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10&10a

3.01 Solving linear equations

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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.

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.

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 operation to both sides of the equation. This way, both sides of the equation will be equal once we solve the equation.

To solve an equation we apply inverse operations to isolate the variable.

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 need to first expand both sets of brackets before collecting like terms. We can then solve the equation by performing inverse operations.

Examples

Example 2

Solve the following equation: 2\left(3x-5\right)+3\left(4x+6\right)=62

Worked Solution
Create a strategy

Expand each set of brackets and use the inverse operations to solve for x.

Apply the idea
\displaystyle 2\left(3x-5\right)+3\left(4x+6\right)\displaystyle =\displaystyle 62Write the equation
\displaystyle 6x-10+12x+18\displaystyle =\displaystyle 62Expand each set of brackets
\displaystyle 18x+8\displaystyle =\displaystyle 62Add like terms
\displaystyle 18x+8-8\displaystyle =\displaystyle 62-8Subtract 8 from both sides
\displaystyle 18x\displaystyle =\displaystyle 54Evaluate
\displaystyle \dfrac{18x}{18}\displaystyle =\displaystyle \dfrac{54}{18}Divide both sides by 18
\displaystyle x\displaystyle =\displaystyle 3Evaluate
Idea summary

For an equation with one set of brackets, we can either expand the brackets first or divide both sides of the equation by the term outside the brackets.

For equations with two sets of brackets, we first expand both sets of brackets before collecting like terms.

We can then solve the equation by performing inverse operations.

Equations with fractions

Similarly, if we have an equation with a fraction such as \dfrac{2x}{5}=4, we can just perform inverse operations to solve, but for an equation involving the addition or subtraction of two, or more, fractions we multiply both sides of the equation by the lowest common multiple of the denominators to eliminate the fractions. We can then proceed as normal to solve.

Examples

Example 3

Solve the following equation: \dfrac{2x}{4}-\dfrac{2x}{3}=-5

Worked Solution
Create a strategy

Multiply both sides of the equation by the lowest common multiple of the denominators.

Apply the idea

The lowest common multiple of 4 and 3 is 12.

\displaystyle \dfrac{2x}{4}-\dfrac{2x}{3}\displaystyle =\displaystyle -5Write the equation
\displaystyle 12 \times \dfrac{2x}{4}-12\times \dfrac{2x}{3}\displaystyle =\displaystyle -5\times12 Multiply both sides by 12
\displaystyle 3\times2x - 4\times2x\displaystyle =\displaystyle -5\times12Cancel out the denominators
\displaystyle 6x-8x\displaystyle =\displaystyle -60Evaluate the multiplications
\displaystyle -2x\displaystyle =\displaystyle -60Simplify like terms
\displaystyle \dfrac{-2x}{-2}\displaystyle =\displaystyle \dfrac{-60}{-2}Divide both sides by -2
\displaystyle x\displaystyle =\displaystyle 30Evaluate
Idea summary

For an equation involving the addition or subtraction of two, or more, fractions we multiply both sides of the equation by the lowest common multiple of the denominators to eliminate the fractions.

Equations with pronumerals on both sides

To solve equations with pronumerals on both sides of the equation, we want to move all pronumerals to one side of the equation. We can then collect like terms and solve using inverse operations.

Examples

Example 4

Solve the following equation: 3x+6=3\left(5x-4\right)+42

Worked Solution
Create a strategy

Expand the brackets and move the pronumerals to one side of the equation.

Apply the idea
\displaystyle 3x+6\displaystyle =\displaystyle 3\left(5x-4\right)+42Write the equation
\displaystyle 3x+6\displaystyle =\displaystyle 15x-12+42Expand the brackers
\displaystyle 3x+6\displaystyle =\displaystyle 15x+30Add like terms
\displaystyle 3x+6-3x\displaystyle =\displaystyle 15x+30-3xSubtract 3x from both sides
\displaystyle 6\displaystyle =\displaystyle 12x+30Subtract like terms
\displaystyle 6-30\displaystyle =\displaystyle 12x+30-30Subtract 30 from both sides
\displaystyle -24\displaystyle =\displaystyle 12xEvaluate the differences
\displaystyle \dfrac{-24}{12}\displaystyle =\displaystyle \dfrac{12x}{12}Divide both side by 12
\displaystyle x\displaystyle =\displaystyle -2Evaluate
Idea summary

To solve equations with pronumerals on both sides of the equation, we move all pronumerals to one side of the equation. We can then collect like terms and solve using inverse operations.

Outcomes

ACMNA235

Solve problems involving linear equations, including those derived from formulas

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