Lesson 4: Solving Linear Equations With Brackets
Lesson 4: Solving Linear Equations With Brackets
Solve multi-step equations that include brackets.
In this lesson you will learn what a linear equation is and what it means to “solve an equation”. In order to solve an equation, we need to complete inverse operations in the reverse BEDMAS order. The solution or root is the value that we get when we solve an equation.
This lesson explains the steps of solving linear equations with brackets. From simplifying the expressions with brackets on both sides of the equal sign; identifying, rearranging and collecting like terms to isolating for the unknown variable by dividing by its coefficient.
In this lesson you will see a concrete example of how to solve an equation that contains brackets, does not initially look like a linear equation, but simplifies to it.
Practice exercises will help you master the skills required for solving linear equations with brackets and multi-step equations.
Solving Linear Equations Short Animation
Key ideas: Solving Linear Equations With Brackets
A linear equation has a variable with an exponent of 1, such as 3(x − 4) = 2x + 5. Solving means finding the value of the variable that makes both sides equal; that value is the solution (or root).
Steps for equations with brackets:
- Expand every bracket using the distributive property.
- Collect like terms on each side.
- Move the variable terms to one side and the constants to the other, using inverse operations (do the same thing to both sides).
- Divide by the coefficient of the variable.
- Check by substituting the solution into the original equation.
If the equation has fractions, multiply every term on both sides by the lowest common denominator first.
Worked examples
Example 1
Solve 3(x − 4) = 2x + 5.
- Expand: 3x − 12 = 2x + 5.
- Subtract 2x from both sides: x − 12 = 5.
- Add 12: x = 17.
- Check: 3(17 − 4) = 39 and 2(17) + 5 = 39 ✓
Example 2
Solve 5(2y + 1) − 3(y − 2) = 4(y + 5).
- Expand: 10y + 5 − 3y + 6 = 4y + 20.
- Simplify: 7y + 11 = 4y + 20.
- Subtract 4y and 11: 3y = 9, so y = 3.
- Check: 5(7) − 3(1) = 32 and 4(8) = 32 ✓
Example 3
Solve (x + 2)/3 = (x − 1)/2.
- Multiply both sides by 6: 2(x + 2) = 3(x − 1).
- 2x + 4 = 3x − 3, so x = 7.
- Check: 9/3 = 3 and 6/2 = 3 ✓
Example 4
How many solutions does each equation have? a) 3y + 18 = 3(y + 6) b) 2(x + 1) = 2x + 5
- a) Expand: 3y + 18 = 3y + 18. Both sides are identical, so the equation is true for every y: infinitely many solutions.
- b) Expand: 2x + 2 = 2x + 5, which leads to 2 = 5. That is never true: no solution.
Common mistakes
- Sign errors when a negative is in front of a bracket: −3(y − 2) = −3y + 6, not −3y − 6.
- Moving a term to the other side without changing its sign.
- Multiplying only some terms by the common denominator.
- Skipping the check: substituting takes 30 seconds and catches most errors.
Questions students ask
What does “no solution” mean?
The variable disappears and you are left with a false statement such as 2 = 5. No value of the variable can make the two sides equal.
Does it matter which side I move the variable to?
No. Many students move it to the side where its coefficient stays positive, to avoid sign errors.
Grade 9 Solving Equations Quiz
Check your understanding
Try the quiz for this lesson. You’ll see right away if each answer is correct.