Here is an example of a lesson I did on linear equations where my objective was to show that they are equations of first degree. The reason I do it this way is because I tend to find that students only see first degree equations as being of the form: mx + b, meaning that they can ONLY look like, say:

2x + 3 = 5.

An equation of the first degree (aka linear equation) with a single variable is defined by the following statement:

mx + b = 0 where $m \neq 0$ and $m, b \in R$

or any other equation that is equal to the above statement.


$(x + 1)(x - 1) = x^2 + 2x + 1$ is an linear equation.


$x^2 - 1 = x^2 + 2x + 1$ $\leftrightarrow -1=2x+1$ $\leftrightarrow 2x + 2 = 0$

Thus: x + 1 = 0

Since I was chided in a previous post for not providing context, I will do so now: I shared this approach with a College Algebra class (first year college students) and it is NOT an honors class. What do you think about my approach? I welcome any comments (positive or negative).

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    $\begingroup$ I shared this approach with a College Algebra class --- As someone who in the past has unfortunately gone down roads like this many times, I STRONGLY RECOMMEND that in classroom teaching you stay far away from excessive focus on terminological issues and discussions of minutiae, especially when initiated by you and not by a student's question or misunderstanding. $\endgroup$ Jun 17 at 6:19
  • $\begingroup$ Would you say because a student may misunderstand it or because it's "too deep" for the subject itself? Just curious. $\endgroup$
    – Wasp
    Jun 17 at 10:57
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    $\begingroup$ I think it's too far from what is truly important in a college algebra course for 1st year college students. I suspect very few (I'd guess below 15%) of these students (U.S. perspective) will even advance as far as 2nd semester calculus (introductory integral calculus -- techniques of integration, applications of integration, basic sequences and series convergence tests, Taylor series expansions), let alone as far as 2nd year courses such as Velleman's How to Prove It or elementary linear algebra where students might appreciate such nuances. $\endgroup$ Jun 17 at 12:37
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    $\begingroup$ For equations like $(x+1)(x-1) = x^2 + 2x + 1$ you could just say that they look like they're quadratic, but this is really a linear equation in disguise. Most (all?) college algebra and precalculus textbooks I've taught from take care of this by defining a linear equation something like: "An equation that can be written in the form $ax+b=0,$ where $a\neq0.$" (exact wording from p. 55 of Swokowski/Cole's 8th edition of Fundamentals of College Algebra, a book I used for 1-2 such courses each semester Fall 1993 to Spring 1996). $\endgroup$ Jun 17 at 12:51
  • $\begingroup$ That is very true. One of the "issues" that often come up for me is that my focus is always quite theoretical and very little emphasis placed on practical applications. Some praise me for it, others say the same as you. It´s something I often think about, so I do welcome your input. Yes, as for "in disguise", I always just end up telling them that. Basically, if they can write it as ax + b = 0, then it counts as linear (I would add the set notation though for all real numbers, etc...not a big deal but just habit for me). $\endgroup$
    – Wasp
    Jun 17 at 12:54

$(x+1)(x−1)=x^2+2x+1$ is an linear equation.

I agree with all above answers. I wouldn't say that it is a linear equation, but it reduces to, or it is solved by or any other phrase you want to use instead. Because, I mean, that's the idea right?

I feel that the relaxed terminology is not quite accurate, and it might generate confusion when they see these ideas in the Cartesian plane. Recall that $ax+b=c$ represents the intersection of two "lines" in the plane (students might associate "lines" to "linear", and that's ok). But $(x+1)(x−1)=x^2+2x+1$ is the intersection of two parabolas.

Finally, if you treat $$(x+1)(x−1)=x^2+2x+1$$ as a linear equation, it is very likely that you have problems when they say that $$(x+1)(x−1)=\color{red}{2x^2}+2x+1$$ is a linear equation as well.

  • $\begingroup$ Yes, correct -- since it reduces to a linear, then we have an equation of the first degree. As to your examples, of course! But if the exercise is something like "identify if the following equations are linear", then by reduction some of them will be, others will not. I do agree, of course, with your example about the parabolas when graphing. $\endgroup$
    – Wasp
    Jun 17 at 11:00
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    $\begingroup$ I have to insist with my point. To me, there is a huuuuuuuuge difference between "to be" and "to reduce into". If the exercise was posed as "identify if the following equations are linear", then my answer would be "the equation $(x-1)(x+1)=x^2+2x+1$ is not linear". $\endgroup$
    – FormerMath
    Jun 17 at 19:28

I think I would tend to disagree with your terminology's bent. The label of the equation speaks to the type of the expression which forms the equation. The type of solution set does not label the equation. Quadratic equations are not differently labeled when they have different solution sets. Furthermore, since there are literally infinitely many equivalent expressions which maintain the same solution set, I think anchoring terminology on the solution set is not a good choice. Your example is quadratic because it is formed with a quadratic expression. Simple as that.

To give some more absurd examples, $$ x^{11} = 0, \qquad x^{8}=0 \qquad x=0, \qquad sinh(x)= 0 \qquad x(\cos^2 x + \sin^2 x+ln(e^x)-x )=0$$ are not all the same equation. Same solution set ? Sure. Same equation, no.

Or, more like your example, is $x^{10}+x = x^{10}-x+2$ a linear equation ? I would say no, however, it allows a reduction to the linear equation $x=-x+2$ hence $2x=2$ and we conclude $x=1$. I want some term other than "same" to communicate that steps were require to bridge the gap from the given equation to the linear equation.

  • $\begingroup$ Thank you. What you are saying is true (I agree that the example given is "quadratic"), but what if the exercise asked you to identify an equation of one degree and had that example? Since the example you give can be reduced to a linear equation, then I would say yes, that it can be because the given example is equal (by reduction) to a linear equation. I do take your point though and agree with your take on this. As for quadratics, I tend to just say their solution is a, b, c (must be real numbers) and you can have either an incomplete or pure quadratic. $\endgroup$
    – Wasp
    Jun 16 at 21:22

I don't approve of teaching students unusual meanings for standard terms. If I correctly understand your example, $\sin x=x$ would be a linear equation. And whether $ax^2+bx+c$ is linear depends on whether $b^2=4ac$. The issue becomes worse with polynomials of higher degree.

It's not clear to me what you mean by two equations being equal (in "or any other equation that is equal to the above statement"). From your example, it appears that you just mean "equivalent". Or maybe "equivalent via very elementary algebraic manipulations".

  • $\begingroup$ Yes - equivalent via the algebraic manipulations. So, basically any linear equation of the form mx + b. The issue certainly would be a bit nastier with polynomials of higher degree, but my objective here was to show them that an equation does not have to "look" linear for it to be linear (of one degree). They often limit their examples (when identifying linear equations) to what they've typically seen: 2x + 1 = 3, but there are many linear equations in disguise. $\endgroup$
    – Wasp
    Jun 16 at 20:39

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