[S2_20151107SLFC] Simultaneous Linear Equations with Fractional Coefficients

Question 

Introduction

     Somebody said, “Dear Algebra, please stop asking me about your  x.  She is not coming back.  Don’t ask me  y!”

     Here we have here linear equations that appear to be more tricky than the usual fanfare.  This is because the coefficients of the unknowns  x  and  y  are fractions.  In school, students learn to solve these using the method of substitution and the method of elimination.  They tend to prefer to do it by the former method, because psychologically it seems easier to accept.  However, the latter matter is generally more effective and yields a shorter solution.  This question is like a fly trap for those who prefer the method of substitution.  You make either  x  or  y  the subject from one of the equations, and then substitute that into the other equation.  This yields a complicated algebraic fractions within algebraic fractions.  The more complicated your solution is, the higher chance there is for making careless mistakes.  It is a good idea that students get out of their comfort zones and adopt a new skill.  So how do we do it?

The smart tactic

     First, we need to clear away the fractions by multiplying through with the LCM of the denominators appearing in each equation.  For the first equation, LCM(4, 8) = 8.  Since  4  is a factor of  8,  8  is like a giant that absorbs the number  4.  OK, so we multiply the first equation through by  8.  For the second equation, we multiply every term by LCM (3, 2) = 6

Solution

Remarks

     As you can see, equations [3] and [4] both contain – 3y  and this can be eliminated via subtraction.  So the value of  x  comes out easily.  Now once  x  is known, one can find the  y!

For another example of simultaneous equations, please refer to this article.

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H10. Simplify the problem

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* Lower Secondary Mathematics (Secondary 2)

* GCE ‘O’ Level “Elementary” Mathematics (revision)

* other syllabuses that involve algebra, linear simultaneous equations

[U_20151029ITCX] Roger Cotes’ Integral of the Reciprocal of x^n – 1

Question

 

Introduction

     The book VisualComplex Analysis by Tristan Needham recounts the story of RogerCotes who considered the above problem.  Without ostensibly using complex numbers, Cotes discovered a geometrical principle that helped to factorise the denominator  xⁿ – 1,  and hence decompose the above integral.

     In this article, I am going to “cheat” by using complex numbers to split up the denominator.  The fact that the denominator splits completely into a product of simple linear factors makes it easy to decompose the integrand into partial fractions.  Once this is done, I can single out the one or two fractions with purely real linear denominators.  Then I can pair up the conjugate fractions to get fractions with real quadratic denominators.  In other words, I apply a divide-and-conquer strategy, splitting up a big problem into smaller problems (Heuristics!).  Then I collect all the partial answers together to form my final answer.

Solution

Remarks

     Note that I have only used real integration, not complex integration.  Complex numbers are used only to derive the various algebraic fractions.

HeuristicsUsed

H02. Use a diagram / model   (mentally: imagine roots of unity in a circle)

H04. Look for pattern(s)

H10. Simplify the problem

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* University / college level calculus

* high school students very strong and interested in calculus and complex numbers

* anybody who loves a challenging calculus problem and complex numbers

[S2_20150520SERC] Simultaneous Equations with Reciprocals

Question

Introduction

     This is a secondary 2 (» grade 8) problem that involves linear simultaneous equations in two unknown.  It looks non-linear and indeed if you try to solve for  x  and  y  directly,  you might get into a big mess with extraneous solutions to boot.  What should our approach be?

Observations

     It is always a good idea to stare at the question for a little while longer before jumping in to try to solve it.  Let us (mentally) reformat the equations a little bit.

     Do you notice any repeated chunks?  Chunking is very useful.  Let use substitutions for those chunks to simplify the equations.

Solution

Final Remarks

     To recap: It is a good learn to make observations first before attempt to solve the problem.  If you see any repeated chunks, it is a good idea to use substitution to simplify the problem.  Once the reciprocals have been substituted, we try to use elimination (which is usually the more effective method).  Also, try as far as possible to avoid fractions.  By multiplying equation [2] by 5,  we get – 30Y, which can be cancelled if we multiply  15Y by 2.  Thus  Y  is eliminated.  The problem becomes easy from this point onwards.
     For another example of simultaneous equations, please refer to thisarticle.

HeuristicsUsed

H04. Look for pattern(s)

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* Lower Secondary Mathematics (Secondary 1)

* GCE ‘O’ Level “Elementary” Mathematics (revision)

* other syllabuses that involve algebra and simultaneous linear equations

[S1_20150516AELI] Apples and Pears on a Table?

