# Tag Archives: system of equations

## Binomial Expansion Variation

Several years ago, I posed on this ‘blog a problem I learned from Natalie Jackucyn:

For some integers A, B, and n, one term of the expansion of $(Ax+By)^n$ is $27869184x^5y^3$.  What are the values of A, B, and n?

In this post, I reflect for a moment on what I’ve learned from the problem and outline a solution approach before sharing a clever alternative solution one of my students this year leveraged through her CAS-enabled investigation.

WHAT I LEARNED BEFORE THIS YEAR

Mostly, I’ve loved this problem for its “reversal” of traditional binomial expansion problems that typically give A, B, and n values and ask for either complete expansions or specific terms of the polynomial.  Both of these traditional tasks are easily managed via today’s technology.  In Natalie’s variation, neither the answer nor how you would proceed are immediately obvious.

The first great part of the problem is that it doesn’t seem to give enough information.  Second, it requires solvers to understand deeply the process of polynomial expansion.  Third, unlike traditional formulations, Natalie’s version doesn’t allow students to avoid deep thinking by using technology.

In the comments to my original post, Christopher Olah and a former student, Bryan Spellman, solved the problem via factoring and an Excel document, respectively.  Given my algebraic tendencies, I hadn’t considered Bryan’s Excel “search” approach, but one could relatively easily program Excel to provide an exhaustive search.  I now think of Bryan’s approach as a coding approach to a reasonably efficient search of the sample space of possible solutions.  Most of my students’ solutions over the years essentially approach the problem the same way, but less efficiently, by using one-case-at-a-time expansions via CAS commands until they stumble upon good values for A, B, and n.  Understandably, students taking this approach typically become the most frustrated.

Christopher’s approach paralleled my own.  The x and y exponents from the expanded term show that n=5+3=8.  Expanding a generic $(Ax+By)^8$ then gives a bit more information.  From my TI-Nspire CAS,

so there are 56 ways an $x^5y^3$ term appears in this expansion before combining like terms (explained here, if needed).  Dividing the original coefficient by 56 gives $a^5b^3=497,664$, the coefficient of $x^5y^3$.

The values of a and b are integers, so factoring 497,664 shows these coefficients are both co-multiples of 2 and 3, but which ones?  In essence, this defines a system of equations.  The 3 has an exponent of 5, so it can easily be attributed to a, but the 11 is not a multiple of either 5 or 3, so it must be a combination.  Quick experimentation with the exponents leads to $11=5*1+3*2$, so $2^1$ goes to a and $2^2$ goes to b.  This results in $a=3*2=6$ and $b=2^2=4$.

WHAT A STUDENT TAUGHT ME THIS YEAR

After my student, NB, arrived at $a^5b^3=497,664$ , she focused on roots–not factors–for her solution.  The exponents of a and b suggested using either a cubed or a fifth root.

The fifth root would extract only the value of a if b had only singleton factors–essentially isolating the a and b values–while the cubed root would extract a combination of a and b factors, leaving only excess a factors inside the radical.  Her investigation was simplified by the exact answers from her Nspire CAS software.

From the fifth root output, the irrational term had exponent 1/5, not the expected 3/5, so b must have had at least one prime factor with non-singular multiplicity.  But the cubed root played out perfectly.   The exponent–2/3–matched expectation, giving a=6, and the coefficient, 24, was the product of a and b, making b=4.  Clever.

EXTENSIONS & CONCLUSION

Admittedly, NB’s solution would have been complicated if the parameter was composed of something other than singleton prime factors, but it did present a fresh, alternative approach to what was becoming a comfortable problem for me.  I’m curious about exploring other arrangements of the parameters of $(Ax+By)^n$ to see how NB’s root-based reasoning could be extended and how it would compare to the factor solutions I used before.  I wonder which would be “easier” … whatever “easier” means.

