The Answers Are Correct (Why Some Students Don’t Realize It)


Many math books post the answers to selected questions in the back of the book. Most of my workbooks include the answers to every question in the back of the book.

A few students get a different form of the same answer, and mistakenly believe that the answer in the back of the book is wrong, when it’s really correct.

One common example occurs when the student gets a squareroot in the denominator, like:

The above answer isn’t in standard form. Most math teachers and authors would rationalize the denominator in order to express the answer in standard form. To rationalize the denominator, multiply both the numerator and denominator by the squareroot of 2, like the example below. Note that the squareroot of 2 times itself equals 2.

If you didn’t realize this, there is a simple way around it. Use a calculator. If you enter 1/sqrt(2) on your calculator, you will get 0.707106781. If you enter sqrt(2)/2 on your calculator correctly, you will get 0.707106781. Now you can see that the two seemingly “different” answers are exactly the same.

Unfortunately, many of the popular algebra solvers that you can find online don’t express their answers in standard form. So if you solve an algebra problem and get 1/sqrt(2) as your answer, an online algebra solver might give you the same answer, 1/sqrt(2), and then you might feel “confident” that the book’s answer, sqrt(2)/2, must be wrong because it’s different. But the book’s answer is actually the same answer. In fact, the book’s answer is better because it is expressed in standard form. It’s “better” in the sense that many teachers prefer for their students to express their answers in standard form.

Here is another common example. Suppose that a student obtains the following answer:

This answer isn’t in standard form. Most math teachers and authors would factor out the perfect square. Note that 8 equals 4 times 2. The number 4 is a perfect square, since the squareroot of 4 is 2 (put another way, 2 squared equals 4). You can factor out the perfect square as follows:

Again, you can check this with a calculator. Whether you enter sqrt(8) or 2*sqrt(2) on your calculator, either way you will get 2.828427125.

As a third example, suppose a student arrives at an answer of 9/6, but the book’s answer is 3/2. Once again, the student sees something different and assumes that one of the two answers must be incorrect. However, 9/6 equals 3/2. The difference is that the fraction 3/2 is in reduced form, whereas the answer 9/6 can be reduced. To see this, divide both the numerator and denominator of 9/6 by 3. You will get 9/6 = (9/3) / (6/3) = 3/2. If you use a calculator, you can check that 9/6 and 3/2 both equal 1.5.

This has been a problem for me with my workbook, Algebra Essentials Practice Workbook with Answers. An occasional review suggests that some of the answers are wrong. However, in every case that a student has mentioned a specific problem, and in every case that a student has contacted me by email to inquire about an answer, it has turned out that the student’s answer was, in fact, equivalent to the answer in the back of the book. The problem has usually been that the student didn’t rationalize a denominator, factor out a perfect square, or reduce a fraction.

As an author, it can be frustrating to know that the answers are correct, yet see a book review on Amazon suggesting otherwise. When I made the updated 2014 edition, an international math guru and I independently checked every single answer in this book with painstaking care. The answers are correct. The challenge is getting students to realize this.

On the product page, I worked out the full solutions, step by step, for some common questions that students have asked about, in order to help show that the answers are, in fact, correct. If students have questions, they are encouraged to contact me and give me an opportunity to teach them something that they might not have realized. Unfortunately, reviews that suggest that the answer key has mistakes discourage some students from using the book. Some customers would believe a random stranger’s review over a teacher with 20+ years classroom experience with a Ph.D. (If you think about it, if the student bought an algebra practice workbook, that student is probably not yet an expert on the subject of algebra.)

If you use any of my math workbooks and come to trust the answer key, please post a review to help other potential customers. If you suspect any of the answers may be wrong, please contact me (there is a contact me button my blog, for example) to ask about the answer. (It would also help if customers who know that the answers are reliable would vote on reviews as being helpful or not helpful.)

