Learning Objectives
- Correctly write and interpret algebraic expressions.
- Recognize common algebraic errors.
- Solve linear equations
Chapter 10 Reading Mathematical Expressions
Section 10.1 So Many Symbols!
As we continue forward into higher levels of mathematics, we will encounter more mathematical notation. Students sometimes start to struggle because they’ve never developed a set of tools to help them interpret those symbols. It’s like trying to read a language when you don’t have a sense of the grammar. You can sometimes piece things together and follow what’s happening, but most of the time you’re completely lost.
We will begin with familiar territory. At some point in their lives, most students learned some version of the order of operations (often called PEMDAS). This is a set of instructions for the correct way to perform a complex calculation by prioritizing certain calculations. Here is how the order of operations is usually presented:
- Parentheses and other grouping symbols.
- Exponents
- Multiplication and division (performed left-to-right)
- Addition and subtraction (performed left-to-right)
This is fine as far as the basics are concerned. And when working with numbers, students are usually able to do this correctly. But it’s not long after we start introducing variables that students can feel overwhelmed by symbols.
The best analogy for understanding this comes from thinking about a complex sentence in English: "The man carrying a large bag containing three apples and two pears bought a blue shirt with six silver buttons from the store on the corner of Main Street and West Park Avenue." There are a lot of details in that sentence, but there’s also the "big picture" of the sentence. What is actually happening in the sentence? Some man bought a shirt from a store. From there we can go into specific details, such as what the man was holding, the color of the shirt, and the location of the store. But those details add information to the big picture without changing the core of the picture.
The focus of this section is not really on algebra, but on helping you familiarize yourself with the language of mathematics. Fortunately, mathematicians have created an entire notation to help you do this. Consider the following polynomial:
\begin{equation*}
x^3 - 3x^2 - 7x + 11
\end{equation*}
At this point, your experience should lead you to interpret this as the sum of four terms. You should be able to visualize the plus and minus signs as being separators for the different terms. Compare this to how we would have to write it if we had to write everything out:
\begin{equation*}
x \cdot x \cdot x + (-3) \cdot x \cdot x + (-7) \cdot x + 11
\end{equation*}
It is much harder for our brains to process this because we have to do a lot of extra work. It’s harder to locate the plus signs from among all of the symbols, and we have to count out all of the products to figure out how many factors of there are.
The two key features here are the use of implied multiplication and exponents. These two bits of notation help us to visually condense the information and make it easier to read. The exponents also mean that we don’t have to count out the products. We also have the convention that subtraction means addition of the opposite, which lets us write the negative coefficients as subtraction.
All of this notation helps us to identify that the "big picture" of this is that we have a sum of four terms. From there, we can choose to look at the details of those terms to see (for example) if there are any like terms that we might want to combine.
Another mathematical notation that was developed is the use of fractions for division. Interestingly, it’s very rare in the modern mathematical world to use the \(\div\) symbol. It is not as rare, but still uncommon to use \(/\) to represent division as well. These tend to cause more confusion than clarity, which is why modern mathematics prefers the use of a fraction. This way, we can clearly distinguish between the numerator and denominator of a fraction.
For example, consider the following expression: \(x^2 + 3/x + 2\text{.}\) The strict (and proper) application of the order of operations is that this result is the same as \(x^2 + \frac{3}{x} + 2\text{.}\) But students will write that to mean \(\frac{x^2 + 3}{x + 2}\text{.}\) If you insist on writing your fractions with a diagonal slash, you will need to use parentheses to separate out the numerator and the denominator: \((x^2 + 3)/(x + 2)\text{.}\)
For this reason, students are strongly urged to not use the diagonal slash for division, but write their division using a long horizontal bar (long enough to span the width of all the terms in the numerator and the denominator) so that it is clear what is in the numerator and what is in the denominator.
Similarly, when using the square root, it is important that the bar of the square root is long enough (and only long enough) to cover the parts inside the square root. For example, \(\sqrt{x + 3}\) is not the same as \(\sqrt{x} + 3\text{,}\) and \(\sqrt{x \,+ \, } \, 3\) doesn’t even make sense. So please be careful with what you’re writing.
Activity 10.1. Identifying the "Big Picture" of an Expression.
