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# An algebra problem to end the week : simultaneous equations

I thought I’d finish the week with a problem from my friends at ‘Quora’.. but then I found most of this weeks problems were difficult questions requiring University Maths! Yes I could follow (most) of it but it was all a little ‘off-piste’ for this diary, So here is a problem with can solve using Simultaneous equations

So I have picked out a wordy problem that might look complicated at first sight, but we re

ally just need to turn it into a couple of equations…  Let’s see Lets call our numbers a and b – That’s always a good start with Algebra.

The first equation gives us the sum. So

a + b = 26.

With the second equation, we need to decide which is the larger number. It doesn’t matter which be choose. Let’s say a is larger.

a =  5 + 2b  (a has a value 5 more than twice b)

This is where we can solve the simultaneous equations. I’ll make a video on all the ways to do this soon, but for now I am going to solve these ‘by substitution’

I am going to substitute into the first equation the expression for a in the second. In the () you will see I have replaced ‘a’ with what a = in the second

(5 + 2b) + b = 26

We can together together terms

5 + 3b = 26

Take 5 from both sides

3b = 21

And from this see that b = 7. And so a = 19 (Twice b plus 5).

As a final check 7 + 19 = 26

So there we have it the two numbers are 19 and 7

For more information on solving simultaneous equations, take a look at this More Help on simultaneous equations

# How we can use ‘Continuity’ to answer questions

In the last post I showed the difference between a Continuous curve and a Discontinuous curve, with a few examples from well known curves.

In this post I am going to show that this can be useful to know in answering a certain sort of question. I’m not going to do this whole exam question.

The first part involves putting the value x = 3 into the given equation. You will get the answer y = -6

Use the value x = 4 and you get the value y = 20.

This is where the fact that this is a continuous line is so important. You know that the line must cross the line y = 0 somewhere between x=3 and x=4 , and that is what we mean by the root.

If you don’t believe that, think of the equation y=1/x.

If x = -2, the y = -1/2
If x = 1 then y = 1.

The same situation; y goes from negative to positive. Does that mean we have a value of x between -2 and 1 for which 1/x = 0?

No it doesn’t, because the curve for y=1/x is not continuous. we can only use that rule for continuous curves.

Note:  Some curves, such as y=1/x in fact, are continuous for much of their length. There is just one place where it is not.  So you can use the rule above if the curve is continuous in the RANGE of numbers you are working in.

# What exactly is a number machine?

This is a question I asked myself when I returned to tutoring Maths a few years ago. Number machines often turn up in questions of Foundation papers. They provide a useful introduction to quite a few things

• Algebra
• Functions
• Computer Programming – Which is what I was doing before I turned to Tutoring. Here is an example Number Machine. The idea is we ‘feed in’ a number on the left, and see what comes out on the right.

A simple starter question would be : If a 4 in entered into this number machine, what would the result be. Number machines work left to right, just like reading. In this case, if the input is 4, we follow the boxes left to right and get  4 x 4 = 12 then + 2 = 14 .

A more advanced question would be – If the output is 8, what is the input.  This asks us to move right to left but also do the operations if reverse. Remember reverse of adding is taking away. The opposite of multiplying is dividing.

8 – reverse of + 2 is -2   so the number between the boxes is 6.  6 Divide by 3 is 2.  So the in number must be 2.

Try putting 2 into the machine and see how this is a reverse of what we did above.

The more complicated questions miss out the instructions. For example. If we put 6 into this number machine we get 6 out. What is missing in the second box?

Clue – Its an subtraction sum.

I’ll show how to answer that in my next post.

# Solving Sim equations by substitution

Today’s post is on a GCSE subject, but quite an advanced one…  Solving Simultaneous  equations by substitution

I say advanced; I think its not often covered, but I think it can be followed;  There might just be some exciting algebra to do

To solve these equations

4x + y =  9
3x – y =  5

we know we need to add the equations together, to get rid of the y.

Sometimes we have to be a bit cleverer, and multiply one of the equations by something. For example

4x + y =  9
3x –2y =  4.

We have to multiply the first equation by 2 so the number in front of the y is the same. Otherwise the y won’t disappear when we add.

This is called ‘solving by addition/subtraction’

It only works sometimes. This new method words more often. It’s called ‘solving by substitution’

Solving By Substitution

Did you solve the first equations? You should have got x = 2 and y = 1

Now let’s change the first equation so its y = something. We can -4x from both sides.  This gives us

y = 9 – 4x.

