Tag Archives: rule of four

Graphing Ratios and Proportions

Last week, some colleagues and I were pondering the difficulties many middle school students have solving ratio and proportion problems.  Here are a few thoughts we developed to address this and what we think might be an uncommon graphical extension (for most) as a different way to solve.

For context, consider the equation \displaystyle \frac{x}{6} = \frac{3}{4}.

(UNFORTUNATE) STANDARD METHOD:

The default procedure most textbooks and students employ is cross-multiplication.   Using this, a student would get

\displaystyle 4x=18 \longrightarrow x = \frac{18}{4} = \frac{9}{2}

While this delivers a quick solution, we sadly noted that far too many students don’t really seem to know why the procedure works.  From my purist mathematical perspective, the cross-multiplication procedure may be an efficient algorithm, but cross-multiplication isn’t actually a mathematical function.  Cross-multiplication may be the result, but it isn’t what happens.

METHOD 2:

In every math class I teach at every grade level, my mantra is to memorize as little as possible and to use what you know as broadly as possible.  To avoid learning unnecessary, isolated procedures (like cross-multiplication), I propose “fraction-clearing”–multiplying both sides of an equation by common denominatoras a universal technique in any equation involving fractions.  As students’ mathematical and symbolic sophistication grows, fraction-clearing may occasionally yield to other techniques, but it is a solid, widely-applicable approach for developing algebraic thinking.

From the original equation, multiply both sides by common denominator, handle all of the divisions first, and clean up.  For our example, the common denominator 24 will do the trick.

\displaystyle 24 \cdot \frac{x}{6} = 24 \cdot \frac{3}{4}

4 \cdot x = 6 \cdot 3

\displaystyle x = \frac{9}{2}

Notice that the middle line is precisely the result of cross-multiplication.  Fraction-clearing is the procedure behind cross-multiplication and explains exactly why it works:  You have an equation and apply the same operation (in our case, multiplying by 24) to both sides.

As an aside, I’d help students see that multiplying by any common denominator would do the trick (for our example, 12, 24, 36, 48, … all work), but the least common denominator (12) produces the smallest products in line 2, potentially simplifying any remaining algebra.  Since many approaches work, I believe students should be free to use ANY common denominator they want.   Eventually, they’ll convince themselves that the LCD is just more efficient, but there’s absolutely no need to demand that of students from the outset.

METHOD 3:

Remember that every equation compares two expressions that have the same measure, size, value, whatever.  But fractions with differing denominators (like our given equation) are difficult to compare.  Rewrite the expressions with the same “units” (denominators) to simplify comparisons.

Fourths and sixths can both be rewritten in twelfths.  Then, since the two different expressions of twelfths are equivalent, their numerators must be equivalent, leading to our results from above.

\displaystyle \frac{2}{2} \cdot \frac{x}{6} = \frac{3}{3} \cdot \frac{3}{4}

\displaystyle \frac{2x}{12} = \frac {9}{12}

2x=9

\displaystyle x = \frac{9}{2}

I find this approach more appealing as the two fractions never actually interact.  Fewer moving pieces makes this approach feel much cleaner.

UNCOMMON(?) METHOD 4:  Graphing

A fundamental mathematics concept (for me) is the Rule of 4 from the calculus reform movement of the 1990s.  That is, mathematical ideas can be represented numerically, algebraically, graphically, and verbally.  [I’d extend this to a Rule of 5 to include computer/CAS representations, but that’s another post.]  If you have difficulty understanding an idea in one representation, try translating it into a different representation and you might gain additional insights, or even a solution.  At a minimum, the act of translating the idea deepens your understanding.

One problem many students have with ratios is that teachers almost exclusively teach them as an algebraic technique–just as I have done in the first three methods above.  In my conversation this week, I finally recognized this weakness and wondered how I could solve ratios using one of the missing Rules: graphically.  Since equivalent fractions could be seen as different representations of the slope of a line through the origin, I had my answer.

