Simultaneous Equations
VIDEO COMING SOON…
In this post, we’re going to be looking at Simultaneous Equations.
This is probably one of my favourite areas of Algebra as you get to solve 2 equations at once!
Simultaneous Equations are where you are given 2 equations and based on each of these equations, you are asked to solve 2 unknowns – these are usually \(x\) and \(y\).
There are two main ways that you can go about solving Simulatenous Equations. There is the way you are mainly taught to do these in school – I would probably try to learn this primarily as this is what your teacher will be expecting to see in your homework. There is, however, an alternative way that I like to use that you may find easier if you are struggling with the first method.
Linear Equations
Let’s take this set of Linear Simultaneous Equations as an example and we can use both methods to show how they can be solved:
\(2x=6-4y\)
\(4x =-3-3y\)
Method 1:
- Put them into the format \(ax+by=c\)
\(2x+4y = 6\)
\(4x+3y=-3\)
- Multiply the top or bottom so that on of the co-efficients match. With these equations, if we multiply the top equation by 2, we will have:
\(4x+8y = 12\)
\(4x+3y=-3\)
- We take the 2 equations away from each other – as we have now made both equations feature a \(4x\), they will cancel each other out only leaving us with \(y\) in the equation:
\(5y=15\)
- Now collect your terms and solve for your first unknown (in this case, it is \(y=5\)).
- Finally, sub your newly found value into either of the starting formulas to get your remaining unknown:
\(2x+4(5)=12\)
\(2x+20=12\)
\(2x=-8\)
\(x=-4\)
This is the most common way to tackle these and the way I was taught to solve simultaneous equations at school. However, I have another way for those that struggle with using this method.
Method 2:
- Rearrange either of the formulas to equal either \(x\) or \(y\):
\(2x=6-4y\) can be rearranged to be \(x=3-2y\).
- Now sub this value for \(x\) into your other equation:
\(4(3-2y)=-3-3y\) → we’ve just replaced the \(x\) with \(3-2y\) per the above.
- Expand the brackets, collect your terms and solve for \(y\):
\(12-8y = -3-3y\)
\(15-8y = -3y\)
\(15=5y\)
\(y=5\)
- This is the same as step 5. in Method 1 → sub in our value for \(y\) into either of the equations and solve to give us our value for \(x\).
As you can see – this is 1. step shorter, although either method works well. It’s good to have a few ways to do things just incase you get stumped in the exam and one method gives you nicer numbers than the other.
Quadratic (non-linear) Simultaneous Equations
These are pretty much the same as Linear Simultaneous Equations, except one of our equations is quadratic. An example of this would be:
\(2x^2-y=3\)
\(7x+y = 1\)
The key with these quadratic simultaneous equations is to make these 2 equations into 1 quadratic equation then solve it using the variety of techniques covered in the Algebra syllabus – these include The Quadratic Formula, Completing The Square and Factorising Quadratics.
In order to make our Quadratic Simultaneous Equations into one Quadratic equation, we need to make \(y\) the subject of our linear equation (i.e. \(7x+y=1\)) and then sub it in for \(y\) in our Quadratic equation – it would look something like this:
\(y=1-7x\)
\(2x^2-(1-7x)=3\)
\(2x^2+7x-1=3\)
\(2x^2+7x-4=0\)
So now we’ve got one quadratic equation that we can look to solve – the important thing to remember here is that we will now have 2 values for \(x\) and \(y\). Let’s go ahead and solve our quadratic to see what we get as values for \(x\):
This will factorise as \((2x-1)(x+4)\), this is because we first look at what multiplies to give us \(4\) then we trial these figures in the brackets until we can expand out our brackets to give us \(2x^2+7x-4 = 0\).
If we make each of these brackets equal $0$ i.e. \(2x-1=0\) and \(x+4 = 0\), we get values of \(x\) equal to \(\frac{1}{2}\) and \(-4\).
Now we’ve got our values for \(x\), we can sub this into our \(y=1-7x\) to give us our 2 values for \(y\). In this case, \(y\) will equal \(29\) and \(-2.5\).
BONUS TIP:
This may come up in your exams and this is probably the best place to cover it. The question will probably give you quite a few marks as it’s likely quite a lot of people will struggle with it. However, the question may ask you to find the points that these 2 simultaneous equations would meet if plotted on a graph.
Now, we know that our \(2x^2-y=3\) will be a parabola as it’s a quadratic and our \(y=1-7x\) will be a straight line so they will intercept at 2 points.
All we need to do is show our 2 values for \(x\) and \(y\) as co-ordinates – so when \(x=\frac{1}{2}\), \(y=-2.5\), we can show this as \((\frac{1}{2},-2.5)\). This is a point that our lines will intercept. Our 2nd and final intercept will be \((-4,29)\) if we follow the logic used to find our first co-ordinates.
Hopefully this all made sense. If you have any questions on this, please head over the forum where you can ask any questions you may have and also see if any of your questions have been answered already.