The first equations that students learn to solve in an algebra are linear equations. These equations can always be solved by a strategy known as isolating the variable. Conceptually, students often think of this strategy as moving all of the numbers to the other side of the equation. Unfortunately, this way of thinking about the strategy often introduces subtle mistakes in the solution process. These mistakes could be avoided with proper attention to order of operations.
On facebook, I have recently seen viral posts like the following.
In this case, the answer is clear if we apply the real definition of subtraction as addition using an additive inverse and order does not matter. The proper interpretation of our formula is \[ 6 + -2 + 4 = ?\] If we start with \(6+-2\), we obtain 4; then adding 4 gives us a total of 8. If we start with \(-2+4\), we obtain 2; then adding 6 again gives us 8. This is simply a manifestation of the associative property: \[ (6+-2)+4 = 6+(-2+4). \]
Real complications arise when we have both operations. Consider the formula \[ 8 - 2 \times 3 = 8 + -2 \times 3 = ?\] Because the two operations are not the same, this is not an example of the associative law. Do we get the same answer regardless of order? Is there some form of mixed associative law? If we show parentheses to emphasize order, we can check both strategies and see what happens. \[ \begin{align*} (8+-2)\cdot 3 &= 6 \cdot 3 = 18 \\ 8+(-2 \cdot 3) &= 8+-6 = 2 \end{align*} \] We see that the order does make a difference.
Historically, every author would establish their own standard for how to interpret a mathematical formula. However, during the last 100 years (I presume due to the development of digital computers as well as the standardization of math textbooks), a standard for the order of operations has been established. Powers (exponents) have the highest priority, multiplication has next highest priority, and addition has the lowest priority. If an operation is desired out-of-order, parentheses or brackets are used. All operations within parentheses must be completed before operations outside the parentheses. In addition, functions apply to the input expression before any other operations. Finally, operations at the same level of precedence are applied from left-to-right.
The acronym PFEMA captures the order.
Parentheses
Functions
Exponents
Multiplication (and division)
Addition (and subtraction)
So let us find the officially accepted answer to the facebook brainteaser, \[4 \times 4 + 4 \times 4 + 4 - 4 \times 4 = ?\] I will show the calculations one step at a time. \[ \begin{align*} &\color{red}{4 \times 4} + 4 \times 4 + 4 - 4 \times 4 \\ &\qquad = 16 + \color{red}{4 \times 4} + 4 - 4 \times 4 \\ &\qquad = 16 + 16 + 4 - \color{red}{4 \times 4} \\ &\qquad = 16 + 16 + 4 \color{red}{- 16} \\ &\qquad = \color{red}{16 + 16} + 4 + -16 \\ &\qquad = \color{red}{32 + 4} + -16 \\ &\qquad = \color{red}{36 + -16} \\ &\qquad = 20 \end{align*} \]
One of the most important principles in mathematics is the idea of transformation, taking one problem and transforming it to a new problem that has an easier solution. In the process of solving equations, the method of transformation consists of transforming one equation into a new equation, with the condition that the solution set is unchanged. The most common technique is to apply the same invertible operation to both sides of the equation.
The idea of applying an operation to both sides of the equation preserves balance. Recall that an equation typically represents the scenario that two different variables (expressions) have the same value. Under the conditions that make the expressions have a common value, adding the same quantity to both expressions would result in two new expressions that still have a common, albeit different value. A similar result happens for any common operation applied to both sides of the equation. This is what we mean by balance—if the expressions are in balance before the operation, then they are in balance after the operation.
However, we also want to ensure that we do not introduce new solutions. Using invertible operations provides this guarantee. Addition is invertible because every number has an additive inverse. (More casually, subtraction and addition are inverse operations.) Multiplication by any non-zero value is invertible because such values have multiplicative inverses. (Division and multiplication are inverse operations.) However, multiplication by 0 (or any expression that could possibly equal zero) can lead to extraneous solutions since this is not invertible.
We are now prepared to focus on the strategy of isolation. We begin with the case that the variable of interest (what we are solving for) appears only once in the equation. The strategy consists of identifying the last operation in the expression involving the variable and applying the inverse operation in a balanced way. Repeating this process, we eventually obtain an equation in which the variable is isolated on one side of the equation.
Example: Solve the equation \(\displaystyle \frac{2x-3}{4} = 5\) for \(x\).
Written out in long-hand, the expression on the left is \((2 \cdot x - 3) \div 4\). The last operation is the division, or rather multiplication by \(\div 4\). The inverse would be to multiply by the multiplicative inverse of \(\div 4\), which is just 4. \[ \begin{align*} \frac{2x-3}{4} \cdot 4 &= 5 \cdot 4 \\ 2x-3 &= 20 \end{align*} \] The new expression involving \(x\) has subtraction as the last operation, or rather addition of \(-3\). So the inverse operation would be to add the additive inverse of \(-3\), which is just 3. \[ \begin{align*} 2x+-3+3 &= 20+3 \\ 2x &= 23 \end{align*} \] Finally, the latest expression involving \(x\) has multiplication by 2. The inverse operation would be to multiply by the multiplicative inverse \(\div 2\), which is simply dividing by 2. \[ \begin{align*} 2x \div 2 &= 23 \div 2 \\ x &= \frac{23}{2} \end{align*} \]
Example: Solve the equation \(\displaystyle \frac{2x-3}{4} = a\) for \(x\).
