# Solving Linear Equations - Variables on Both Sides

Now we are becoming familiar with solving for a variable (like x) on one side of an equation, we can move onto solving for a variable on both sides of an equation.

Any time we make a step forward into new mathematical realms, it may seem daunting looking at a problem that is unfamiliar. However, all we need to do is remember our methods we have learned, understand what the problem is asking us, and then apply those methods! With solid mathematical method understand comes the tools to tackle most problems.

In this question below, we will quickly go over some of the methods we know, and that we will later use to solve problems.

## Question 2

We have the equation 4(x - 2) + 3x = 5(x + 2)

Apply the Distributive Property and any other properties (commutative & or associative) to the equation above to write an equivalent equation.

The distributive property states a (b + c) = a(b) + a(c)

We can see we have parts of the equation above we can apply to this property,

4(x - 2) and 5(x + 2)

Remember, do not get confused, we are not changing anything about the equation yet. We are just writing it differently; it is not like the Inverse Operations where we do the opposite operations to both sides...yet!

So, applying the distributive property to the equation,

4(x - 2) + 3x = 5(x + 2)

Do not forget to include the (-) signs when multiplying out,

4(x) - 4(2) + 3x = 5(x) + 5(2)

4x - 8 + 3x = 5x + 10,

Commutative Property tells us this is the same as,

4x + 3x - 8 = 5x + 10

7x - 8 = 5x + 10

Now we have our version of the equation. Note, it looks slightly easier to deal with now.

Use the new equation to solve the equation for x, then check it is true for the original equation,

When we have our variable, x , on both sides of the equation, before we do any Inverse Operations, we want to “get our x just on one side of the equation”, this is our new method that we will explore in this lesson,

How do we do this? Use the same methods we have been using, the Additive Properties of Equality and the Multiplicative Properties of Equality. These properties still hold true even when subtracting, adding, multiplying, or dividing something with a variable in it, like 5x.

Simply put, these properties state we can do anything do it mathematically, as long as its done to both sides,

In the same way, 2 = 2 is still true when you subtract one from both sides 2 - 1 = 2 -1, 1 = 1,

It’s the same for subtracting, adding multiplying, or dividing variables from both sides

For example, if 5 = 5, then adding “something we don't know” to both sides is still true,

5 + “something we don't know” = 5 + “something we don't know”,

Now, we can use these same properties to manipulate equations so that we can solve them, take,

7x - 8 = 5x + 10,

We can rearrange this to get a variable just on one side,

We do this by subtracting one of the variables & their coefficients, from both sides,

We can either subtract 7x from both sides, or 5x from both sides, lets choose 5x (both would work)

(7x - 8) - 5x = (5x + 10) - 5x

7x - 8 - 5x = 5x + 10 - 5x

Rearranging,

7x - 5x - 8 = 5x - 5x + 10

Notice the 5x on the right side will cancel out, leaving us with our x on one side, which we can solve!

[7x - 5x = 2, 5x - 5x = 0]

2x - 8 = 0 + 10

2x - 8 = 10

This should now look familiar and like something we can solve,

Remember our Inverse Operations,

2x - 8 + 8 = 10 + 8

2x = 18

Now divide by 2,

(2x)/ 2 = (18)/ 2

x = 9,

Now last step is to sub this back into the original equation, 4(x - 2) + 3x = 5(x + 2)

4((9) - 2) + 3(9) = 5((9) + 2)

4(7) + 3(9) = 5(11)

28 + 27 = 55

55 = 55,

There we can see x = 9 is the solution for the equation, so to go over what we did,

1) We used our three properties to see if we can rearrange to make it easier to solve

2) Subtracted 5x from both sides so give us our x variable on one side only

3) Used Inverse Operations to solve for x

We will become quite familiar with the method in algebra

With the two larger problems below, apply the same methods used above to solve for x.

