Algebra Part One |

Algebra! There's I've said it, and I'm going to write about it!

Get ready for fun, excitement and interesting puzzles. No i promise you, this is mathematics, but not as you learned it at school.

I’ve never really understood all the hysterical confusion about algebra. People wail that they dont understand what all these x’s and y’s mean and that they couldn’t possibly being to comprehend how to manipulate equations. I’ve no doubt that some people genuinely find it difficult – but I often wonder if it’s a combination of fear and bad teaching rather than any lack of ability that generates this fear of x's and y's.

The thing is, I find algebra to a thing of some beauty; the ability to reduce a complicated problem down to a single concise statement and then logically deduce a result is a puzzling pastime that genuinely fills me with joy. In puzzle solving that's what algebra is for, simplifying problems. Everything we do to an algebraic equation has a perfectly sound basis in common sense.

Before we get started some notes about notation. We use x’s, y’s and z’s in algebra for two reasons. First of all a very skilled mathematician called Rene Descarte decided we should – and since he more or less invented the subject it only fair to follow his lead.

The other reason is a little more practical, the x’s and y’s in an equation are called variables and you can put different numbers in place of the x and the y as long as you make the overall equation add up correctly. These pairs of values for x and y can be plotted onto a graph, matching up the x values with the x coordinate and the y values with the y coordinate. Each equation therefore has a graph, a way of visualising how one value changes with the other.

There are all kinds of properties of equations that can help you find a suitable solution to a problem. The simplest of these boil down to one thing. If you do something to one part of an equation, you must do it to all parts.

x + y = 4 is exactly the same as 2x + 2y = 8, both sides of the equation were multiplied by two, and all elements on each side were treated the same way.

x + y = 4 is also the same as x + y + z = 4 + z, this time the same element was added to each side of the equation. Think of it as an old fashioned set of balance scales, if you double one side, you must double the other to keep it balanced. If you add something to one side you must add exactly the same to the other side to keep it balanced.

No doubt in your time at school you came across the dreaded simultaneous equations. Pairs of equations for which, for some specific value of x and y, both equations work. Now of course you can see that those specific values, are the points where the graphs of the two equations cross. To solve a pair of simultaneous equations you can always just draw the graph and note down the answer; but there are other simpler ways. Take a look at this example where we first create equations, then manipulate them.

No doubt in your time at school you came across the dreaded simultaneous equations. Pairs of equations for which, for some specific value of x and y, both equations work. Now of course you can see that those specific values, are the points where the graphs of the two equations cross. To solve a pair of simultaneous equations you can always just draw the graph and note down the answer; but there are other simpler ways. Take a look at this example where we first create equations, then manipulate them.

**A Coffee Conundrum**I was sitting on a bench observing a coffee bar the other day and trying to work out if I had enough change for a cappuccino. I’m not a big coffee drinker and I’m really shy so there was no way I was going to look at the prices and risk having to walk away.

A couple came up to the counter and ordered one cappuccino and one latte; it came to £5. I baulked, I only had a pound coin and some other bits of change, certainly not £2.50!

The next customer was a harassed looking office type with a list. He wanted four cappuccinos and a pair of lattes. That came to £14, which did make me wonder whether the office was footing the bill. It also made it clear that it wasn’t just £2.50 a coffee, as six times £2.50 is £15 not £14.

All this being said, I now had enough information to work out the cost of my prospective cappuccino.

I effectively had a pair of simultaneous equations; two different equations that will work only for specific values – in this case the costs of a cappuccino and a latte.

So, let’s use the classic X’s and Y’s, which, apart from tradition, are much easier to spell than cappuccino and latte, to write down our information.

Let x = the cost of a single cappuccino, let y = the cost of a single latte.

One cappuccino and one latte came to £5 so:

1. x + y = 5

Four cappuccinos and two lattes came to £14 so:

2. 4x + 2y = 14

It doesn’t matter any more what x and y mean, we can look that up at the end and just work with the nice short letters.

Instead of writing x + x + x + x + y + y; we condense them down to 4x and 2y; grouping the x’s and the y’s will help us later.

Now what we need to do is find either the value of just x or y; from that we can then work out the other one, so somehow we need to eliminate either x or y from those equations. We cant just get rid of them however, the equations still have to remain mathematically sound.

As long as the equations stay in the same proportions we can multiply and divide them by numbers. In practical terms this means we must multiply each of the individual terms, the x’s, the y’s and the numbers by the same number each time.

Let’s multiply equation number 1 by four:

1. x + y = 5 multiplied by 4 equals

3. 4x + 4y = 20

Now we can use another key rule of simultaneous equations; you can add them and subtract them, as long as you do it to all the different terms x’s, y’s and numbers at once. That’s one whole equation, plus or minus another whole equation.

3. 4x + 4y = 20 minus

2. 4x + 2y = 14 equals

4. 0x + 2y = 6

Hold on a minute, ‘0x’ isn’t that just the same as saying zero multiplied by x? Anything multiplied by zero is zero – so we can get rid of those x’s all together from equation 4!

4. 2y = 6

That's a nice simple equation, we can divide both sides by two to get the first half of our answer.

5. y = 3

So y equals 3; we can plug that number into any of our previous equations with x’s and y’s involved to find out what x is. Let’s use equation 1

1. x + y = 5 y is equal to 3 so

6. x + 3 = 5 subtract 3 from both sides

7. x = 2

Finally we go back to our original statements, x is the cost of a cappuccino and y is the cost of a latte so the solution is.

Cappuccino £2

Latte £3

As it turned out I only had £1.95 in change so I bought I diet coke instead.

So whenever you come across a puzzle that looks like it might be simultaneous equations; grab a pad and paper and just work methodically.

1. Identify the equations

2. State what your variables are what what symbols you will use.

3. Get an equation which has just one term and a number by either

a. Adding or subtracting whole equations

b. Multiplying or dividing by whole numbers

c. Adding or subtracting whole numbers or terms.

4. State the solution in terms of the variable symbols.

5. State the solution in terms if the variables themselves.

5. y = 3

So y equals 3; we can plug that number into any of our previous equations with x’s and y’s involved to find out what x is. Let’s use equation 1

1. x + y = 5 y is equal to 3 so

6. x + 3 = 5 subtract 3 from both sides

7. x = 2

Finally we go back to our original statements, x is the cost of a cappuccino and y is the cost of a latte so the solution is.

Cappuccino £2

Latte £3

As it turned out I only had £1.95 in change so I bought I diet coke instead.

So whenever you come across a puzzle that looks like it might be simultaneous equations; grab a pad and paper and just work methodically.

1. Identify the equations

2. State what your variables are what what symbols you will use.

3. Get an equation which has just one term and a number by either

a. Adding or subtracting whole equations

b. Multiplying or dividing by whole numbers

c. Adding or subtracting whole numbers or terms.

4. State the solution in terms of the variable symbols.

5. State the solution in terms if the variables themselves.

If this is still causing you trouble then please, don't panic, lots of people genuinely do have trouble getting a feel for algebra in puzzles. We'll be using it lots over the coming weeks so there's plenty of time to practice.

Here's an extra worked example via the awesome media of diamonds and youtube.

Here's an extra worked example via the awesome media of diamonds and youtube.

**The Apprentice Jeweller**This is is a video puzzle, the question, solution and working are all contained in one video so get ready to pause the video if you want to work out the answer for yourself.

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