A basic skill when learning to read mathematical notation is to mentally "group together" things to be able to process what’s happening. For example, the polynomial from before can be seen as the sum of four terms.
\begin{equation*}
x^3 - 3x^2 - 7x + 11
\longrightarrow \boxed{x^3} + \boxed{(-3x^2)} + \boxed{ (-7x) } + \boxed{11}
\longrightarrow \boxed{A} + \boxed{B} + \boxed{C} + \boxed{D}
\end{equation*}
Similarly, we can think of fractions as being basically a numerator divided by a denominator.
\begin{equation*}
\frac{x^2 + 3}{x + 2}
\longrightarrow \frac{\boxed{x^2 + 3}}{\boxed{ x + 2} }
\longrightarrow \frac{ \boxed{A} }{\boxed{B}}
\end{equation*}
Try it!
Consider the expression \(\frac{x + 3}{x - 2} + 3x\text{.}\) Describe the "big picture" perspective of the expression and put boxes around the terms as appropriate.
Solution.
\begin{equation*}
\boxed{ \frac{ x + 3 }{ x - 2 } } + \boxed{ 3x }
\end{equation*}
This is the sum of a fraction and a monomial.
After seeing the "big picture" we can then break things down further. Once we’re inside of the numerator or the denominator of the fraction, and that may consist of multiple terms.
\begin{equation*}
\frac{x^2 + 3}{x + 2}
\longrightarrow \frac{\boxed{x^2 + 3}}{\boxed{ x + 2} }
\longrightarrow \frac{\boxed{ \boxed{x^2} + \boxed{3} }} {\boxed{ \boxed{x} + \boxed{2} } }
\longrightarrow \frac{ \boxed{\boxed{A} + \boxed{B} }}{\boxed{\boxed{C} + \boxed{D}}}
\end{equation*}
And we can go deeper into the individual terms and break them up into the coefficient (which is sometimes an unwritten and the variable part. And the variable part can be broken down into individual variables. But at this point, it’s like focusing on the number of apples in the bag and you’ve lost sight of the big picture.
But so far, we’ve only focused on the basic arithmetic operations. The next large piece of language you will encounter are functions. We aren’t going to go into the details of what functions are here. We’re just going to use the notation to help you understand how to think about it when you get to them.
The basic shape of a function is \(f(x)\text{.}\) This represents a number just like the variable \(y\) would represent a number. The only difference is that with a function, there is a "rule" that tells you how to calculate the number. When we write \(f(x)\) we are saying to use the rule associated with \(f\) to do the calculation. In addition to the letters \(f\) and \(g\) (which are used for generic functions), there are also a number of special functions that have names. Here are a few examples:
| Name | Notation |
| sine | \(\sin()\) |
| cosine | \(\cos()\) |
| tangent | \(\tan()\) |
| common logarithm | \(\log()\) |
| natural logarithm | \(\ln()\) |
| exponential | \(\exp()\) |
The empty parentheses indicate that there is a value to "plug into" the function. So there would normally be some expression inside of those parentheses, which can be a number (such as with \(\ln(2)\)), a variable (such as with \(\sin(x)\)) or a variable expression (such as with \(\exp(-x^2 + 1)\)).
At this point, it doesn’t matter what these functions represent, other than they represent a method for calculating some specific number. So it doesn’t matter that \(\ln(2) \approx 0.693\text{,}\) only that \(\ln(2)\) is some number that your calculator can calculate for you.
At least in terms of algebra, functions behave very similarly to variables. Here are some examples:
\begin{equation*}
\begin{aligned}
2f(x) + 3f(x) \amp = 5f(x) \\
x \ln(3) - x \amp = (\ln(3) - 1)x
\end{aligned}
\end{equation*}
The important note is that the function name cannot be separated from its argument (the parentheses and everything inside of the parentheses). All of that notation should be seen as a single object. That is, you cannot factor out the \(f\) in the first example:
\begin{equation*}
2f(x) + 3f(x) \overset{\times}{=} f(2(x) + 3(x))
\end{equation*}
It may help for you to put a box around the function and its argument.
\begin{equation*}
2f(x) + 3f(x) = 2 \, \boxed{ f(x) } + 3 \, \boxed{ f(x) }
\end{equation*}
The error of breaking apart the function from its argument is similar to other errors where students "distribute" incorrectly:
\begin{equation*}
\begin{aligned}
\frac{1}{a + b} \amp \overset{\times}{=} \frac{1}{a} + \frac{1}{b} \\
(x + y)^2 \amp \overset{\times}{=} x^2 + y^2 \\
\sqrt{n + m} \amp \overset{\times}{=} \sqrt{n} + \sqrt{m} \\
\sqrt{n^2 + m^2} \amp \overset{\times}{=} n + m \\
\sin(\theta + \phi) \amp \overset{\times}{=} \sin(\theta) + \sin(\phi)
\end{aligned}
\end{equation*}
It all comes down to not understanding what the symbols mean or how they behave.
