, we know that if we can get either of the terms of parentheses to equal zero, the whole expression on the left side of the equals sign will equal zero as well. Thus, 3 and 2 are two answers to the equation.