As a ‘blog topic for another day, I’ve learned much by sharing this particular problem with several teachers over the years.  In particular, the initial “not enough information” feel of the problem statement actually indicates the presence of some variations that lead to multiple solutions.  If you think about it, NB’s root variation of the solution suggests some direct paths to such possible formulations.  As intriguing as the possibilities here are, I’ve never assigned such a variation of the problem to my students.

As I finish this post, I’m questioning why I haven’t yet taken advantage of these possibilities.  That will change. Until then, perhaps you can find some interesting or alternative approaches to the underlying systems of equations in this problem.  Can you create a variation that has multiple solutions?  Under what conditions would such a variation exist?  How many distinct solutions could a problem like this have?

## From a Square to Ratios to a System of Equations

Here’s another ratio problem from @Five_Triangles, this time involving triangle areas bounded by a square.

Don’t read further until you’ve tried this for yourself. It’s a fun problem that, at least from my experience, doesn’t end up where or how I thought it would.

INITIAL THOUGHTS

I see two big challenges here.

First, the missing location of point P is especially interesting, but is also likely to be quite vexing for many students. This led me to the first twist I found in the problem: the introduction of multiple variables and a coordinate system. Without some problem-solving experience, I don’t see that as an intuitive step for most middle school students. Please don’t interpret this as a knock on this problem, I’m simply agreeing with @Five_Triangle’s assessment that this problem is likely to be challenging for middle school students.

The second challenge I found emerged from the introduction the coordinate system: an underlying 2×2 system of equations. There are multiple ways to tackle a solution to a linear system, but this strikes me as yet another high hurdle for younger students.

Finally, I’m a bit surprised by my current brain block on multiple approaches for this problem. I suspect I’m blinded here by my algebraic bias in problem solving; surely there are approaches that don’t require this. I’d love to hear any other possibilities.

POINT P VARIES

Because I was given properties of point P and not its location, the easiest approach I could see was to position the square on the xy-plane with point B at the origin, $\overline{AB}$ along the y-axis, and $\overline{BC}$ along the x-axis. That gave my point P coordinates (x,y) for some unknown values of x & y.

The helpful part of this orientation is that the x & y coordinates of P are automatically the altitudes of $\Delta ABP$ and $\Delta BCP$, respectively. The altitudes of the other two triangles are determined through subtraction.

AREA RATIOS BECOME A LINEAR SYSTEM

From here, I used the given ratios to establish one equation in terms of x & y.

$\displaystyle \frac{\Delta ABP}{\Delta DAP} = \frac{\frac{1}{2}*12*x}{\frac{1}{2}*12*(12-y)} = \frac{3}{4}$

Of course, since all four triangles have the same base lengths, the given area ratios are arithmetically equivalent to corresponding height ratios.  I used that to write a second equation.

$\displaystyle \frac{\Delta BCP}{\Delta CDP} = \frac{y}{12-x} = \frac{1}{3}$

Simplifying terms and clearing denominators leads to $4x=36-3y$ and $3y=12-x$, respectively.

A VERY INTERESTING insight at this point is that there is an infinite number of locations within the square at which each ratio is true.  Specifically, the $\Delta ABP : \Delta DAP = 3:4$ ratio is true everywhere along the line 4x=36-3y.  This problem constrains us to only the points within the square with vertices (0,0), (12,0), (12,12), and (0,12), but setting that aside, anywhere along the line 4x=36-3y would satisfy the first constraint.  The same is true for the second line and constraint.

I think it would be very interesting for students to construct this on dynamic geometry software (e.g., GeoGebra or the TI-Nspire) and see the ratio remain constant everywhere along either line even though the triangle areas vary throughout.

Together, these lines form a 2×2 system of linear equations with the solution to both ratios being the intersection point of the two lines.  There are lots of ways to do this; I wonder how a typical 6th grader would tackle them. Assuming they have the algebraic expertise, I’d have work them by hand and confirm with a CAS.

