Maths - Investigate how many people can be carried in each type of vessel.

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Maths Coursework

By Jeffrey Li

The teacher has presented me with a mathematical problem in which I must solve and investigate. It is as follows:

  • The ‘Great Outdoors Boating Company’ has 20 large canoes, 5 small canoes and 10 motorboats for hire.
  • The ‘New Canoe Company’ has 10 large canoes, 15 small canoes and 6 motorboats for hire.
  • The ‘Ship-u-Like Company’ has 8 large canoes, 4 small canoes and 2 motorboats for hire.
  • The ‘Great Outdoors Boating Company’ can carry a maximum of 225 people.
  • The ‘New Canoe Company’ can carry a maximum of 179 people.
  • The ‘Ship-u-Like Company’ can carry a maximum of 92 people.

Investigate how many people can be carried in each type of vessel.

When we first consult this problem, the first thing to look at is what the problem tells us and what I can determine from it. It is easy to see that it is a problem which contains three undetermined variables. Therefore, I am quite convinced that the route to solving this certain problem lies in the depths of Algebra. However, in order to solve algebra; we must take the numbers from the realistic world into the abstract world of mathematics. Hence, form equations from the problem. Firstly, I will let large canoes be called “x”, let small canoes be called “y” and let motorboats be called “z”. We can then create a set of three equations with this information:

20x + 5y + 10z = 225

10x + 15y + 6z = 179

8x + 4y +2z = 92

Of all the different ways in which I am familiar with in terms of solving simultaneous equations, there are:

  • Elimination
  • Substitution
  • Trial and Improvement
  • Graphical Solution
  • Matrices

Of these possible methods, I believe that I will use a combination of both elimination and substitution to solve this problem. This is because I believe that in order to get the best possible result in the fastest possible manner, one must utilise as many techniques in effective combination as available. This technique was taught to me in Year 8 by a mathematics teacher and I plan to make full use of it.

Elimination and Substitution

I will begin this problem by simplifying all three equations, thus make life far easier for myself.

(i) 20x + 5y +10z = 225 => 4x + y +2z = 45 (divided by 5)

(ii) 10x + 15y + 6z = 179 => (cannot be simplified)

(iii) 8x + 4y + 2z = 92 => 4x + 2y + z = 92 (divided by 2)

After doing so, I will then subtract (i) from (iii) since they both have a “4x” factor in it, hence I can eliminate the factor of “x”. This as a result gives me an equation of:

4x + y + 2z = 45

4x + 2y +z = 92

y – z = 1 (I will call this equation (iv))

Now that I have formed one equation with “y” and “z”, I now need to form another on using (i) and (ii). I will multiply (ii) by 2 in order to make the factor of ‘10x’ common to both equations and then subtract and eliminate.

20x + 5y + 10z = 225

10x + 15y + 6z = 179 => 20x + 5y + 10z = 225

25y + 2z = 133 (I will call this equation (v))

We now take the newly formed equations (iv) and (v) and we will now eliminate one of the factors from them.  

25y + 2z = 133

X2)                y – z = 1 =>        2y – 2z = 2

                                27y = 135

                                y = 5

Now that we know one of the factors, we will now start to utilise the technique of substitution in our problem. We will begin by taking one of the equations that was formed earlier (equation (iv)) and will substitute y = 5 into the equation.

y – z = 1

5 – z = 1

-z = -4

z = 4

Now we have already solved two variables of the equation, since y and z have been solved, we can substitute their values into and of the equations we’ve used before in order to solve ‘x’.

4x + y + 2z = 45

4x + 5 + 8 = 45 => 4x = 32

x = 8

Now, the equation has been solved and we can complete our work by proving these. We will substitute them back into the equations that we began with.

Join now!

(20 X 8) + (5 X 5) + (10 X 4) = 225

(10 X 8) + (15 X 5) + (6 X 4) = 179

(8 X 8) + (4 X 5) + (2 X 4) = 92

Thankfully, the numbers substitute perfectly back into the equation and as a result, we can conclude that our method has worked and thus:

x = 8

y = 5

z = 4

Matrices

After solving the mathematical problem using one method, it is best to solve it via another method in order to enhance one’s mathematical ...

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