Students learning ratios and proportions may not seen slope yet and may or may not have seen an xy-coordinate grid, so I’d avoid initial use of any formal terminology.  I labeled my vertical axis “Top,” and the horizontal “Bottom”.  More formal names are fine, but unnecessary.  While I suspect most students might think “top” makes more sense for a vertical axis and “bottom” for the horizontal, it really doesn’t matter which axis receives which label.

In the purely numeric fraction in our given problem, \displaystyle \frac{x}{6} = \frac{3}{4}, “3” is on top, and “4” is on the bottom.  Put a point at the place where these two values meet.  Finally draw a line connecting your point and the origin.

ratio1

The other fraction has a “6” in the denominator.  Locate 6 on the “bottom axis”, trace to the line, and from there over to the “top axis” to find the top value of 4.5.

ratio2

  Admittedly, the 4.5 solution would have been a rough guess without the earlier solutions, but the graphical method would have given me a spectacular estimate.  If the graph grid was scaled by 0.5s instead of by 1s and the line was drawn very carefully, this graph could have given an exact answer.  In general, solutions with integer-valued unknowns should solve exactly, but very solid approximations would always result.

CONCLUSION:

Even before algebraic representations of lines are introduced, students can leverage the essence of that concept to answer proportion problems.  Serendipitously, the graphical approach also sets the stage for later discussions of the coordinate plane, slope, and linear functions.  I could also see using this approach as the cornerstone of future class conversations and discoveries leading to those generalizations.

I suspect that students who struggle with mathematical notation might find greater understanding with the graphical/visual approach.  Eventually, symbolic manipulation skills will be required, but there is no need for any teacher to expect early algebra learners to be instant masters of abstract notation.

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Exploring Sequences and Lines

Here’s another favorite problem that could be used for any middle or high school students who’ve been exposed to both arithmetic sequences and linear equations.

There is a family of lines, Ax+By+C=0, for which \left\{ A, B, C \right\} can be any arithmetic sequence.  What do all members of this family have in common?

As with most great math problems, the problem is simply stated and can be approached from many different perspectives.  In the end, this one isn’t that difficult to crack, but the solution (at least to me) is not intuitively obvious from the problem statement. If you’ve not seen this before, please try it before reading further.

WARNING:  Problem Solution Follows

MOST COMMON STUDENT APPROACH:  Given the problem’s phrasing, most begin by writing out a few possible equations using different arithmetic sequences, typically with positive common differences.  After a few tries, most get a little frustrated as no obvious pattern emerges from the equations.

It is this point that is typically the most difficult for me as a teacher.  I want to help anyone who struggles, but “help” often means allowing others the room to struggle and to find ways of solving problems on their own.  Only when a student or group gets really frustrated do I sometimes ask, “Are there any other ways you can look at this problem or at your work?”

Eventually, most decide to graph their equations to see if anything pops out.  The following graph was submitted this past semester by one of my precalculus students using the free online Desmos calculator.

ALines1

Two lines intersecting in a point is common.  Three or more in a single point almost always indicates something interesting.  Seven lines through a single point is screaming for attention!  From this graph, all lines in this family apparently contain the point (1,-2).  That seems a bit random until you investigate further, but pattern recognition is just half of the battle for a mathematician.  Now that something interesting has been discovered, a mathematician wants to know if this is a fluke or something inherent to all members of the family.

NOTE:  All graphs my students have produced over the years have always contained specific equations.  I don’t know that it’s any more enlightening, but I did create on Desmos a slider version of the graphs of this family with one slider for the initial term (A) and a second for its common difference (d).

UNIQUE SOLUTION METHODS FROM STUDENTS:

All successful solutions eventually rewrite the sequence \left\{ A,B,C \right\} to \left\{ A,A+d,A+2d \right\} where d is the common difference for a generic arithmetic sequence with initial term A.