The question asks for the area of $\Delta ABP = \frac{1}{2}*12*x = 6*8 = 48$.

PROBLEM VARIATIONS

Just two extensions this time.  Other suggestions are welcome.

1. What’s the ratio of the area of $\Delta BCP : \Delta DAP$ at the point P that satisfies both ratios??
It’s not 1:4 as an errant student might think from an errant application of the transitive property to the given ratios.  Can you show that it’s actually 1:8?
2. If a random point is chosen within the square, is that point more likely to satisfy the area ratio of $\Delta ABP : \Delta DAP$ or the ratio of $\Delta BCP : \Delta CDP$?
The first ratio is satisfied by the line 4x=36-3y which intersects the square on the segment between (9,0) and (0,12).  At the latter point, both triangles are degenerate with area 0.  The second ratio’s line intersects the square between (12,0) and (0,4).  As the first segment is longer (how would a middle schooler prove that?), it is more likely that a randomly chosen point would satisfy the $\Delta ABP : \Delta DAP$  ratio.  This would be a challenging probability problem, methinks.

FURTHER EXTENSIONS?

What other possibilities do you see either for a solution to the original problem or an extension?

## Chemistry, CAS, and Balancing Equations

Here’ s a cool application of linear equations I first encountered about 20 years ago working with chemistry colleague Penney Sconzo at my former school in Atlanta, GA.  Many students struggle early in their first chemistry classes with balancing equations.  Thinking about these as generalized systems of linear equations gives a universal approach to balancing chemical equations, including ionic equations.

This idea makes a brilliant connection if you teach algebra 2 students concurrently enrolled in chemistry, or vice versa.

FROM CHEMISTRY TO ALGEBRA

Consider burning ethanol.  The chemical combination of ethanol and oxygen, creating carbon dioxide and water:

$C_2H_6O+3O_2 \longrightarrow 2CO_2+3H_2O$     (1)

But what if you didn’t know that 1 molecule of ethanol combined with 3 molecules of oxygen gas to create 2 molecules of carbon dioxide and 3 molecules of water?  This specific set coefficients (or multiples of the set) exist for this reaction because of the Law of Conservation of Matter.  While elements may rearrange in a chemical reaction, they do not become something else.  So how do you determine the unknown coefficients of a generic chemical reaction?

Using the ethanol example, assume you started with

$wC_2H_6O+xO_2 \longrightarrow yCO_2+zH_2O$     (2)

for some unknown values of w, x, y, and z.  Conservation of Matter guarantees that the amount of carbon, hydrogen, and oxygen are the same before and after the reaction.  Tallying the amount of each element on each side of the equation gives three linear equations:

Carbon:  $2w=y$
Hydrogen:  $6w=2z$
Oxygen:  $w+2x=2y+z$

where the coefficients come from the subscripts within the compound notations.  As one example, the carbon subscript in ethanol ( $C_2H_6O$ ) is 2, indicating two carbon atoms in each ethanol molecule.  There must have been 2w carbon atoms in the w ethanol molecules.

This system of 3 equations in 4 variables won’t have a unique solution, but let’s see what my Nspire CAS says.  (NOTE:  On the TI-Nspire, you can solve for any one of the four variables.  Because the presence of more variables than equations makes the solution non-unique, some results may appear cleaner than others.  For me, w was more complicated than z, so I chose to use the z solution.)

All three equations have y in the numerator and denominators of 2.  The presence of the y indicates the expected non-unique solution.  But it also gives me the freedom to select any convenient value of y I want to use.  I’ll pick $y=2$ to simplify the fractions.  Plugging in gives me values for the other coefficients.

Substituting these into (2) above gives the original equation (1).

VARIABILITY EXISTS

Traditionally, chemists write these equations with the lowest possible natural number coefficients, but thinking of them as systems of linear equations makes another reality obvious.  If 1 molecule of ethanol combines with 3 molecules of hydrogen gas to make 2 molecules of carbon dioxide and 3 molecules of water, surely 10 molecule of ethanol combines with 30 molecules of hydrogen gas to make 20 molecules of carbon dioxide and 30 molecules of water (the result of substituting $y=20$ instead of the $y=2$ used above).