For my newest workbooks, like 50 Challenging Algebra Problems (Fully Solved), I’m now working out the full solution to every problem in the book so that students can see how I arrived at each answer. In many cases, I also offer alternative forms of the same answer (but it’s impossible to list every possible answer that a student might obtain). I hope that the full solutions will be helpful to students, and help them realize that the book’s answers are correct.

I have two new workbooks on the subject of calculus coming soon.


Copyright © 2018 Chris McMullen, author of the Improve Your Math Fluency series of math workbooks

Newest releases (in math):

  • 50 Challenging Algebra Problems (Fully Solved)
  • Fractions Essentials Workbook with Answers
  • 300+ Mathematical Pattern Puzzles

Improve Your Math Fluency. Build fluency in:

  • arithmetic
  • long division
  • fractions
  • algebra
  • trigonometry
  • graphing
  • physics
  • calculus (coming August, 2018)

Challenging Algebra Problems


This article includes the following sections:

  • Ways to Challenge Advanced Students
  • How to Make Challenging Algebra Problems
  • A Couple of Sample Problems (from my latest book)
  • 50 Challenging Algebra Problems – Fully Solved (my latest workbook)


Some students catch on quickly, while others struggle to understand.

So how can you engage and challenge the better problem-solvers without blowing the minds of the other students?

  • If you’re a teacher, you can include a challenging bonus question on an assignment, quiz, or exam.
  • If you’re a parent, you can search for supplemental material with challenging problems.
  • If you’re a student, you can try solving harder unassigned problems, reading ahead in your textbook, or looking for supplemental material.


For teachers (or very knowledgeable parents) who would like to make challenging problems, here are some ideas for how to go about it:

  • Disguise the quadratic equation. One way is to make more than the usual three terms and put some on both sides of the equation, so that students need to combine like terms. Another way is to substitute y = xc, where c is a power of your choice. For example, if you let c = 2, your equation will look like a quartic, but if students make the substitution y = x2, the quartic will turn into a quadratic. Another example is to let c = 1/2 and replace y = x1/2 with a squareroot. When students see that squareroot, it won’t seem like a quadratic. Tip: Think of what you want the answers to be, like x = 2 and x = 4/3, and then f.o.i.l. this into the quadratic, like (x – 2)(x – 4/3) = 0. (In this example, I would multiply both sides by 3 to get rid of the fraction.)
  • Make a system of equations look different from normal. Students are used to seeing things like 3x + 2y = 21 and 4x – 5y = 5. Instead, you might write the second equation like xy = 15, or y = 15/x. Another variation is to replace x with 1/x and y with 1/y. In my example, you would get 3/x + 2/y = 21 and 4/x – 5/y = 5. If students define t = 1/x and u = 1/y, they could then solve for t and u like usual, and then solve for x and y from t and u. Tip: When making a system, first decide on the answers, like x = 9 and y = 12, next make the coefficients, and lastly determine the constants.
  • Put squareroots in problems they should know how to solve, but which usually don’t have squareroots in them. For example, if you have squareroot(3) and 1/squareroot(3) in the same equation, the terms can be combined by rationalizing the denominator to rewrite 1/squareroot(3) as squareroot(3) / 3. (Multiply the numerator and denominator by the squareroot of 3, and use the rule that the squareroot of 3 times itself equals 3.) Then factor out squareroot(3) to combine the terms.
  • Put variables inside of squareroots. For example, you could write something like squareroot(x + 8) = x + 2. When you square both sides, you f.o.i.l. out the (x + 2) squared. There are many other ways to write a solvable equation with a variable in a squareroot. You could have things like squareroot(x) or 1/squareroot(x).
  • Include an extra variable that cancels out. If students count equations and unknowns, it will seem like the problem can’t be solved. They’re partly right: The variable that cancels out will be indeterminate. But since one variable cancels, in this case it will be possible to solve for the other unknown(s).
  • Use fractional exponents like 2/3 or 3/4. For example, if you isolate x to the power of 2/3, you can solve for x by raising both sides of the equation to the power of 3/2. You can make the numbers work out so that a calculator isn’t needed for students who understand the fractional powers. For example, if you raise 8 to the power of 2/3, you can do this in your head, since the cube root of 8 equals 2 (the power of 1/3 means to take the cube root, and you can think of 2/3 as involving a square and a cube root in any order). Of course, you can easily check your answer with a calculator, just to be sure.
  • Put the unknown in a denominator. It’s amazing how many students freeze up over this, yet it’s not uncommon in science for variables to appear in the denominators of formulas. The progression of this occurs when students need to make a common algebraic denominator to solve for the unknown.
  • Give problems that involve inequalities, especially when there are minus signs. For example, make a system of inequalities that involves substitution.
  • Look for applications of algebra in higher-level math, physics, engineering, chemistry, economics, astronomy, and other subjects. (Sometimes, the problem isn’t really that hard, but it just seems hard because it’s out of context. But there are also challenging applied algebra problems in many subjects.)
  • Make word problems that apply algebra. There is great variety in what can be done in the way of algebraic word problems.
  • Prepare an elaborate solution that has an intentional mistake and ask students to figure out what is wrong. Ideally, the mistake would seem subtle, and would be a common mistake that many students tend to make on their own (like not distributing a minus sign correctly).