Method I: After discovering the common point of intersection, most plug it into the left side of the equation and simplify to get

Ax+By+C=A\cdot 1+\left( A+d\right)\cdot -2+\left( A+2d \right)=0.

Because the left side reduces to zero for all generic arithmetic sequences, \left\{ A,A+d,A+2d \right\}, (1,-2) must be common to all members of this family.

A few students aren’t comfortable dealing with 0=0, so these tend to plug in x=1 and solve for y to get y=-2, proving that the y-coordinate for x=1 for all members of this family is always -2.

Method II:  A few students every year work algebraically from properties of arithmetic sequences.  For any arithmetic sequence, \left\{ A,B,C \right\}, \frac{A+C}{2}=B.  This rewrites to 1\cdot A-2\cdot B+C=0, so whenever \left( x,y \right)=\left(1,-2 \right), then Ax+By+C=0 is a fundamental property of all arithmetic sequences.

Personally, I think this method gets closest to explaining why the point (1,-2) is the common characteristic of this family.

Method III:  This year, I had a student take an approach I’d never seen before.  She defined one arithmetic sequence as \left\{ a,a+d,a+2d \right\} and another as \left\{ m,m+n,m+2n \right\} for any real values of a,d,m, and n.  This leads to a system of equations: a\cdot x+(a+d)\cdot y+(a+2d)=0 and m\cdot x+(m+n)\cdot y+(m+2n)=0 .  If you have some younger students or if all the variables make you nervous, the solution is available via Wolfram Alpha.

Still, this system is pretty easy to manipulate.  If you multiply the first equation by m and the second by a, the x-terms will eliminate with subtraction, giving

m\cdot((a+d)\cdot y+(a+2d))-a\cdot((m+n)\cdot y+(m+2n))=0.

Solving for y shows that all of the coefficients simplify surprisingly easily.

((ma+md)-(am+an))\cdot y=-(ma+2md)+(am+2an)
(md-an)\cdot y = -2\cdot (md-an)\Longrightarrow y=-2

From here, determining x=1 is easy, proving the relationship.

SOLUTIONS and APPROACHES NOT YET OFFERED BY STUDENTS:

Approach A:  High school students don’t often think about ways to simplify problem situations, especially at the beginning of problems.  One way I did that for this problem in later class discussions was to recognize that it one of the terms in the arithmetic sequence was 0, you didn’t need to deal with nearly as many terms. For example, if your sequence was {1,0,-1}, the linear equation would be x-1=0.  Similarly, the sequence \left\{ 01,2 \right\} leads to y+2=0.  Obviously, the only thing these two lines have in common is the point (1,-2).  A proof of the property must still be established, but this is one of the fastest ways I’ve seen to identify the central property.

Approach B:  A purely algebraic approach to this problem could redefine the arithmetic sequence as \left\{ a,a+d,a+2d\right\}as before, giving:

a\cdot x+(a+d)\cdot y+(a+2d)=0

Collecting like terms gives

(x+y+1)\cdot a+(y+2)\cdot d=0.

The values of a and d must remain as parameters to include all possible arithmetic sequences.  Because the equation always equals 0, the coefficients of a and d are both 0, making y=-2 (for the coefficient of d) and therefore x=1.

EXTENSION:

We once had a test question at the end of the unit containing this exercise.  Basically, it reminded students that they had discovered that all lines Ax+By+C=0, for which \left\{ A, B, C \right\} can be any arithmetic sequence contained the point (1,-2).  It then asked for an equation of a family of linear functions using the same arithmetic \left\{ A, B, C \right\} that all contained the point (1,2).

The two most common responses we’ve seen involve a reflection or a vertical translation.  (1,-2) can become (1,2) by reflecting over the x-axis, so making the y-values negative would do the trick:  Ax-By+C=0.  Similarly, (1,-2) can become (1,2) by translating up 4 units, giving Ax+B(y-4)+C=0.