In this problem, there are two variables, \(x\) and \(a\). However, the problem informs us that we are solving for \(x\), so we treat \(a\) as any other number. The steps are identical to the previous problem. \[ \begin{align*} \frac{2x-3}{4} &= a \\ \frac{2x-3}{4} \cdot 4 &= a \cdot 4 \\ 2x-3 &= 4a \\ 2x+-3+3 &= 4a+3 \\ 2x &= 4a+3 \\ 2x \div 2 &= (4a+3)\div 2 \\ x &= \frac{4a+3}{2} \end{align*} \] Notice that when we multiplied both sides by \(\div 2\), we needed to add parentheses around \(4a+3\) in order to maintain the correct order of operations. Furthermore, notice that this problem actually solves the previous example if we were to use the value \(a=5\).
Example: Solve the equation \(\displaystyle \frac{10}{3x-1} = 2\) for \(x\).
The expression involving \(x\) is written \(10 \div (3x-1)\), so that we see the
last operation is multiplication of an expression \(\div(3x-1)\) by 10. So we
multiply both sides by \(\div 10\).
\[
\begin{align*}
10 \div (3x-1) \div 10 &= 2 \div 10 \\
\div(3x-1) &= \frac{1}{5}
\end{align*}
\]
If this looks strange, just remember that we are using ÷
to represent
multiplicative inverse (or reciprocal). The advantage to using multiplicative
inverses instead of division is that multiplication commutes, but division does not.
The multiplicative inverse is an operation of its own right and happens to be its own inverse.
\[
\begin{align*}
\div(\div(3x-1)) &= \div \frac{1}{5} \\
3x-1 &= \div \frac{1}{5} \\
3x-1 &= 5 \\
3x+-1+1 &= 5+1 \\
3x &= 6 \\
3x \div 3 &= 6 \div 3 \\
x &= 2
\end{align*}
\]
If the variable of interest appears in multiple locations, then we should determine if it is possible to transform the equation into a new equation with the variable appearing only once. Then we apply the strategy of isolation. If this is not possible, then we need to attempt alternative strategies. One of the most important of these alternatives is the method of factoring, which is addressed in the next section.
Algebra courses often talk about combining like terms
. I don't actually like
this phrase because it de-emphasizes the mathematical operation in favor of a concrete
analogy that does not necessarily always work. For example, in algebra we might learn
that \(2x+3x\) can be combined to form \(5x\), using an analogy something like
2 apples plus 3 apples equals 5 apples,
treating the variable \(x\) as though
it were a physical object rather than an abstract representation of a number.
In reality, we are applying the distributive property in reverse—factoring. If we show the multiplication more carefully, the formula \(2 \cdot x + 3 \cdot x = (2+3) \cdot x\) and \(2+3=5\). Thinking in terms of grouping and factoring rather than combining like terms allows us to deal with the more complicated looking \(2x+ax = (2+a)x\).
Example: Solve the equation \(5x+3 = 3x-1\) for \(x\).
The big picture at first is to apply operations that group the terms with \(x\) on the left and all other terms on the right. \[ \begin{align*} 5x+3 &= 3x - 1 \\ 5x + 3 + -3 &= 3x + -1 + -3 \\ 5x &= 3x + -4 \\ 5x + -3x &= 3x + -4 + -3x \\ (5+-3)x &= -4 \\ 2x &= -4 \\ 2x \div 2 &= -4 \div 2 \\ x &= -2 \end{align*} \]
Example: Solve the equation \(5x+3 = ax-1\) for \(x\).
This example is intentionally similar to the previous example, except that the symbol \(a\) is introduced where the number \(3\) was previously. Notice that the steps are essentially the same. \[ \begin{align*} 5x+3 &= ax - 1 \\ 5x &= ax + -4 \\ 5x - ax &= -4 \\ (5-a)x &= -4 \\ (5-a)x \div (5-a) &= -4\div(5-a) \\ x &= \frac{-4}{5-a} \end{align*} \]
Notice that this time, because we do not know the value of \(a\), there was one step that may not have been valid. In the special case that \(a=5\), the expression \(5-a=0\) has no multiplicative inverse. In that case, the equation prior to this step would have been \[0 = -4\] which is never true. Geometrically, the equation \(5x+3=5x-1\) corresponds to finding where two parallel lines intersect (never!). A complete solution to this problem indicates this possibility, \[ x= \frac{-4}{5-a}, \quad \hbox{if $a \ne 5$.} \] Again, the solution to the previous example, which corresponds to \(a=3\) is obtained by using the formula, \[ x = \frac{-4}{5-3} = -2, \quad \hbox{if $a = 3$}.\]
The next section focuses on the strategy of
factoring for solving
an equation.
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