There are two types of problems that especially appeal to me:

  • One kind where the problems look harder than they really are. I love it when students feel convinced that the problem is impossible based on what they know, but when they eventually figure it out on their own. This happens sometimes when a problem appears out of context, involves an application of algebra, or when a simple substitution makes something seem a lot more complicated than it really is. Word problems often fall in this category, too.
  • The other kind is where the problem looks much easier than it really is. The structure might be so simple that at first glance you expect it to be a piece of cake, but there is some subtle aspect to it that it turns out to be much harder than it seemed. The catch is that the problem still needs to be reasonably solvable, so that the top students can still figure it out given enough time and sufficient background. It’s surprising how many mathematical problems look easy, but turn out not to be nearly as easy as they look.


Here are a couple of sample problems from my latest math workbook.


My latest math workbook includes 50 challenging algebra problems, space to try and solve them (in the paperback edition), and full solutions on the page following each problem (you won’t be able to see the full solutions until you turn the page).

There is a healthy variety of problems, involving a range of techniques. There is also a good range in the level of difficulty from problem to problem.

50 Challenging Algebra Problems (Fully Solved)

Paperback ISBN: 19416912344

Kindle ASIN: B07C7WHV3R


Copyright © 2018 Chris McMullen, author of the Improve Your Math Fluency series of math workbooks

Newest releases (in math):

  • 50 Challenging Algebra Problems (Fully Solved)
  • Fractions Essentials Workbook with Answers
  • 300+ Mathematical Pattern Puzzles

Improve Your Math Fluency. Build fluency in:

  • arithmetic
  • long division
  • fractions
  • algebra
  • trigonometry
  • graphing
  • physics

How to Use Algebra in Arithmetic



It’s interesting that algebra can be useful in arithmetic.

Since arithmetic is a prerequsite to learning algebra, most algebra courses don’t focus on how to apply algebra to do arithmetic.

However, you can apply your algebra skills to arithmetic problems.

Following is an example.

Suppose that you would like to multiply 58 times 58, but don’t have a calculator (or cell phone) handy.

One way to do this, which is often taught in elementary school, is basically to multiply 8 times 58, then multiply 50 times 58, and add these together.

Algebraically, what your elementary school teacher taught you is that 58 × 58 = (50 + 8) × 58 = 50 × 58 + 8 × 58.

You may not have realized it at the time, but your teacher was applying the distributive property of mathematics.

When you solve the problem this way, first you multiply 8 × 58 to get 464. Then you multiply 50 × 58 to get 2900. Finally, you add these together to get 3364.

Now let’s see how algebra can supply an alternative solution.

I’d prefer to work with nice round numbers. Hey, 60 is a round number close to 58. If I could work with a round number, like 60, and a small number, like 2, that would make the arithmetic much simpler to do without a calculator.