You could even let $y=1$ to get $z=\frac{3}{2}$, $w=\frac{1}{2}$, and $x=\frac{3}{2}$.  Shifting units, this could mean a half-mole of ethanol and 1.5 moles of hydrogen make a mole of carbon dioxide and 1.5 moles of water.  The point is, the ratios are constant.  A good lesson.

ANOTHER QUICK EXAMPLE:

Now let’s try a harder one to balance:  Reacting carbon monoxide and hydrogen gas to create octane and water.

$wCO + xH_2 \longrightarrow y C_8 H_{18} + z H_2 O$

Setting up equations for each element gives

Carbon:  $w=8y$
Oxygen:  $w=z$
Hydrogen:  $2x=18y+2z$

I could simplify the hydrogen equation, but that’s not required.  Solving this system of equations gives

Nice.  No fractions this time.  Using $y=1$ gives $w=8$, $x=17$, and $z=8$, or

$8CO + 17H_2 \longrightarrow C_8 H_{18} + 8H_2 O$

Simple.

EXTENSIONS TO IONIC EQUATIONS:

Now let’s balance an ionic equation with unknown coefficients a, b, c, d, e, and f:

$a Ba^{2+} + b OH^- + c H^- + d PO_4^{3-} \longrightarrow eH_2O + fBa_3(PO_4)_2$

In addition to writing equations for barium, oxygen, hydrogen, and phosphorus, Conservation of Charge allows me to write one more equation to reflect the balancing of charge in the reaction.

Barium:  $a = 3f$
Oxygen:  $b +4d = e+8f$
Hydrogen:  $b+c=2e$
Phosphorus:  $d=2f$
CHARGE (+/-):  $2a-b-c-3d=0$

Solving the system gives

Now that’s a curious result.  I’ll deal with the zeros in a moment.  Letting $d=2$ gives $f=1$ and $a=3$, indicating that 3 molecules of ionic barium combine with 2 molecules of ionic phosphate to create a single uncharged molecule of barium phosphate precipitate.

The zeros here indicate the presence of “spectator ions”.  Basically, the hydroxide and hydrogen ions on the left are in equal measure to the liquid water molecule on the right.  Since they are in equal measure, one solution is

$3Ba^{2+}+6OH^- +6H^-+2PO_4^{3-} \longrightarrow 6H_2O + Ba_3(PO_4)_2$

CONCLUSION:

You still need to understand chemistry and algebra to interpret the results, but combining algebra (and especially a CAS) makes it much easier to balance chemical equations and ionic chemical equations, particularly those with non-trivial solutions not easily found by inspection.

The minor connection between science (chemistry) and math (algebra) is nice.

As many others have noted, CAS enables you to keep your mind on the problem while avoiding getting lost in the algebra.

I’m teaching Algebra 2 this summer for my school.  In a recent test on quadratic functions, I gave a question I thought would be a little different, but still reachable for those willing to make connections or exert a little creativity.

Write a system of quadratic functions that has exactly one solution:  (1,1).

Their handheld graphing calculators were allowed.  Some students definitely had difficulty with the challenge, some gave a version of the answer I expected, and one adopted a form I knew was possible, but doubted anyone would actually find during a test situation.

I show my students’ solutions below.  But before you read on, can you give your own solution?

WHAT I EXPECTED

We’ve had many discussions in class about the power of the Rule of 4–that math ideas can be expressed numerically, graphically, algebraically, and verbally.  When you get stumped in one representation, being able to shift to a different form is often helpful.  That could mean a different algebraic representation, or a different Rule of 4 representation altogether.

