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2.2 D) Linear Equations: Solving when Expanding Brackets – Part 1
2.2 D) Linear Equations: Solving when Expanding Brackets – Part 1
Sometimes we will be given an equation that contains brackets. There are a few different methods to find the value of unknowns when we are given equations that contain brackets. During this section, we will see a few different ways that we can solve these types of equations. The method that you can use to solve the equation depends on personal preference and the equation that we are given (some methods work better for certain equations).
Example 1
Find the value of x.
Find the value of x.
Method 1
One method to find the value of x is to expand the brackets on the left side of the equation. We do this by multiplying every term in the bracket by 4.
One method to find the value of x is to expand the brackets on the left side of the equation. We do this by multiplying every term in the bracket by 4.
We are now able to solve the question in exactly the same way as we did in the previous sections; we get all of the unknowns to one side and all of the numbers to the other side. There are more x’s on the left side, so let’s get all of the x’s to the left and all of the numbers to the right. This means that we need to move the -24 from the left side of the equation to the right, which we do by doing the opposite; we add 24 to both sides.
We want to find the value of x and not 12x, so we divide both sides by 12.
Therefore, x is 3.
We can check that we have obtained the correct value for x by subbing in x as 3 into the original equation.
We can check that we have obtained the correct value for x by subbing in x as 3 into the original equation.
This equation works, which means that we have the correct value for x; x is 3.
Method 2
Another way to find the value of x is to divide both sides by the coefficient of the bracket. For this equation, we would divide both sides by 4. This works quite nicely for this equation because dividing 12 by 4 gives us 3. However, it will not always work as well for other equations.
Method 2
Another way to find the value of x is to divide both sides by the coefficient of the bracket. For this equation, we would divide both sides by 4. This works quite nicely for this equation because dividing 12 by 4 gives us 3. However, it will not always work as well for other equations.
We now need to get all of the x’s on one side of the equation and all of the numbers on the other side. There are more x’s on the left, so I am going to have all of the x’s on the left and all of the numbers on the right. This means that we need to move the -6 that is currently on the left to the right. We are able to do this by doing the opposite, and the opposite of taking 6 is adding 6. Therefore, we add 6 to both sides of the equation.
We want to find the value of x and not 3x. Therefore, we divide both sides of the equation by 3.
We can see that both of these methods give us x as 3.
Example 2
Find the value of y.
Find the value of y.
There are two brackets in this equation. The best way to answer this question is to expand both of the brackets on the left, and then solve in the usual way. When we are multiplying out the brackets, we need to be very careful with the second bracket because we are multiplying the second bracket by -3; we need to be careful with the signs.
I am now going to collect like terms on the left side of the equation.
We now need to decide which side to get the unknowns to and which sides to get the numbers to. We get the unknowns to the side that has more unknown. Currently there are -19y on the left side and 0y on the right side. 0 is greater than -19. Therefore, we want to get all of the y’s to the right and all of the numbers to the left. This means that we need to move the -19y from the left side to the right, which we are able to do by adding 19y to both sides of the equation.
The next step is to move the -1 from the right side to the left, which we are able to do by adding 1 to both sides of the equation.
We want to find the value of y not 19y, so we divide both sides by 19.
Therefore, y is equal to 1. We can check this by subbing y as 1 into the first equation.
2(3 – 2y) – 3(5y – 4) = -1
2(3 – 2(1)) – 3(5(1) – 4) = -1
2(3 – 2) – 3(5 – 4) = -1
2(1) – 3(1) = -1
2 – 3 = -1
-1 = -1
2(3 – 2(1)) – 3(5(1) – 4) = -1
2(3 – 2) – 3(5 – 4) = -1
2(1) – 3(1) = -1
2 – 3 = -1
-1 = -1
This equation works, which means that we have found the correct value for y; y is -1.