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9.08 Variables on both sides

Lesson

Introduction

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?

Move 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.

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.

Examples

Example 1

Solve the equation: 4x+24=x-9

Worked Solution
Create a strategy

Apply reverse operations to rearrange the equation such that all of the x-terms are on the same side.

Apply the idea
\displaystyle 4x+24\displaystyle =\displaystyle x-9Write the equation
\displaystyle 4x+24-x\displaystyle =\displaystyle x-9-xSubtract x from both sides
\displaystyle 3x+24\displaystyle =\displaystyle -9Simplify
\displaystyle 3x+24-24\displaystyle =\displaystyle -9-24Subtract 24 from both sides
\displaystyle 3x\displaystyle =\displaystyle -33Evaluate
\displaystyle \dfrac{3x}{3}\displaystyle =\displaystyle \dfrac{-33}{3}Divide by 3 on both sides
\displaystyle x\displaystyle =\displaystyle -11Evaluate
Reflect and check

In the example above we chose to reverse the addition of x, gathering all the variable terms to the left hand side of the equation. We could have also moved all of our variable terms to the right hand side.

Idea summary

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.

Equations with brackets and variables on both sides

Suppose we want to solve the equation:8(6x-4)=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(-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:

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

\displaystyle 8(6x-4)\displaystyle =\displaystyle 7(-8x+6)+134
\displaystyle 8(6x-4) - 7(-8x+6)\displaystyle =\displaystyle 7(-8x+6)+134 - 7(-8x+6)Subtract 7(-8x+6) from both sides
\displaystyle 48x-32+56x-42\displaystyle =\displaystyle 134Use the distributive law
\displaystyle 104x-32-42\displaystyle =\displaystyle 134Combine the x-terms
\displaystyle 104x-74\displaystyle =\displaystyle 134Combine the constant terms
\displaystyle 104x\displaystyle =\displaystyle 208Reverse the subtraction
\displaystyle x\displaystyle =\displaystyle 2Reverse 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(6x-4) to the right hand side as our first step which would involve almost identical steps to the working shown directly above.

Examples

Example 2

Solve this equation for x by first expanding the brackets:8(8x+5)=3(6x+8)+108

Worked Solution
Create a strategy

Apply the distributive law and then reverse operations so that all of the x-terms are on the same side.

Apply the idea
\displaystyle 8(8x+5)\displaystyle =\displaystyle 3(6x+8)+108
\displaystyle 64x+40\displaystyle =\displaystyle 18x+24+108Use the distributive law
\displaystyle 64x-18x+40\displaystyle =\displaystyle 18x-18x+24+108Subtract 18x from both sides
\displaystyle 46x+40\displaystyle =\displaystyle 132Simplify
\displaystyle 46x+40-40\displaystyle =\displaystyle 132-40Subtract 40 from both sides
\displaystyle 46x\displaystyle =\displaystyle 92Evaluate
\displaystyle \dfrac{46x}{46}\displaystyle =\displaystyle \dfrac{96}{46}Divide by 46 on both sides
\displaystyle x\displaystyle =\displaystyle 2Evaluate
Idea summary

To solve equations with brackets and variables on both sides we can either apply the distributive law to expand all the brackets in the equation first, or we can move one set of brackets to the other side first so that the variables are on the same side.

Cross-multiply to solve

It should also be noted that for equations like:\dfrac{6x+4}{8 }=\dfrac{-3x+12}{3}

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(6x+4)=8(-3x=12)

This is called cross-multiplication.

Examples

Example 3

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

Worked Solution
Create a strategy

Apply cross-multiplication and then the distributive law.

Apply the idea
\displaystyle \dfrac{5x-4}{3}\displaystyle =\displaystyle \dfrac{3x+4}{5}Write the equation
\displaystyle 5(5x-4)\displaystyle =\displaystyle 3(3x+4)Apply cross-multiplication
\displaystyle 25x-20\displaystyle =\displaystyle 9x+12Apply the distributive law
\displaystyle 25x-20-9x\displaystyle =\displaystyle 9x+12-9xSubtract 9x from both sides
\displaystyle 16x-20\displaystyle =\displaystyle 12Evaluate
\displaystyle 16x-20+20\displaystyle =\displaystyle 12+20Add 20 to both sides
\displaystyle 16x\displaystyle =\displaystyle 32Evaluate
\displaystyle \dfrac{16x}{16}\displaystyle =\displaystyle \dfrac{32}{16}Divide by 16 on both sides
\displaystyle x\displaystyle =\displaystyle 2Evaluate
Idea summary

When cross multiplying fractions, we remove the denominators by multiplying them to the opposite numerators, as shown below:

A diagram demonstrating cross multiplication. Ask your teacher for more information.

Which gives us:

A diagram demonstrating cross multiplication. Ask your teacher for more information.

After performing this step, we can then distribute or move the variables to one side and solve the equation.

Outcomes

MA4-10NA

uses algebraic techniques to solve simple linear and quadratic equations

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