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Equation Solving
Hectare is the eighth in a series of fifteen puzzle caches of increasing difficulty based on solving equations. It is recommended that one start with Abacus (GC6A006) and continue in alphabetical sequence through Brackets (GC6AR81), Compass (GC6AWD0), Decagon (GC6C4EE), Ellipse (GC6CFW3), FOIL (GC6HDNE), Googol (GC751JK), Hectare (GC75BD3), Index (GC75EAM), Jargon (GC75X8W), Kite (GC76010), LessThan (GC763HV ), Median (GC7DH17), Nonagon (GC7DWC4) and Obelus (GC7DWCD).
The Name - Hectare - Irrelevant for this Puzzle
A “hectare” is the metric “equivalent” to the “acre” - both measurements of area usually applied to land or small bodies of water. We put quotation marks around “equivalent” because the hectare is a much larger area: there are about 2.47 acres in a hectare.
Formally defined, a hectare is 10 000 m2 most easily visualized, probably, as a square of 100 m per side. Although Canadians have gradually adapted to many of the metric measurements (well, we’re 45 years into the official general use of “the metric system” now!), the “hectare” seems to be one of the least used. One reason for this is that in the pioneering days in pre-Canada, grants were frequently made of “quarter sections” of land. A “quarter section” measured half a mile by half a mile - i.e. 880 yards by 880 yards or 774 400 sq. yd. That sort of area was easy to fit into a grid of roads which were a mile (or multiples thereof) apart. For example, if two north-south roads were one mile apart and two east-west roads were a mile and half apart, then six quarter sections just fitted into the space.
Since an acre was defined as 1 chain (66') by 1 furlong (660'), a little arithmetic will show that a quarter section was comprised of 160 acres. Our farms were measured out in this way and that fact helps to perpetuate the use of the “acre” as a useful measurement of land although more and more now, tracts such as provincial parks and conservation areas are described as having areas measured in hectares.
So much for that interesting tid-bit!
Hectare - the Cache
The point of this cache is that you can do anything you want to one side of an equation . . . provided that you do the same thing to the other side at the same time and this allows us to solve equations which are impossible to solve “by inspection” as we have done so far!
To Show It Works!
The use of the questions marks in the middle of the following statements means that we’re not sure yet (we’re just checking it out) that the two sides are actually equal.
We can all accept the “equation” that 15 = 15.
Now, let us add a number, say 7, to each side of that:
15 + 7 ? 15 + 7
or
22 = 22 . . . hey, that worked!
Now let us subtract something from each side of that original “equation:”
15 - 8 ? 15 - 8
or
7 = 7 . . . and that worked too!
Could we multiply both sides by 2, say, and get similar results? Let’s try:
15 X 2 ? 15 X 2
or
30 = 30 . . . so that works alright!
Similarly, let’s divide each side by 5:
15 ÷ 5 ? 15 ÷ 5
and
3 = 3 . . . which we can all accept!
The important thing to notice is that we can do any of those four operations to both sides of an equation and get a statement that is true - another equation - but we must be careful to follow the other rules of mathematics in so doing.
Another Illustration
Let us consider multiplying both sides of the following by 3 (although there is no actual reason to do so except to illustrate my point):
2x - 5 = x + 2
If you substitute (insert) a 7 in place of the x, you will see that the statement is true because you get 9 = 9; that means that the the solution to this equation is, in fact, that x = 7.
Let us multiply both sides by 3:
3(2x - 5) = 3(x + 2) (note the brackets to ensure that all of each side gets multiplied by the 2.
The result is:
6x - 15 = 3x + 6.
Again, test to see if substituting 7 for x gives us a true statement . . . and it does - i.e. 27 = 27.
So this demonstrates, in an actual algebraic (oh, scary word!) equation, that it is legitimate to multiply both sides of the equation by a number and the value of x is not affected. And this is true - trust me - of the other three operations we tried above - addition, subtraction and division.
Why Would You Want to Do This Anyway?
Well, to make life easier when life means solving equations! To illustrate this, let us look at a real little problem.
A “Problem”
“I have a number which, when 8 is subtracted from it, then that result is divided by 4 and then that second result is multiplied by 5, yields the answer 20; what is my number?”
Forming an Equation from the Problem
First of all, we use a time-honoured phrase, “Let x [or any other letter] represent [take the place of] the number [or whatever it is we are trying to find].”
Then, we gradually translate the statement from words to mathematical symbols:
(x - 8) . . . “number which, when 8 is subtracted from it”
(x - 8)/4 . . . “then that result is divided by 4"
5(x - 8)/4 . . . “then that second result is multiplied by 5"
5(x - 8)/4 = 20 . . . “yields the answer 20.”
If you write that on paper, you would put a horizontal line under the (x - 8) and the 4 under it to indicate the division by 4. Unfortunately, we do not know how to do that here. So, do that and show the rest of the equation and let’s see what comes next!
Solving the Equation
Now look at what you have written on the page in front of you and try to imagine solving that “by inspection” as we have done in previous caches in this series - that is, by looking at it and, more or less by trial and error, coming up with an answer. It looks a bit tricky as, indeed, it is . . . and this is a relatively simple equation as equations go. Before long, we’ll be solving equations much more difficult than this! But from what we learned above, we can solve it with no difficulty.
Remember that what we want to end up with is:
x = something . . . right?
So, what we do is essentially get rid of all the other things that are in the way of making that simple statement. Let’s start by getting rid of that fraction!