Activity 10.2. Hiding Information in Boxes.
Students often feel intimidated by the notation. One trick is to think of the the function as being hidden inside of a box. This will "hide" the details so that you can focus on the big picture.
\begin{equation*}
\begin{aligned}
x \ln(3) + 4 \amp = \ln(6) \\
x \cdot \boxed{ A } + 4 \amp = \boxed{ B } \amp \eqnspacer \amp \text{Substitute} \\
x \cdot \boxed{ A } \amp = \boxed{ B } - 4 \amp \amp \text{Subtract $4$ from both sides} \\
x \amp = \frac{ \boxed{ B } - 4}{\boxed{A}} \amp \amp \text{Divide both sides by $\boxed{A}$} \\
x \amp = \frac{ \ln(6) - 4}{\ln(3)} \amp \amp \text{Substitute}
\end{aligned}
\end{equation*}
Try it!
Solve the equation \(x \sin(1) + 5 = 3x - \cos(2)\text{.}\) Do it once using a substitution similar to the example above, and then do it without that substitution. Use a complete presentation both times.
Solution.
\begin{equation*}
\begin{aligned}
x \sin(1) + 5 \amp = 3x - \cos(2) \\
x \cdot \boxed{ A } + 5 \amp = 3x - \boxed{ B } \amp \amp \text{Substitute} \\
x \cdot \boxed{ A } \amp = 3x - \boxed{ B } - 5 \amp \amp \text{Subtract $5$ from both sides} \\
x \cdot \boxed{ A } - 3x \amp = - \boxed{ B } - 5 \amp \amp \text{Subtract $3x$ from both sides} \\
\left( \boxed{ A } - 3 \right) x \amp = - \boxed{ B } - 5 \amp \amp \text{Factor out the $x$} \\
x \amp = \frac{ - \boxed{B} - 5}{\boxed{A} - 3} \amp \amp \text{Divide by $\boxed{A} - 3$} \\
x \amp = \frac{ - \cos(2) - 5}{\sin(1) - 3} \amp \amp \text{Substitute}
\end{aligned}
\end{equation*}
\begin{equation*}
\begin{aligned}
x \sin(1) + 5 \amp = 3x - \cos(2) \\
x \sin(1) \amp = 3x - \cos(2) - 5 \amp \amp \text{Subtract $5$ from both sides} \\
x \sin(1) - 3x \amp = - \cos(2) - 5 \amp \amp \text{Subtract $3x$ from both sides} \\
\left( \sin(1) - 3 \right) x \amp = - \cos(2) - 5 \amp \amp \text{Factor out the $x$} \\
x \amp = \frac{ - \cos(2) - 5}{\sin(1) - 3} \amp \amp \text{Divide by $\sin(1) - 3$}
\end{aligned}
\end{equation*}
Activity 10.3. Solving for an Expression.
When we solve equations, sometimes we’re not solving for the variable, but some expression involving the variable. This is often the case when the variable is wrapped up in a function. The same type of process can be done by making a substitution similar to the ones above. It is important to remember to substitute back to the original variable if you do this.
\begin{equation*}
\begin{aligned}
2 \sin(x) + \sqrt{2} \amp = 0 \\
2 y + \sqrt{2} \amp = 0 \amp \eqnspacer \amp \text{Substitute $y = \sin(x)$} \\
2 y \amp = -\sqrt{2} \amp \amp \text{Subtract $\sqrt{2}$ from both sides} \\
y \amp = - \frac{\sqrt{2}}{2} \amp \amp \text{Divide both sides by $2$} \\
\sin(x) \amp = - \frac{\sqrt{2}}{2} \amp \amp \text{Substitute back to the original variable}
\end{aligned}
\end{equation*}
Try it!
Solve the equation \(4 \exp(x) + 6 = \ln(4)\) for Do it once using a substitution similar to the example above, and then do it without that substitution. Use a complete presentation both times.
