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2.2 E) Linear Equations: Solving when Expanding Brackets – Part 2
This section follows on from the previous section. Make sure that you have covered the previous section before coming on to this section (click here to be taken through to the previous section). 

Example 1
Find the value of c.
We could expand the bracket on the left side of the equation. However, it would get quite messy with a lot of fractions. Therefore, when we have an equation like this, the best way to solve it is to multiply both sides of the equation by the denominator of the fraction that is outside the bracket. By doing this we eliminate the fraction from the outside of the bracket, thus making it easier to solve the equation.
 
The denominator of the fraction outside the bracket on the left is 5, so we multiply both side of the equation by 5. 
Multiplying a bracket by 1 results in the bracket staying the same. 
We now want to have all of the unknowns on one side and the numbers on the other side. Let’s have the left side as unknowns and the right side as numbers. This means that we need to move the -2 from the left to the right, which we do by adding 2 to both sides. 
Currently we have the value of 3c, but we want to find the value of c. Therefore, we divide both sides by 3. 
​We can check we have the correct value by subbing in c as 4 into the first equation. 
​Therefore, we have the correct value for c. 

Example 2
Find the value of d​. 
Like the question before, I am going to multiply the left and the right side of the equation by the denominator in the fraction next to the brackets. Therefore, I am multiplying both sides by 3. 
The next step is to multiply out the brackets on the left of the equation (we could divide both sides by 2; I will go through this method at the end)
Therefore, d is equal to 23.
 
The other way that we could have obtained the answer is to divide both sides by 2 after we had multiplied by 3. 
We divide both sides by 2 because we have 2 lots of the bracket on the left hand side. 
This method gives us d as 23, which is exactly the same value that we found for the previous method. 
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