So let’s write 58 as 60 minus 2.

Then the problem is 58 × 58 = (60 – 2) × (60 – 2).

Use the f.o.i.l. method from algebra. Recall that the f.o.i.l. method stands for “first,” “outside,” “inside,” “last.”

One example of the f.o.i.l. method from algebra is (x + y)(x + y) = x² + xy + xy +y² = x² + 2xy +y².

Let’s apply this to the arithmetic problem. We can think of (60 – 2) as (x + y), where x = 60 and y = –2.

58 × 58 = (60 – 2) × (60 – 2) = 60 × 60 + 2 (60)(–2) + (–2) × (–2).

Look what we did. We wrote the original problem, 58 × 58, in terms of three simple multiplication problems: 60 × 60, 2 × 60 × (–2), and (–2) × (–2).

58 × 58 = 3600 – 240 + 4 = 3364.

Note that (–2) × (–2) = +4 because the two minus signs cancel out.


Copyright © 2015 Chris McMullen, author of the Improve Your Math Fluency series of math workbooks

Newest release:

  • 300+ Mathematical Pattern Puzzles

Improve Your Math Fluency. Build fluency in:

  • arithmetic
  • long division
  • fractions
  • algebra
  • trigonometry
  • graphing

Tips for Finding the Slope of a Straight Line



Slope is a measure that indicates how steep or shallow a straight line is:

  • A line with greater slope is steeper.
  • A line with less slope is shallower.
  • A horizontal line has zero slope.

Slope Sign

Slope can be positive or negative:

  • A line with positive slope slants upward.
  • A line with negative slope slants downward.
  • A line with zero slope is horizontal.

Slope Sign B

The slope of a straight line equals rise over run.

  • The rise between two points is vertical. It’s the change in y.
  • The run between two points is horizontal. It’s the change in x.

Slope Eqn


From the graph of a straight line, determine the slope as follows:

  • Mark two points on the line.
  • Read the x- and y-coordinates of the two points, (x1, y1) and (x2, y2).
  • Subtract y2 – y1 to get the rise.
  • Subtract x2 – x1 to get the run.
  • Divide the rise by the run.

Slope Triangle

Here are a few tips:

  • When choosing the two points, try to find points where both x and y are easy to read without interpolating. This isn’t always possible: In that case, at least one coordinate should be easy to read without interpolating.
  • Choose two points far apart. This reduces the relative error in interpolating.
  • Make sure that both points lie on the straight line. Don’t choose a point that’s close to the line, but doesn’t lie on it.


Example: Find the slope of the straight line in the graph below.

Slope Example

Solution: First, choose two points on the line. Ideally, these points should be far apart and easy to read.

In this case, it’s easy to read both the x- and y-coordinates for the leftmost and rightmost points shown in the graph. So let’s choose those.

Slope Example 2

  • The leftmost point has coordinates (0, 3).
  • The rightmost point has coordinates (10, 8).

Subtract the y-values to determine the rise:

y2 – y1 = 8 – 3 = 5

Subtract the x-values to determine the run:

x2 – x1 = 10 – 0 = 10

(In coordinate graphing, recall that x is horizontal and y is vertical.)

Divide the rise by the run to find the slope:

Slope Example 3

The slope of the line is 0.5.

Check: You can check your answer as follows.

Look at the graph. Starting from (0, 3), the next point that’s easy to read is (2, 1).

From (0, 3) to (2, 1), the line goes one unit up (vertically) and 2 units over (horizontally).

The ratio of the rise to the run is 1 to 2. Divide the rise (of 1) by the run (of 2). The slope is 0.5. ♦


Copyright © 2015 Chris McMullen, author of the Improve Your Math Fluency series of math workbooks

Newest release:

  • Basic Linear Graphing Skills Practice Workbook

Related books:

  • Trigonometry Essentials Practice Workbook with Answers
  • Learn or Review Trigonometry Essential Skills
  • Algebra Essentials Practice Workbook with Answers
  • Systems of Equations: Simultaneous, Substitution, Cramer’s Rule

Fibonacci Sequence & a Cool Pattern

Image from ShutterStock.