The question is phrased verbally asking for an algebraic answer.  But it asks about a solution to a system of equations.  I hoped my students would recall that the graphical version of a system solution is equivalent to the point(s) where the graphs of the equations intersected.  In my mind, the easiest way to do this is to write quadratic functions with coincident vertices.  And this is most easily done in vertex form.  The cleanest answer I ever got to this question was

A graphical representation verifies the solution.

Another student recognized that if two parabolas shared a vertex, but had different “slopes”, their only possible point of intersection was exactly the one the question required.  Here’s a graphical version of her answer.

From these two, you can see that there is actually an infinite number of correct solutions.  And I was asking them for just one of these!  🙂

WHAT I KNEW, BUT DIDN’T EXPECT

Another way to solve this question makes use of the geometry of quadratic graphs.  If two quadratics have the same leading coefficients, they are the same graph, intersect exactly once, or never intersect.  This is a very non-trivial idea for most students.  While I’m not convinced the author of the following solution had this in mind when he answered the question, his solution works because of that fact.  Here’s what J wrote on last week’s test and its graph.

J used more equations than he needed, but had he restricted himself to just two equations, I’m not sure the lovely pattern would have been so obvious.

This is a very different (and super cool) answer than what I expected my students to produce.  Lesson re-learned:  Challenge your students, give them room to express creativity and individuality, and be prepared to be amazed by them.

NEXT STEPS

J’s answer actually opens the door to other avenues of exploration.

1. Can you generalize the form of all of J’s equations, essentially defining a family of quadratics?  Can you prove that all members of your generalization satisfy the question posed and that no other answers are possible?
2. Can you find forms of other generalized families of quadratic functions whose only solution is (1,1)?
3. Notice that there were two types of solutions above:  A) those with coincident vertices and different lead coefficients and B) those with identical lead coefficients and different vertices.  Are these the only types of quadratics that can answer this question?  That is, is there a system of quadratics with (1,1) as the only solution that have identical vertices and lead coefficients?  Could both be different and (1,1) be the only solution?
4. If I relax the requirement that the quadratics be functions, what other types of quadratics are possible?  [This could be a very nice calculus question!]

For my part, I’m returning to some of these questions this week to stretch and explore my student’s creativity and problem-solving.

I’d love to hear what you or your students discover.

## Circle and Square

Here’s another great geometry + algebra problem, posed by Megan Schmidt and pitched by Justin Aion to some students in his Geometry class.

Following is the problem as Justin posed it yesterday.

Justin described the efforts of three of his students’ on his his ‘blog.  Following is my more generalized approach.  Don’t read further if you want to solve this problem for yourself!

My first instinct in any case like this is build it in a dynamic geometry package and play.  Using my TI-Nspire, without loss of generality, I graphed a circle centered at the origin, constructed a tangent segment at the bottom of the circle centered on the y-axis, and then used that segment to construct a square.  I recognized that the locus of the upper right corners of all such squares would form a line.

That made it clear to me that for any circle, there was a unique square that intersected the circle three times as Megan had posed.

Seeing this and revealing its algebraic bias, my mind conceived an algebraic solution.  Assuming the radius of the circle is R, the equation of my circle is $x^2+y^2=R^2$ making the lower y-intercept of the circle $(0,-R)$.  That made $y=2x-R$ the locus line containing the upper right corner of the square.

To find generic coordinates of the upper right corner of the square in terms of R, I just needed to solve the system of equations containing the circle and the line.  That’s easy enough to compute by hand if you can handle quadratic algebra.  That manipulation is not relevant right now, so my Nspire CAS’s version is:

The output confirms the two intersections are $(0,-R)$ and the unknown at $\displaystyle \left( \frac{4R}{5} , \frac{3R}{5} \right)$.

Because of the horizontal symmetry of the square with respect to the y-axis, the system solution shows that the generic length of the side of the square is $\displaystyle 2\left( \frac{4R}{5} \right) = \frac{8R}{5}$.  The circle’s y-intercept at $(0,-R)$ means the generic diameter of the circle is $2R$.