Solution.
\begin{equation*}
\begin{aligned}
4 \exp(x) + 6 \amp = \ln(4) \\
4 y + 6 \amp = \ln(4) \amp \amp \text{Substitute $y = \exp(x)$} \\
4 y \amp = \ln(4) - 6 \amp \amp \text{Subtract $6$ from both sides} \\
y \amp = \frac{\ln(4) - 6}{4} \amp \amp \text{Divide both sides by $4$} \\
\exp(x) \amp = \frac{\ln(4) - 6}{4} \amp \amp \text{Substitute back to the original variable} \\
\end{aligned}
\end{equation*}
\begin{equation*}
\begin{aligned}
4 \exp(x) + 6 \amp = \ln(4) \\
4 \exp(x) \amp = \ln(4) - 6 \amp \amp \text{Subtract $6$ from both sides} \\
\exp(x) \amp = \frac{\ln(4) - 6}{4} \amp \amp \text{Divide both sides by $4$}
\end{aligned}
\end{equation*}
Section 10.2 Worksheets
PDF Version of these Worksheets
1
external/worksheets/10-Worksheets.pdfWorksheet Worksheet 1
1.
Consider the expression \((x + 2)(x - 5)\text{.}\) Describe the "big picture" perspective of the expression and put boxes around the terms as appropriate.
2.
Consider the expression \(x + 2(x - 5)\text{.}\) Describe the "big picture" perspective of the expression and put boxes around the terms as appropriate.
3.
Determine whether you think the following equation is valid. Explain your reasoning.
\begin{equation*}
\sin(x + y) = \sin(x) + \sin(y)
\end{equation*}
4.
Solve the equation \(x \ln(2) + 3 = -4\text{.}\) Do it once using a substitution for then do it without a that substitution. Use a complete presentation both times.
Worksheet Worksheet 2
1.
Consider the expression \(2x(x - 3) + 4\text{.}\) Describe the "big picture" perspective of the expression and put boxes around the terms as appropriate.
2.
Evaluate the expression \(2x(x - 3) - 5\) when \(x = 2\text{.}\) Use a complete presentation.
3.
Check the presentation for errors. If you find one, circle it and describe the mistake in words.
\begin{equation*}
\begin{aligned}
3x + f(4) \amp = f(10) \\
3x + 4 \amp = 10 \amp \eqnspacer \amp \text{Cancel out the $f$} \\
3x \amp = 6 \amp \amp \text{Subtract $4$ from both sides} \\
x \amp = 2 \amp \amp \text{Divide both sides by $3$}
\end{aligned}
\end{equation*}
4.
Check the presentation for errors. If you find one, circle it and describe the mistake in words.
\begin{equation*}
\begin{aligned}
\exp(x) + 3 \amp = 8 \\
\exp(x) \amp = 5 \amp \eqnspacer \amp \text{Subtract $3$ from both sides} \\
x \amp = \frac{5}{\exp} \amp \amp \text{Divide both sides by $\exp$} \\
\end{aligned}
\end{equation*}
Worksheet Worksheet 3
1.
Consider the expression \((x + 1)^2 - (x - 1)^2\text{.}\) Describe the "big picture" perspective of the expression and put boxes around the terms as appropriate.
2.
Check the presentation for errors. If you find one, circle it and describe the mistake in words.
\begin{equation*}
\begin{aligned}
(x + 4)^2 - (x - 3)^2 \amp = (x^2 + 16) - (x^2 - 9) \amp \eqnspacer \amp \text{Distribute the square} \\
\amp = x^2 - x^2 + 16 + 9 \amp \amp \text{Rearrange the terms} \\
\amp = 25 \amp \amp \text{Combine like terms}
\end{aligned}
\end{equation*}
3.
Simplify the expression \((x + 1)^2 - (x - 1)^2\) using a complete presentation.
4.
Solve the equation \(2 \tan(x) - 5 = -3\) for Do it once using a substitution for then do it without that substitution. Use a complete presentation both times.
Worksheet Worksheet 4
1.
Check the presentation for errors. If you find one, circle it and describe the mistake in words.
\begin{equation*}
\begin{aligned}
6 \exp(2) \amp = 10 \\
x \amp = \frac{10}{\exp(2)} \amp \eqnspacer \amp \text{Divide both sides by $\exp(2)$} \\
x \amp = \frac{5}{\exp(1)} \amp \amp \text{Reduce}
\end{aligned}
\end{equation*}
2.
Solve the equation \(ax + b = c\) for the variable using a complete presentation.
3.
