Section B — Chapters 23–39

Algebra: Verifying Solutions to Linear Equations

Junior Cycle — 1st Year

✓By the end of this lesson students will be able to understand what a linear equation is.
✓By the end of this lesson students will be able to substitute given values into algebraic expressions and equations.
✓By the end of this lesson students will be able to verify if a given pair of values is a solution to a system of two linear equations.
✓By the end of this lesson students will be able to solve simple linear equations in one variable.
✓By the end of this lesson students will be able to formulate and solve simple word problems leading to linear equations.

Key concepts

Linear Equation

A linear equation is an equation where the highest power of any variable is 1. For example, x + 5 = 10 or 2x + 3y = 7 are linear equations. Equations like x² = 9 are not linear because the variable x is raised to the power of 2.

Solution to an Equation

A solution to an equation is a value (or set of values) for the variable(s) that makes the equation true when substituted. For example, in the equation x + 3 = 7, the solution is x = 4 because 4 + 3 = 7.

Verifying a Solution

Verifying a solution means checking if a given value or set of values actually satisfies an equation. You do this by substituting the given values into the equation and seeing if the left-hand side (LHS) equals the right-hand side (RHS).

Simultaneous Equations (Introduction)

Simultaneous equations are a set of two or more equations that share the same variables. We look for values for these variables that satisfy *all* the equations at the same time. In 1st Year, we focus on verifying given solutions. The algebraic methods for finding these solutions (e.g., elimination or substitution methods) are covered in later years of the Junior Cycle.

Solving Simple Linear Equations

To solve a simple linear equation in one variable, you need to isolate the variable on one side of the equals sign. You do this by performing the same operation (addition, subtraction, multiplication, division) to both sides of the equation to maintain balance.

Key facts to remember

1A linear equation has variables raised to the power of 1.
2A solution makes an equation true when substituted.
3To verify a solution, substitute the values and check if LHS = RHS.
4For simultaneous equations, a solution must satisfy *all* equations.
5To solve a linear equation, use inverse operations to isolate the variable.
6Whatever you do to one side of an equation, you must do to the other side.

Worked examples

Example 1

Verify if x = 5 is a solution to the equation 3x - 7 = 8.

ISubstitute x = 5 into the equation:

IILHS = 3(5) - 7

IIILHS = 15 - 7

IVLHS = 8

VRHS = 8

VISince LHS = RHS, x = 5 is a solution.

Answer

x = 5 is a solution.

Always show both sides of the equation when verifying.

Example 2

Verify if (x = 2, y = 3) is a solution to the following pair of equations:\nEquation 1: x + y = 5\nEquation 2: 2x - y = 1

IFor Equation 1: x + y = 5

IISubstitute x = 2 and y = 3:

IIILHS = 2 + 3

IVLHS = 5

VRHS = 5

VISince LHS = RHS, (2, 3) satisfies Equation 1.

VII

VIIIFor Equation 2: 2x - y = 1

9Substitute x = 2 and y = 3:

10LHS = 2(2) - 3

11LHS = 4 - 3

12LHS = 1

13RHS = 1

14Since LHS = RHS, (2, 3) satisfies Equation 2.

16As (2, 3) satisfies both equations, it is a solution to the pair of simultaneous equations.

Answer

(x = 2, y = 3) is a solution to the pair of equations.

For simultaneous equations, the solution must work for *all* equations.

Example 3

Solve the equation 4x + 6 = 18.

I4x + 6 = 18

IISubtract 6 from both sides to isolate the term with x:

III4x + 6 - 6 = 18 - 6

IV4x = 12

VDivide both sides by 4 to find x:

VI4x / 4 = 12 / 4

VIIx = 3

Answer

x = 3

Remember to perform the same operation on both sides to keep the equation balanced.

Example 4

A bag contains some red and blue marbles. There are 5 more red marbles than blue marbles. If there are 12 red marbles, how many blue marbles are there?

ILet 'b' represent the number of blue marbles.

IIThe problem states there are 5 more red marbles than blue marbles. So, Red marbles = Blue marbles + 5.

IIIWe are given that there are 12 red marbles.

IVSo, we can write the equation: 12 = b + 5

VTo find 'b', subtract 5 from both sides:

VI12 - 5 = b + 5 - 5

VII7 = b

Answer

There are 7 blue marbles.

Define your variables clearly before setting up the equation.

Common mistakes

✗Making calculation errors during substitution.
✗Not checking the proposed solution in *all* equations when verifying simultaneous equations.
✗Forgetting to perform the same operation on both sides of the equation when solving.
✗Confusing addition/subtraction with multiplication/division when isolating a variable.
✗Not defining variables clearly in word problems.

Exam tips

★Always show your substitution steps clearly when verifying a solution.
★When solving equations, write down each step of your working to avoid errors and gain partial marks.
★For word problems, read the question carefully and identify what you need to find. Define your variables.
★After solving an equation, substitute your answer back into the original equation to check if it works.

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