Therefore, the generic ratio of the circle’s diameter to the square’s side length is

$\displaystyle \frac{diameter}{side} = \frac{2R}{(8R)/5} = \frac{5}{4}$.

And this is independent of the circle’s radius!  The diameter of the circle is always $\frac{5}{4}$ of the square’s side.

CONCLUSION:

For Megan’s particular case with a side length of 20, that gives a circle diameter of 25, confirming Justin’s students’ solution.

Does anyone have a different approach?  I’m happy to compile and share all I get.

AN ASIDE:

While not necessary for the generalized solution, it was fun to see a 3-4-5 right triangle randomly appear in Quadrant 1.

## Systems of lines

Here’s an interesting variation of a typical (MS) problem I found by following the Five Triangles ‘blog: http://fivetriangles.blogspot.com/2013/09/97-no-triangle.html .

(Note:  If you sign up on this or other ‘blogs, you can get lots of problems emailed to you every time they are added.)

INITIAL SOLUTION

I know this question can absolutely be solved without using technology, but when a colleague asked if it was appropriate to use technology here (my school is one-to-one with tablet laptops), I thought it would be cool to share with her the ease and power of Desmos.  You can enter the equations from the problem exactly as given (no need to solve for y), or you can set up a graph in advance for your students and email them a direct link to an already-started problem.

If you follow this link, you can see how I used a slider (a crazy-simple addition on Desmos) to help students discover the missing value of a.

FOLLOW-UP

I suggest in this case that playing with this problem graphically would grant insight for many students into the critical role (for this problem) of the intersection point of the two explicitly defined lines.  With or without technology support, you could then lead your students to determine the coordinates of that intersection point and thereby the value of a.

Keeping with my CAS theme, you could determine those coordinates using GeoGebra’s brand new CAS View:

Substituting the now known values of x and y into the last equation in the problem gives the desired value of a.

NOTE:  I could have done the sliders in GeoGebra, too, but I wanted to show off the ease of my two favorite (and free!) online math tools.

CONCLUSION

Thoughts?  What other ideas or problems could be enhanced by a properly balanced use of technology?

As an extension to this particular problem, I’m now wondering about the area of triangle formed for any value of a.  I haven’t played with it yet, but it looks potentially interesting.  I see both tech and non-tech ways to approach it.

## Non-Calculus approach to Invariable Calculus Project

I shared my posts (here and here) on the Invariable Calculus Project in the AP Calculus Community.  Gary Litvin posted a response within the Community offering there a great non-calculus alternative solution to the original problem of the area of the triangle formed by the x- and y-axes and any Quadrant I tangent line to $\displaystyle \frac{1}{x}$.  Here’s a paraphrase of Gary’s approach.

Let $\displaystyle \frac{x}{a} + \frac{y}{b}=1$ be any Quadrant I tangent line to $\displaystyle y=\frac{1}{x}$.  (In case you don’t recognize it, this tangent equation uses the intercept form of a line–a is the x-intercept and b is the y-intercept.)  Because the line intersects the parabola in a single point, we can find that point by solving the system of equations defined by the two equations.  Substituting for y gives

$\displaystyle \frac{x}{a} + \frac{\frac{1}{x}}{b}=1$.

This is equivalent to $x^2 - a \cdot x+\frac{a}{b}=0$, a quadratic.  We could determine the value of x using the quadratic formula.  Because there is only one solution to this equation (there is only one point of intersection, the point of tangency), the discriminant must be zero.  That means

$\displaystyle (-a)^2 - 4\cdot \left( \frac{a}{b} \right)=0$

which can be rearranged to give $ab=4$ ($a=0$ is extraneous).  Therefore, the area of the triangle formed by the tangent line to $\displaystyle y=\frac{1}{x}$ and the coordinate axes is $\displaystyle Area=\frac{1}{2} ab=2$ no matter what the point of tangency.

Shiny.