Image from ShutterStock.


The Fibonacci sequence adds consecutive terms:

  • 1
  • 1
  • 1 + 1 = 2
  • 2 + 1 = 3
  • 3 + 2 = 5
  • 5 + 3 = 8
  • 8 + 5 = 13
  • 13 + 8 = 21
  • 21 + 13 = 34
  • 34 + 21 = 55
  • 55 + 34 = 89

Since the last two terms were 55 and 89, we would add these together to get 89 + 55 = 144.

Then you would add 144 and 89 to make 233, and so on.

I saw a cool pattern involving the Fibonacci sequence recently at the Mathemagical Site:

Fibonacci Triples via Mathemagical Site

This involves Fibonacci triples.

A Fibonacci triple consists of three consecutive numbers from the Fibonacci sequence, such as:

  • 1, 1, 2
  • 1, 2, 3
  • 2, 3, 5
  • 3, 5, 8
  • 5, 8, 13
  • 8, 13, 21
  • 13, 21, 34

As shown on the Mathemagical Site, the square of the middle number is always one less or one more than the product of the first and third numbers:


Here are a few examples:

  • (2, 3, 5): 3 x 3 = 2 x 5 – 1
  • (3, 5, 8): 5 x 5 = 3 x 8 + 1
  • (5, 8, 13): 8 x 8 = 5 x 13 – 1

Curious about this, I’ve been working through the algebra, and finally came up with an algebraic proof, which follows.

My proof is algebraic and not necessarily obvious, but since I worked it out, I thought I would share it here. 🙂

We begin with two facts about the Fibonacci sequence:


These are two different ways of saying that if you add two consecutive numbers from the Fibonacci sequence, you get the next number in the sequence.

Now solve for xn in each sequence:


Multiply these together:


Foil this out:


Recall that


Plug this into the first term on the right-hand side of the previous equation:




Two of these terms cancel (the remaining terms are rearranged):


Believe it or not, this basically concludes the proof. The remainder is basically interpreting this result.

This is a recursion relation that relates the square of the n-th term to the square of the previous term (xn-1 times itself).

Following is the Fibonacci sequence, labeling values of n:

  • n = 1 is 1.
  • n = 2 is 1.
  • n = 3 is 2.
  • n = 4 is 3.
  • n = 5 is 5.
  • n = 6 is 8.
  • n = 7 is 13.
  • n = 8 is 21.

Let’s plug in n = 3 and see what happens:


If instead you plug in n = 4, you get:


Now just plug in these two expressions (x3x3 – x4x2 and x4x4 – x5x3) into the previous recursion relation and you can prove that all of the Fibonacci triples satisfy one of the following relations:


That is, if x3x3 – x4x2 = 1 and x4x4 – x5x3 = -1, the previous recursion relation gives similar expressions for x5x5 – x6x4, x6x6 – x7x5, and so on.


Copyright © 2015 Chris McMullen, author of the Improve Your Math Fluency series of math workbooks

  • Algebra Essentials Practice Workbook with Answers
  • Trigonometry Essentials Practice Workbook with Answers
  • Learn or Review Trigonometry: Essential Skills

Algebra with Roman Numerals

Algebra Roman Numerals


Aren’t you glad you don’t have to do algebra with Roman numerals?

The problem is: Xx + X = CX.

This translates to 10x + 10 = 110.

Subtract X from both sides: Xx = CX – X = C.

Translation from Roman numerals: 10x = 110 – 10 = 100.

Now divide both sides by the coefficient: x = C/X = X.

Not expressed in Roman numerals, it looks like this: x = 100/10 = 10.


Copyright © 2015 Chris McMullen, author of the Improve Your Math Fluency series of math workbooks