Simultaneous Equations - Don't get in a Pickle, Save it for Sandwiches
Simultaneous equations scare more students than they should.
In reality they are less scary than having your sandwiches cut in the wrong shape. (Triangle cut is the best, right?)
So, let's have a look at them from a different angle and hopefully clear them up for you.
The name itself is a little weird and might be putting you off.
Simultaneous, loosely means: At the same time.
Usually you solve one equation, with one unknown letter to work out, like this:
\[x+4\ =\ 10\]
You rearrange it and find that x is 6.
Simultaneous equations usually have 2 equations and 2 unknown letters that you need to find out, something like this:
\[4x\ +3y\ =\ 23\]
\[5x\ +\ 3y\ =28\]
And they are named simultaneous because you need to work with both and solve both in order to find the answers. The solution to x is the same for both equations and the solution to y is the same for both equations.
Where do you start?
Although you have to find both x and y, you actually start by finding one first.
The way you do that is either by removing one letter through addition or subtraction, or by rearranging and substituting - I know that sounds like a lot of confusing work, but stay with me and we will do it in small steps.
You can choose, but there are some hints to look for as to which is best to start with.
In this example, 3y is in both equations. I'm also going to number the equations to make this easier.
\[Equation\ 1:\ 4x\ +3y\ =\ 23\]
\[Equation\ 2:\ 5x\ +\ 3y\ =\ 28\]
Equation 2 is 'bigger' than equation 1 as it has more x and 28 is higher than 23.
So if we do Equation 2 minus Equation 1, this will get rid of the 3y and leave us with one equation and just x to solve.
5x - 4x = x
3y-3y = 0
28 - 23 = 5
So that leaves us with:
\[x\ =\ 5\]
So we've just found what x is - wohoo, half way there!
Now we need to find y.
All you need to now is choose either equation 1 or 2, it doesn't matter which, and replace the x with 5 and this should tell you what y is.
\[Equation\ 1:\ 4\ times\ 5\ +\ 3y\ =\ 23\]
\[Equation\ 1:\ 20\ +\ 3y\ =\ 23\]
\[Equation\ 1:\ 3y=23-20\]
\[Equation\ 1:\ 3y=3\]
\[Equation\ 1:\ y=1\]
And that's it! x is 5 and y is 1.
Handy hint:
If one equation had -3y and the other had 3y, we would add them together to cancel out the y.
Method 2: Substitution
Unfortunately, they aren't always that easy - you knew I was going to say that, right?
You can't always get rid of one letter by adding or subtracting.
When that's the case you have to rearrange to get x= or y = and then you put that back into one of the equations. So with this method you actually end up with 3 equations.
Let's do one together:
\[Equation\ 1:\ 5x\ +5y\ =0\]
\[Equation\ 2:\ x\ -\ y\ =\ 2\]
With this one, there is no way to remove x or y by adding or taking away. We would end up with 6x or 4x, and 4y or 6y.
But what we can do is rearrange equation 2 to give:
\[Equation\ 3:\ x=2+y\]
How does this help us?
Its almost like we've kind of solved x or at least have a way to work with x and we can put it back into equation 1 or 2.
This can get tricky and look confusing...a good time to eat a biscuit.
Let's rewrite equation 1, swapping x for 2+y:
\[5\left(2+y\right)\ +\ 5y\ =\ 0\]
\[10\ +\ 5y\ +5y\ =\ 0\]
\[10\ +\ 10y\ =\ 0\]
\[10\ =-10y\]
\[1=-y\]
\[y=-1\]
Then if we put y = -1 into equation 2, we get:
x - -1= 2
x+1 = 2
Which means x = 1
Let's check if that is correct by putting the values of x and y into equation 1:
5x1 + 5x-1 = 0
5 - 5 = 0
That looks good to me :)
I hope that has helped you with understanding simultaneous equations.
It is super important that you understand how to do them for GCSE Maths exams. Please drop a comment below if you need more help or have any other questions.
Or, just let me know what your favourite sandwich is - I'm loving the M&S Cheddar Ploughman at the moment ;)