## Question 3a

5(x - 1) - 3(x + 5) = 7(x - 2) + x

Again, this is a slight step up from what we have done before, stick to the same methods and what we know,

See, if we can apply our distributive, commutative and associative properties to either side of the equation to make it simpler and easier to solve,

First, distributive,

5(x - 1) - 3(x + 5) = 7(x - 2) + x

Be careful! There are lots of steps and negative signs involved, so be careful to include them or cancel them where appropriate,

In particular, the … -3(x + 5)... part of this equation, this is the same as ...+ (-3)(x + 5)... so multiply out the (-3), we will emphasize this on this equation, but in the future will assume awareness of the (-) sign, so quickly re writing the equation as,

5(x - 1) + (-3) (x + 5) = 7(x - 2) + x

5(x) - 5(1) + (-3) (x) + (-3) (5) = 7(x) - 7(2) + x

5x - 5 - 3x - 15 = 7x - 14 + x

Applying our properties again,

5x - 3x - 5 - 15 = 7x + x - 14

2x - 20 = 8x - 14

First, let’s get our variable on one side only, so we can subtract 2x or 8x from both sides, lets subtract 8x,

(2x - 20) - 8x = (8x - 14) - 8x

2x - 8x - 20 = 8x - 8x - 14

[2x - 8x = -6x & 8x - 8x = 0]

-6x - 20 = 0 - 14

Now Inverse Operations,

(-6x - 20) + 20 = (-14) + 20

-6x = 6

Divide -6 on both sides,

(-6x) / -6 = (6) / -6

x = -1

Sub this back into our original equation to make sure it is correct,

Sub x = -1 into,

5(x - 1) - 3(x + 5) = 7(x - 2) + x

Be very aware of the negatives and double negatives cancelling out!

5((-1) - 1) - 3((-1) + 5) = 7((-1) - 2) + (-1)

Do the addition and subtraction in the brackets first,

5(-2) - 3(4) = 7(-3) + (-1)

Multiply out,

-10 - 12 = -21 - 1

-22 = -22

x = -1 is correct

## Question 3b

6 - 3(x + 2) = 4(x - 2) - 20

Start by using our three properties to try and simplify,

Use the distributive to multiply out the brackets, remember the negatives,

6 -3(x) - 3(2) = 4(x) -4(2) - 20

6 -3x - 6 = 4x - 8 - 20

Group like terms together,

6 - 6 - 3x = 4x - 8 - 20

0 - 3x = 4x - 28

-3x = 4x - 28

Now we get our variable on one side, paying attention to the negative signs, we can subtract 4x from both sides of subtract (-3x) [ the same as adding 3x] from both sides, lets subtract 4x,

(-3x) - 4x = (4x - 28) - 4x

-7x = -28

Now Inverse Operations, we only have one step this time, divide by -7,

(-7x)/ -7 = (-28) / -7

x = 4

Now sub x = 4 into 6 - 3(x + 2) = 4(x - 2) - 20,

6 - 3((4) + 2) = 4((4) - 2) - 20

Do the brackets first,

6 - 3(6) = 4(2) - 20

6 - 18 = 8 - 20

-12 = -12

x = 4 is correct

## Question 4

Solve the equation below for x and determine if this has a solution. Hint, an equation has no solution if you are left with a contradictory statement.

6(x + 2) + 1 = 2(3x + 1)

So, use the distributive property to expand and simplify first,

6(x) + 6(2) + 1 = 2(3x) + 2(1)

6x + 12 + 1 = 6x + 2

6x + 13 = 6x + 2

Now let’s get our x-value on one side, subtract both sides by 6x,

6x + 13 = 6x + 2

6x + 13 - 6x = 6x + 2 - 6x

0 + 13 = 0 + 2

13 = 2,

This is a false and contradictory statement; therefore the equation has no solutions.

## Recap

We learned how to solve more complicated linear equations - those having variables on both sides of the equation. So, to expand on the steps, we learned to:

1. Use the Distributive, Associative and Commutative Properties - To simplify long complicated equations
2. Get our variable terms on one side of the equation - Example, if we have 4x + 1 = 3x + 4, subtract the 3x from both sides to give us x + 1 = 4
3. Identify the order of operations - make a list of the order of operations that you would do to the variable
4. Inverse order of operations and reverse each operation itself - reverse the order of operations, and do the inverse operation in each step (i.e. subtract 2 becomes add 2, divide by 3 becomes multiply by 3)
5. Apply the Inverse operations and solve for the variable - Follow all the correct steps and you will be left with your variable isolated giving you and answer.