NSW Year 8 - 2020 Edition 5.04 Variables on both sides
Lesson

When solving equations we often aim to isolate the variable or pronumeral we are solving for. We also know that if we have multiple like terms on one side of the equation we can combine them into a single term.

So what happens if we have variables on both sides of the equation?

### Moving the variables

In the same way that we apply reverse operations to isolate our variables, we can also use them to gather variable terms together.

#### Worked Example

Solve the equation: $5p+7=2p+19$5p+7=2p+19

Think: To gather all the $p$p-terms to one side of the equation, we can reverse the addition of $2p$2p. After combining the $p$p-terms into a single term, we can then isolate $p$p to solve the equation.

Do: We can reverse the addition of $2p$2p by subtracting $2p$2p from both sides of the equation. This will give us:

$5p+7-2p=19$5p+72p=19

Combining the $p$p-terms gives us:

$3p+7=19$3p+7=19

From here, we can solve the equation as per usual to get the solution:

$p=4$p=4

Reflect: In the example above we chose to reverse the addition of $2p$2p, gathering all the variable terms to the left hand side of the equation. This is an arbitrary choice and it doesn't matter which side we chose to move all our variable terms to.

If we instead wanted to move all of our variable terms to the right hand side, our working would instead look like this:

 $5p+7$5p+7 $=$= $2p+19$2p+19 $7$7 $=$= $2p+19-5p$2p+19−5p Subtract $5p$5p from both sides of the equation $7$7 $=$= $-3p+19$−3p+19 Combine the $p$p-terms $-12$−12 $=$= $-3p$−3p Reverse the addition $4$4 $=$= $p$p Reverse the multiplication

Whenever we want to solve equations that have variables on both sides, our main goal is to move all the variable terms to one side so that we can combine them, then use reverse operations to finish solving the equation.

While the side we move the variable terms to doesn't affect the final answer, it can introduce a lot of negative signs that will need to be cancelled out later. For this reason, it is best to move the smaller variable term over as to avoid negative coefficients.

#### Practice question

##### Question 1

Solve this equation for $x$x by rearranging the equation:
$5x-3=-4x+33$5x3=4x+33

Enter each line of working as an equation.

### When do we move the variables?

Suppose we want to solve the equation:

$8\left(6x-4\right)=7\left(-8x+6\right)+134$8(6x4)=7(8x+6)+134

We have a couple of choices for our first step; we can either apply the distributive law to expand all the brackets in the equation, or we can move the whole expression $7\left(-8x+6\right)$7(8x+6) to the left hand side.

If we choose to apply the distributive law first, the working out to solve the equation will look like this:

 $8\left(6x-4\right)$8(6x−4) $=$= $7\left(-8x+6\right)+134$7(−8x+6)+134 $48x-32$48x−32 $=$= $-56x+42+134$−56x+42+134 Use the distributive law to expand the brackets $48x+56x-32$48x+56x−32 $=$= $42+134$42+134 Reverse the subtraction of $56x$56x $104x-32$104x−32 $=$= $42+134$42+134 Combine the $x$x-terms together $104x$104x $=$= $42+134+32$42+134+32 Reverse the subtraction $104x$104x $=$= $208$208 Perform the addition $x$x $=$= $2$2 Reverse the multiplication

If we choose to move all the variables to the left hand side first, the working out to solve the equation will look like this:

 $8\left(6x-4\right)$8(6x−4) $=$= $7\left(-8x+6\right)+134$7(−8x+6)+134 $8\left(6x-4\right)-7\left(-8x+6\right)$8(6x−4)−7(−8x+6) $=$= $134$134 Subtract $7\left(-8x+6\right)$7(−8x+6) from both sides of the equation $48x-32+56x-42$48x−32+56x−42 $=$= $134$134 Use the distributive law to expand the brackets /> $104x-32-42$104x−32−42 $=$= $134$134 Combine the $x$x-terms together $104x-74$104x−74 $=$= $134$134 Combine the constant terms together $104x$104x $=$= $208$208 Reverse the subtraction $x$x $=$= $2$2 Reverse the multiplication

As we can see, both methods take the same number of steps and are both about as difficult as each other. We could also choose to move the expression $8\left(6x-4\right)$8(6x4) to the right hand side as our first step which would involve almost identical steps to the working shown directly above.

#### Practice question

##### Question 2

Solve this equation for $x$x by expanding the brackets:
$8\left(8x+5\right)=3\left(6x+8\right)+108$8(8x+5)=3(6x+8)+108

Enter each line of working as an equation.

### Cross-multiplying to solve

It should also be noted that for equations like:

$\frac{6x+4}{8}=\frac{-3x+12}{3}$6x+48=3x+123

The division of the numerators by the denominators will need to be reversed first. The result of this is that each numerator gets multiplied by the denominator from the other side of the equation, like so:

$3\left(6x+4\right)=8\left(-3x+12\right)$3(6x+4)=8(3x+12)

This is called cross-multiplication.

Cross-multiplication

When cross multiplying fractions, we remove the denominators by multiplying them to the opposite numerators, as shown below: Which gives us: After performing this step, we can then distribute or move the variables to one side and solve the equation.

#### Practice question

##### Question 3

Solve this equation for $x$x by cross-multiplying and then expanding the brackets:
$\frac{5x-4}{3}=\frac{3x+4}{5}$5x43=3x+45

Enter each line of working as an equation.

### Outcomes

#### MA4-10NA

uses algebraic techniques to solve simple linear and quadratic equations