Question

Michael has enough money to buy either 12 pears or 36 apples. if he intends to buy equal number of pears and apple, how many of each fruit can he buy with the money

Introduction

     This is an interesting algebra problem meant for secondary 1 students (» grade 7) that has the potential to lead to a system of complicated simultaneous equations in two variables.  Fortunately, there are some interesting and simpler approaches.  I present three of them below.

Solution 1 (using a Table and Unit Costs)

Solution 2 (Insight from proportion)

     This is perhaps the intended algebraic approach.  Observe that since  12 pears are as expensive as  36  apples,  1  pear is “equivalent” to  3  apples.  Michael’s budget is  36  apples-worth of money.  If there are  n  pears and  n  apples,  the  n  pears can be exchanged for  3n  apples.  We arrive at an equation as in solution  1.

  

Solution 3 (Acting it Out)

     You can role-play this with your friend using toy-fruits.  Your friend is the fruit-seller and you are the buyer.  No toy-fruits?  Well, use some counters, bottle caps, Lego bricks, ... whatever to represent the apples and pears.  If you are a lonely person, or if your friend is too busy, or your mother has thrown away all your toys since you are (sort of) more grown up already, maybe just do a thought experiment.  Imagine, at first, you took  36  apples to the check-out counter.  Then you saw some pears, and you grabbed  12 of them, ditching the apples.  You figure out that  1  pear is equivalent to  3  apples.  Just then, you change your mind.  You decide that you want an equal number of apple and pears.  OK, so you replace  1  pear with  3 apples.  You get  11  pears,  3  apples.  Keep on swapping pears for apples.   You get  10  pears,  6  apples.  Then  9  pears,  9 apples.  Bingo!

Final Remarks

     There is no magic potion for mathematics (although heuristics is a start).  There is also no one fixed method for you to memorise to solve problems.  As they say, you have more than one way to skin the cat.

Heuristics Used

H01. Act it out

H02. Use a diagram / model

H03. Make a systematic list

H04. Look for pattern(s)

H05. Work backwards

H06. Use before-after concept

H07. Use guess and check

H08. Make suppositions

H09. Restate the problem in another way

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* Lower Secondary Mathematics (Secondary 1)

* GCE ‘O’ Level “Elementary” Mathematics (revision)

* other syllabuses that involve ratios or algebra

[IBHL_SOTA201304_1B10c] Quadratic Discriminants

Question

Important Reminders

Solution

Suitable Levels

* GCE ‘O’ Level Additional Mathematics

*  IB Mathematics HL / SL

* other syllabuses that involve quadratics

[S2_20150402RLD] Fifty balls left behind

Question

This problem can be solved with Primary School knowledge using ratios.  The famous Singapore bar diagramming method can be used to model the situation, but I prefer my own Distinguished Ratio Units.  The former method is good for visualisation for beginners, while the latter is faster if you want to solve it quickly without fussing around drawing the perfect diagram.  My DRU method is also visual in another way, and it works with big numbers as well as small numbers.  Alternatively, this can be solved using algebra via simultaneous equations.

Solution 1 (Using my Distinguished Ratio Units method) [H02]

Explanation: Since the number of white balls is a multiple of  3,  I let “triangle” 3 represent the number of white balls.  I let “heart” 1 represent the number of red balls.  There are 50 more white balls than red balls.  [H04]  When the white balls are removed three at a time, the number of groups of three would be one-third of the number of white balls, i.e. 1 triangle unit.  [H04]  This number is less than the number of red balls (1 heart unit) by 50.  So 1 triangle unit plus 50 gives 1 heart unit.  [H04]  Following on from the heart to the “triangle” 3, one realises that 2 “triangle” units is the same as  100.  [H05]  So one triangle unit is  50.  [H11]  From here we can solve the rest of the problem.

Solution 2 (using Algebra)  [H13, H05]

                     w = k + 50 = 3h             –––––––––– [1]

                     r  = k         =   h + 50     –––––––––– [2]

for some unknown  k  and  h.  And then [1] – [2] gives  [H10]

                     w – r = 50 = 2h – 50

so                      2h = 50 + 50

                            h = 50

This quickly leads to

                           w = 150

and                       r = 100.

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* Primary School Mathematics (“Ratio”)

* Lower Secondary School (“Simultaneous Linear Equations”)

* other syllabuses that involve ratio or algebra