Check the presentation for errors. If you find one, circle it and describe the mistake in words.
\begin{equation*}
\begin{aligned}
x \log(5) \amp = \log(7) \\
x \amp = \frac{\log(7)}{\log(5)} \amp \eqnspacer \amp \text{Divide both sides by $\log(5)$} \\
x \amp = \frac{7}{5} \amp \amp \text{Cancel out the $\log$}
\end{aligned}
\end{equation*}
4.
Solve the equation \(2 \log(x) = \log(x) + 5\) for Do it once using a substitution for then do it without that substitution. Use a complete presentation both times.
Worksheet Worksheet 5
1.
Solve the equation \(ax + b = cx + d\) for the variable using a complete presentation.
2.
Solve the equation \(x\sin(1) + \cos(2) = x \ln(3) - f(4)\text{.}\) Use a complete presentation.
3.
Solve the equation \(3\ln(x) + \ln(4) = 8\) for Use a complete presentation.
4.
Solve the equation \(3x + \log(6) = \exp(3)\text{.}\) Use a complete presentation.
Section 10.3 Deliberate Practice: Solving for Variables and Function Expressions
Algebra is a skill, which means it requires practice to become proficient. But it will take more than rote repetition to get there. Deliberate practice is the thoughtful repetition of a task. For each of these sections, you will be given a list of specific skills or ideas to focus on as you practice thinking through the problems.
Focus on these skills:
- Write the original expression.
- You do not need to use a substitution to replace the function expressions, but it might help.
- Avoid inappropriate cancellations and other algebraic errors involving functions.
- Present your work legibly.
Worksheet Worksheet
Instructions: Solve for the indicated quantity.
1.
Solve \(\ln(3) x + 5 = \ln(2) x - 7\) for \(x\text{.}\)
2.
Solve \(\pi x + \sqrt{3} = 2x + \cos(2)\) for \(x\text{.}\)
3.
Solve \(\exp(4) x + 3 = \exp(2) x + \exp(5)\) for \(x\text{.}\)
4.
Solve \(\sqrt{5} x - \log(4) = \pi x + \sqrt{7}\) for \(x\text{.}\)
5.
Solve \(2 \sin(x) + \sqrt{3} = 4 \sin(x)\) for \(\sin(x)\text{.}\)
6.
Solve \(5 \cos(x) + \sqrt{2} = 3 \cos(x) - 2\) for \(\cos(x)\text{.}\)
7.
Solve \(4 \ln(x) + 3 = -2 \ln(x) + \ln(4)\) for \(\ln(x)\text{.}\)
8.
Solve \(5 \exp(x) - \sqrt{6} = 2 \exp(x) - \cos(1)\) for \(\exp(x)\text{.}\)
9.
Solve \(3 f(x) - 4 f(2) = g(3) + 6\) for \(f(x)\text{.}\)
10.
Solve \(5 \sqrt{x} + \ln(3) = 4 \exp(2) - 3\) for \(\sqrt{x}\text{.}\)
Section 10.4 Closing Ideas
There’s an internet math meme that makes its way around social media every now and then. There are a few versions of it, but they all look something like this:
\begin{equation*}
\text{Calculate: } 6 \div 2(1+2)
\end{equation*}
People will argue for hours over whether the correct answer is or Some people will insist on the "rule" that division and multiplication and done left-to-right, but others will note that the implied multiplication is usually given priority. For example, in writing \(5 \div 2x\text{,}\) there’s virtually no context in which we would separate the 2 from the \(x\text{.}\)
But all of that arguing misses the point. The "answer" is that the person writing the problem did not communicate effectively. The person who wrote the expression did not use notation in a clear manner. And that’s an important lesson. Mathematics is not about arcane rules, but about communicating ideas.
One of the ongoing challenges for math courses is that students can keep advancing forward with a weak foundation. They can often do well enough to pass the new material while there are weaknesses from previous sections that remain unaddressed. And this can continue for a while, sometimes multiple years, until there’s a certain moment where things suddenly make no sense at all.
Sometimes, students will get to that spot and try to brute force their way through it. It is not often successful from the educational perspective, though with enough effort students they may still be able to "pass" the material (even if just barely).
The first part of this book has been focused on shoring up a number of key algebraic ideas and concepts. This last section is an important step towards true algebraic fluency. As soon as you are able to "see through" all of the notation and break down complicated expressions to their more basic components, you have access to a much wider range of algebraic thought processes. Going back to the language analogy, this is the point where you cross over from speaking in broken sentence fragments into the early stages of fluency.