Now, this is where the ‘substitution’ comes in.  We change the y in the other equation –

3x – (9-4x) = 5

We re-arrange this, remembering that – -4x = + 4x!
3x – 9 +4x = 5
7x = 14; x = 2..  and y = 1

Now that looks similar but more complicated. In fact I would recommend solving this equation the way we knew before.

So why am I showing you ‘substitution’?

To show the result is the same, and to get ready for when the other method doesn’t work.

For example
2x2  + y2  =11
y – x = 2

Here we have ‘squared’ in the mix, and this makes it hard to see how we can get rid of terms by adding or subtracting to two equations. This is where substitution comes into its own

Change the 2nd equation to say y = x + 2 (I’ve just added the x to both sides)
Substitution y = x + 2 into the first

2x2 + (x + 2)2 = 11

We have to a little work here, multiplying the brackets out and rearranging the terms
2x2 + x2 + 4x + 4 = 11
3x2 + 4x – 7 = 0  (I’ve combined the x2 terms and taken 11 from both sides)
Factorising this gives
(3x  + 7 ) (x  – 1)

That could be quite a hard factorisation to see straight off, though it helps that not many pairs of numbers multiply together to make -7. A little ‘trial and error’ may help.
So x = 1 or x = -7/3

Now remember y = x + 2 so

When  x = 1, y = 3..  That’s quite easy to see

When x = -7/3, y = -1/3. Just add on 2;

# Working out Recurring Decimals

A clever trick that sometimes comes up in GCSE exams is to how to turn a repeating decimal back into a fraction.

At first glance the question looks difficult… but actually it’s not really.

Write the decimal 0.393939393939 (recurring)  as a decimal.

So how do we do this?

Well we start with a step familiar from algebra – We use a letter!

Let r = 0.393939393939

We need to get rid of that long train of decimals – se we use the same method as when we want to lose something in solving simultaneous equations – we do a subtract

But how will that work?

Well think what 100r is

r = 0.393939393939  so 100r = 39.3939393939

[Can you see where we are going with this now?]

Now we can do a subtraction and all the numbers in the recurring chain cancel out
39.3939393939    _
0.3939393939

= 39

What we have done here is 100r – r   so this is 99r

99r = 39

So r = 39/99…  A fraction!  We’ve done it!

Just two more things to note

1. We should always ‘cancel down’ fractions is we can to use the lowest numbers at top an bottom.
So I should have finished this question by writing as  13/33.
2. How did I know to multiply by 100? Would this work for every example?  Well, no, a careful study of what we have done shows that the key is to ‘cancel out’ all the recurring decimals. To do that, we need to ‘line up’ the same numbers in the subtraction. To do this we need to count how many numbers in the repeating pattern.. we then need that number of 0s in the ‘multiplying factor’

So if we have 0.123123123123123…   we have 3 numbers in the pattern

r = 0.123123123123123

1000r = 123.123123123123..

999r = 123

R = 123/999  = 41/333

Here are three more to work on,,,

1. Write 0.777777777777… as a decimal
2. Write 0.787878787878.. as a decimal

3. Write 0.0200200200200..  as a decimal  (That’s a bit tougher)

0.7777 = 7/9
0.78787878 = 78/99 = 26/33
0.0200200200 = 20/999

Lets work through to the answer.

The first thing to do is simplify the situation by taking the clown out of the picture.  We know the clown + the pin is worth 7…
so clown = 7 – pin.

Now we replace the clown in the 2nd equation with ‘7 – Pin’

From the first equation  the pineapple = 7 – 2 pins

We can change the 2nd equation to be just about Pins.

3 x ( 7 – 2 Pins) + 7 – Pin = 14

28 -7 Pins = 14

so 7 Pins are worth 14, and 1 pin is worth 2. We can find from the other two equations. I’ve solved this by a ‘substitution method’ I could also have done this by adding or subtracting equations.  But the key is to eliminate some of the variables until we have one, then substitute back.

The final stage relies on knowing we should do the multiplication first, so the answer is

2 + 5 x 3 = 17.

I know getting the better of equations is one of the skills that can take more work if you are not a natural at Maths.

That’s why I’ve been fascinated with how this has become a craze on Social Media.  Give someone a picture with bottles of beers and balloons and it becomes a problem worth solving.

Here is one of my own –  OK, I’ve gone with clowns, Pineapples and bowling pins. Can you work out what the ? is worth? Clues –

1 –  Try to eliminate the clown from your enquiries
2- This should help you find the value of the Pineapple, then bowling pin, and then clown.
3, Remember the BODMAS rule – This is the mistake people most often make.

(pssst  – The solution will be posted in three days)