You will remember from the dim, perhaps distant, past that if you have a number in the numerator (top) of a fraction and the same one in the denominator (bottom) you can “cancel them out.” So, we are going to get rid of that 4 in the denominator by multiplying the left side of the equation by 4 . . . and that means, of course, that we have to do the same thing to the right hand side at the same time. So, let’s do it:
(4)(5)(x - 8)/4 = (20)(4)
It’s hard to see here, but on paper you will see that the two 4s on the left hand side of the equation can be cancelled out because one is in the numerator and the other in the denominator . . . and at the same time we can multiply by the 4 on the right hand side . . . and that results in:
5(x - 8) = 80
Now multiply the 5 by what’s in the brackets - a math teacher would tell you to “expand.”
5x - 40 = 80
Remember, we want “x = something” and that - 40 on the left hand side is a nuisance . . . so let’s get rid of it by making it zero . . . and to get to zero from - 40, you just add 40! But we have to do that to both sides:
5x - 40 + 40 = 80 + 40
or, once simplified,
5x = 120
Now, we’re almost there and you could do that easily “by inspection” now, but let’s complete it in this new more formal way. We want “x = something” so we don’t want that 5 on the left hand side. Remember that “5x” means (5)(x) or 5 X x, so, we can eliminate the 5 by dividing . . . but on both sides of the equation remember!
5x ÷ 5 = 120 ÷ 5 and when that is simplified, we have:
x = 24 . . . and you can check that by replacing the x in the original equation with the 24 and you will find that you have a true statement!
And you have now “solved an equation” - perhaps for the first time in a long time! We hope you understand the idea because we will be doing that in future parts of this series.
Some Other Equations for You to Work Through
(1)
2x + 14 = 154
We don’t want the (+ 14) on the left side so we add (- 14) to make it zero but we do it to the right side too:
2x + 14 - 14 = 154 - 14
Simplify:
2x = 140
We want x, not 2x, so we divide (both sides, remember!) by 2:
2x/2 = 140/2
Simplify:
x = 70
(2)
30 - 3x = 16 - x
We don’t want 30 on the left side nor the (- x) on the right side, so we add (- 30) and (+ x) to both sides:
30 - 3x - 30 + x = 16 - x - 30 + x
Simplify:
- 2x = - 14
We want x, not - 2x so we divide both sides by (- 2):
- 2x/- 2 = - 14/- 2
And we get:
x = 7
(3)
3(20 - x) + 5x = 84
”Expand” the 3(20 - x)
60 - 3x + 5x = 84
Simplify:
60 + 2x = 84
We don’t want 60 on the left side so we add (- 60) to both sides:
60 + 2x - 60 = 84 - 60
Simplify: 2x = 24
Divide both sides by 2 and the result is
x = 12
(4) . . . just one more!!
2x - (5x - 4)/6 = 7 - (1 - 2x)/5
Please write that on paper showing the 6 and the 5 each below a horizontal line underneath (5x - 4) and (1-2x) respectively - sorry I am incapable of doing that here, I’m afraid - and you will see four terms: 2x, (5x - 4) over 6, 7 and (1 - 2x) over 5. We don’t want the denominators so we multiply each term by the smallest (only for convenience as it doesn’t have to be “the smallest”) number into which both the denominators divide evenly - in this case, 30.
(30)(2x) - (30)(5x - 4)/6 = (30)(7) - (30)(1 - 2x)/5
The denominators will “divide out” and you will have:
(30)(2x) - 5(5x - 4) = (30)(7) - 6(1 - 2x)
Expand:
60x - 25x + 20 = 210 - 6 + 12x
Simplify:
35x + 20 = 204 + 12x
Get rid of the 20 on the left and the 12x on the right as we did before:
35x + 20 - 12 x - 20 = 204 + 12x - 12 x - 20
Simplify:
23x = 184
Divide both sides by 23 and
x = 8
Remember that in any of these equations - or any other - you can substitute for x in the original equation the value you got for your answer and you will find that the equals sign holds true. If it does not, then there’s an error - either in your equation solution or in your checking process!
And Now . . . You’re on Your Own!
Solve Each Equation for the Unknown
(1)
4x + 16 = 188 (Record your answer as A B = ___ ___)
(2)
3p - 4 = p + 42 (Record your answer as C D = ___ ___)
(3)
m - 52 = - 3m + 88 (Record your answer as E F = ___ ___)
(4)
5x/2 - 5x/4 = 9/4 - (3 - x)/2 (Use horizontal lines where I have a “/” and record your answer as G = ___)
(5)
(60 - x)/7 = x/8 (Use horizontal lines as in (4) and record your answer as H J = ___ ___)
And Now, Calculate the Co-ordinates
A = ___ , B = ___ , C = ___ , D = ___ , E = ___ ,
F = ___ , G = ___ , H ___ , J = ___ .
The co-ordinates of the cache are N 44̊ ab.cde’ and W 078̊ fg.hij’ where none of the lower case letters is related to the upper case letters above except as defined below:
a = F - B = ___ ; b = (C)(J) = ___ ; c = A - G = ___ ; d = E2 - G = ___ ;
e = H = ___ ; f = 3D - F = ___ ; g = H + C = ___ ; h = (D)(H) = ___;
i = F - (E + J) = ___ ; j = A ÷ (B - C) = ___ .
You Will Find the Cache at:
N 44̊ __ __ . __ __ __’ and W 078̊ __ __ . __ __ __’
Additional Comments
- Please bring your own writing utensil;
- Note the parking areas provided - one in the “winter” months, three more in the summer months;
- There is a parking machine near the permanent parking - $4 for the day; however one might wish first to visit the office - 80 m away - and purchase an annual pass to all "Kawartha Conservation conservation areas" for $84.75 - 20% less for seniors;
- Please stay on the trails until you are as close as you can get to GZ without leaving them;
- Be aware of the possibility of others on the trail;
- Hours of operation in this area: May to October: 07:00 - 21:00; November to April: 08:00 - 18:00;
- Please feel free to confirm your answer on GeoChecker.com.