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Stage 5.1-2

2.04 Solving linear equations

Lesson

Inverse operations

To solve equations using algebra, the most important rule to remember is that if we apply operations to one side of the equation, we must also apply it to the other.

When applying operations to equations, we always apply the same step to both sides of the equation. This way, both sides of the equation will be equal once we solve the equation.

Making sure to follow this rule, we can isolate the pronumeral in an equation by applying operations to both sides of the equation which reverse the operations applied to the pronumeral.

To do this, we need to know which operations are inverses of each other.

OperationInverse operationExample
\text{Addition} \\ \text{Subtraction}\text{Subtraction} \\ \text{Addition}x+4-4=x
\text{Multiplication} \\ \text{Division}\text{Division} \\ \text{Multiplication}y\times4\div4=y
\text{Powers} \\ \text{Roots}\text{Roots} \\ \text{Powers}\sqrt{x^{2}}=x

As we are only dealing with linear equations at this stage, powers and roots will not be involved but have been included for reference.

Examples

Example 1

Solve the equation: -x-7=7

Worked Solution
Create a strategy

Use the inverse operations to isolate x on the left hand side of the equation.

Apply the idea
\displaystyle -x-7 + 7\displaystyle =\displaystyle 7 + 7Add 7 to both sides
\displaystyle -x\displaystyle =\displaystyle 14Evaluate the addition
\displaystyle \dfrac{-x}{-1}\displaystyle =\displaystyle \dfrac{14}{-1}Divide both sides by -1
\displaystyle x\displaystyle =\displaystyle -14Simplify
Reflect and check

The original operations applied to the pronumeral were cancelled out by the inverse operations which we applied to both sides of the equation, leaving just x on the left hand side.

Idea summary

When applying operations to equations, we always apply the same step to both sides of the equation. This way, both sides of the equation will be equal once we solve the equation.

Equations with brackets

If we have an equation with one set of brackets such as 3\left(x-5\right)=9 we can either expand the brackets before solving or, in this case as 3 is a factor of 9, divide both sides of the equation by 3. But in cases where we have two sets of brackets, we will first want to expand both sets of brackets before combining like terms. We can then solve the equation by performing inverse operations.

Examples

Example 2

Solve this equation for x: 2\left(2x+5\right) + 3\left(4x+6\right)=76

Worked Solution
Create a strategy

Use the distribitive law and inverse of operations to isolate x on the left hand side of the equation.

Apply the idea
\displaystyle 2\left(2x+5\right) + 3\left(4x+6\right)\displaystyle =\displaystyle 76Write the equation
\displaystyle 4x+10+12x+18\displaystyle =\displaystyle 76Expand the brackets
\displaystyle 16x+28\displaystyle =\displaystyle 76Add like terms
\displaystyle 16x+28-28\displaystyle =\displaystyle 76-28Subtract 28 from both sides
\displaystyle 16x\displaystyle =\displaystyle 48Evaluate subtraction
\displaystyle \dfrac{16x}{16}\displaystyle =\displaystyle \dfrac{48}{16}Divibe both sides by 16
\displaystyle x\displaystyle =\displaystyle 3Simplify
Idea summary

To solve equations with brackets, expand the brackets using the distributive law, and then use inverse operations to solve for x.

Equations with pronumerals on both sides

To solve equations with variables on both sides of the equation, we want to use inverse operations (usually by adding or subtract terms) to eliminate the variables from one side of the equation. We can then combine like terms and solve using inverse operations.

Examples

Example 3

Solve this equation for x: 5x=x+8

Worked Solution
Create a strategy

Use inverse operations to get all x terms on one side of the equation and solve for x.

Apply the idea
\displaystyle 5x-x\displaystyle =\displaystyle x-x+8Substract x from both sides
\displaystyle 4x\displaystyle =\displaystyle 8Simplify like terms
\displaystyle \dfrac{4x}{4}\displaystyle =\displaystyle \dfrac{8}{4}Divide both sides of the equation by 4
\displaystyle x\displaystyle =\displaystyle 2Evaluate

Example 4

Solve this equation for x: \dfrac{3x+8}{7} = \dfrac{-3x+12}{3}

Worked Solution
Create a strategy

Multiply both sides of the equation by the product of the denominators and use the distributive law to expand the brackets.

Apply the idea

The product of the denominators 7 and 3 is 21.

\displaystyle 21 \times \dfrac{3x+8}{7}\displaystyle =\displaystyle 21 \times \dfrac{-3x+12}{3}Multiply both sides by 21
\displaystyle 3\left(3x+8\right)\displaystyle =\displaystyle 7\left(-3x+12\right)Cancel out the denominators
\displaystyle 9x +24\displaystyle =\displaystyle -21x+84Expand the brackets
\displaystyle 9x+21x+24\displaystyle =\displaystyle -21x +21x +84Add 21x to both sides
\displaystyle 30x+24\displaystyle =\displaystyle 84Simplify like terms
\displaystyle 30x+24-24\displaystyle =\displaystyle 84-24Subtract 24 from both sides
\displaystyle 30x\displaystyle =\displaystyle 60Evaluate subtraction
\displaystyle \dfrac{30x}{30}\displaystyle =\displaystyle \dfrac{60}{30}Divide both sides by 30
\displaystyle x\displaystyle =\displaystyle 2Evaluate
Idea summary

To solve equations with variables on both sides of the equation, we want to use inverse operations to eliminate the variable from one side of the equation. We can then combine like terms and solve using inverse operations.

Outcomes

MA5.2-8NA

solves linear and simple quadratic equations, linear inequalities and linear simultaneous equations, using analytical and graphical techniques

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