What you need to know
Solving Equations
Linear equations are a type of equation that appear all over the place in maths. They can look quite simple, like x+2=5, or they can look a little more complicated, such as
5p-7=3p+3.
The golden rule when solving these equations is that whenever you apply some operation to one side of the equation (e.g you add 5 to it, or multiply it by 3, etc), you must also do the same thing to the other side.
To solve an equation, we need the unknown (letter) on one side of the equals sign and everything else on the other. To do this, we will collect terms of the same type together on the same side of the equals sign, a process known as collecting like terms.
Take Note:
The equations we will solve on this page are linear equations meaning they are straight forward and don’t contain any powers or roots. In order to be confident with this topic it is best to make sure you are happy with collecting like terms first. Once happy with solving linear equations you can start to apply these fundamental skills to more difficult topics:
These are just a few of the topic areas that require the fundamental skills learnt whilst solving equations, there are many, many more that these skills apply to.
Example 1: Solving Linear Equations
Solve the equation 5p-7=3p+3.
Our first step will be to collect the p terms on one side. To do this, we are going to get rid of the positive 3p on the right-hand side by subtracting 3p from both sides of the equation:
5p-7-3p=3p+3-3p
Collecting the like terms means the equation becomes
2p-7=3
We now collect the constant terms on the other side. To do this, we will get rid of the -7 on the left-hand side by adding 7 to both sides of the equation:
2p-7+7=3+7
Collecting the like terms means the equation becomes
2p=10
The final step is to get rid of the 2. Since p is being multiplied by the 2, we will divide both sides by 2:
2p\div2=10\div2
2p\div2 just becomes p – this was the aim – and then 10\div2=5, so we have
p=5
The key workings are shown below. This is how to set out your workings when solving linear equations.
Example 2: Solving Linear Equations
Solve the equation 3a+1=5-7a.
To get rid of -7a do the opposite and add 7a to both sides of the equation. 3a+7a=10a, so the equation becomes
10a+1=5
Do the opposite of +1 and subtract 1 from both sides of the equation. Doing so, we get
10a=5-1=4
As a is being multiplied by 10, we will do the opposite and divide by 10:
a=\dfrac{4}{10}
Simplify
a\dfrac{2}{5}
If we were to answer this using the lines method, it would look like:
Example Questions
1) Solve the equation 2x+1=2
To get the constant terms (numbers) only on one side of the equation, we will get rid of the +1 by subtracting 1 from both sides. Then, the equation becomes
2x=1
Then, all that remains to get x on its own by dividing both sides by 2 to get,
x=\dfrac{1}{2}
2) Solve the equation \dfrac{1}{2}x-3=7
To get the constant terms (numbers) only on one side of the equation, we will get rid of the -3 by adding 3 to both sides. Then, the equation becomes
\dfrac{1}{2}x=10
Then, all that remains to get x on its own by multiplying both sides by 2 to get,
x=20
3) Solve the equation 3x+4=13
So, to get the constant terms (numbers) only on one side of the equation, we will get rid of the +4 by subtracting 4 from both sides. Then, the equation becomes
3x=9
Then, all that remains to get x on its own is to get rid of the 3, and since it is being multiplied by this 3, we will divide both sides by 3 to get
x=9\div3=3
4) Solve the equation 5m-2=16-4m
Firstly, we want to get all the m terms on one side so we will add 4m to both sides of the equation,
9m-2=16
To get the constant terms on the other side of the equation, we must get rid of the -2 on the left. We will do this by adding 2 to both sides of the equation, which leaves us with
9m=16+2=18
So, now all that remains is to get rid of the 9. As it is being multiplied by the x, we will divide both sides by 9, to get
m=18\div9=2
5) Solve the equation 4(z-2)=-20
This requires one additional step as we first need to expand the bracket. So, multiplying 4 by everything inside the bracket, the equation becomes
4z-8=-20
To get all constant terms over to the other side of the equation, we must get rid of the -8 on the left, and we will do this by adding 8 to both sides. So, our equation becomes
4z=-20+8=-12
So, now all that remains is to get rid of the 4. As it is being multiplied by the z, we will divide both sides by 4, to get
z=(-12)\div4=-3
Worksheets and Exam Questions
(NEW) Solving Equations (Foundation) Exam Style Questions - MME
Level 4-5Algebra Forming And Solving Equations - Drill Questions
Level 4-5Solving Linear Equations - Drill Questions
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