The next major step in forward development is contextualized practice. It’s not enough to simply see these algebraic ideas in isolation. They need to be brought into perspective through the lens of other mathematical ideas. And that is where the college level math course will take over.
The rest of this book is about backwards development. In other words, it’s about finding and filling some of the earlier holes that create problems for students. In a real sense, we will be working backwards from concepts found in high school algebra down to elementary school arithmetic. The content is full of a number of simple ideas that are sometimes missed by students as they’re coming up through the educational system. And those are sometimes the concepts that help to support higher levels of mathematical thinking.
Section 10.5 Going Deeper: Your Calculator and You
Teachers used to say that you wouldn’t be carrying around a calculator in your pocket all the time. They were wrong. We do this all the time. But it’s surprising that even though we have these calculators, most people can’t actually use them properly. Most people can use them for simple arithmetic, but will make errors for even slightly complex situations.
From what I can tell, the basic challenge is that calculators behave in a way that doesn’t always match up with how students think the calculator should behave. Here is an example:
\begin{equation*}
\frac{2 + 6}{3 + 1}
\end{equation*}
When students read this in their minds, they see it as plus divided by plus and proceed to press the following buttons:
If you do the calculation by hand, you can see that the result will be 2. But the calculator will give the answer 5. Most students won’t even recognize that there’s something wrong because they have learned to trust their calculators. The problem here is exactly the same as the problem with the \(x^2 + 3/x + 2\) example from earlier in this section. The calculator is trying to apply the order of operations based on what you’ve entered, but that results in the calculation of something other than what you had intended.
The problems only get worse as you get further along in math. There are other situations where different calculators are programmed to interpret expressions differently. A common test for calculators is to see how it handles implied multiplication. Consider the following key presses:
Some calculators will interpret this as \(\frac{1}{2\sqrt{2}}\) (giving 0.5356), others will treat this as \(\frac{1}{2} \sqrt{2}\) (giving 0.7071), and still there’s a whole other set of calculators that might just give you the number 0.5.
Your instinct may be to ask the following question: "Which one is right?" And the reason you might do this is because you’re used to there always being a right answer. But that’s simply not the case. Just as with the math meme in Section 10.4, this isn’t a problem of right and wrong. It’s a problem of communication. Did the buttons communicate the right instruction to the calculator? In this case, you don’t know what the intention was, and neither does the calculator. It’s just blindly following its instructions.
Fortunately, most modern calculators give you a way around this problem. They give you parentheses so that you can be explicit with the calculator what you want it to do. And this is a habit that you will want to develop for any time you use a computational device to perform a calculation. It also doesn’t hurt to be explicit about multiplication calculations.
The problems continue to get worse when you start using functions. Some calculators evaluate the function when you press the button, and others wait for you to type in your full expression before it decides what it wants to calculate. And depending on which type of calculator you have, you’ll need to do things in different orders to get the result that you want. If you didn’t read the margin comment about the square root function, you should read that now.
When you get into trigonometry, you will also need to know whether your calculator is in degree mode or radian mode. And failing to switch to the appropriate mode will lead to even more errors. Basically, you’re picking between different units for numbers. It’s like if someone were to tell you that the length of an object is 3. On its own, you don’t know what units of measurement they used, and so it’s impossible to know if they meant 3 inches or 3 miles.
Ultimately, it’s impossible to give clear guidance for how your specific calculator will behave, because there’s so much variation between them. You just need to know how to use it right. There are a couple techniques you can use. The first was mentioned earlier, which is to use parentheses to explicitly tell the calculator the order you want it to calculate everything. Another technique is to break the calculation down into smaller steps, so that you’re only putting in one calculation at a time, and then writing down that number on your paper before going to the next calculation. This is especially helpful if you have a calculator that evaluates functions immediately. This saves you from the mental gymnastics of trying to plan ahead to make sure that you’ve worked out the correct order for typing everything into your calculator.
In some ways, even if your calculator has parentheses it’s preferable for you to do it step by step and write out the results of each set of calculations. The reason is that this is easier to work with when you check your work. For longer expressions, you can end up with four or five sets of parentheses, and it can be a lot of work to make sure they’re all correct. It also helps you to mentally think through the steps of the calculation to make sure that you really understand what you’re doing. But the precise bounds on when to switch is going to depend on the amount of practice you have with your calculator.
