Linear Equations in One Variable — Class 8 IMO

Variables on Both Sides & Parentheses

What is a Linear Equation in One Variable? A linear equation in one variable is an equation that can be written in the form \(ax + b = 0\), where \(a\) and \(b\) are constants, \(a \neq 0\), and \(x\) is the variable. The highest power of the variable is 1.

Examples:

\(2x + 3 = 7\) (linear)
\(5x - 4 = 6x + 2\) (linear)
\(x^2 + 3 = 5\) (not linear – power is 2)

What does it mean to "solve" an equation? Solving an equation means finding the value(s) of the variable that make the equation true. This value is called the solution or root of the equation.

Rules for Solving Linear Equations:

Rule	Explanation	Example
Addition/Subtraction Property	Add or subtract same number from both sides	\(x - 5 = 10 \Rightarrow x = 15\)
Multiplication/Division Property	Multiply or divide both sides by same non-zero number	\(3x = 12 \Rightarrow x = 4\)
Combining Like Terms	Simplify each side before solving	\(2x + 3x = 5 \Rightarrow 5x = 5\)
Transposition	Move terms from one side to the other (change sign)	\(x + 4 = 10 \Rightarrow x = 10 - 4\)

Step-by-Step Solving Process:

Simplify both sides (combine like terms, remove brackets)
Collect variable terms on one side, constant terms on the other
Isolate the variable using multiplication/division
Check your answer by substituting back

Example 1: Solve 2x + 3 = x + 8.

2x − x = 8 − 3, so x = 5.

Example 2: Solve 3(x − 2) = 9.

3x − 6 = 9, so 3x = 15 and x = 5.

Example 3: Solve: \(3x - 7 = 14\)

- Add 7 to both sides: \(3x - 7 + 7 = 14 + 7\) - \(3x = 21\) - Divide both sides by 3: \(\frac{3x}{3} = \frac{21}{3}\) - \(x = 7\) - Check: \(3(7) - 7 = 21 - 7 = 14\) ✓ - **Answer:** \(x = 7\) *Example 2: Solve: \(5x + 3 = 2x + 15\) Solution: - Subtract \(2x\) from both sides: \(5x - 2x + 3 = 2x - 2x + 15\) - \(3x + 3 = 15\) - Subtract 3 from both sides: \(3x + 3 - 3 = 15 - 3\) - \(3x = 12\) - Divide both sides by 3: \(x = 4\) - Check: LHS = \(5(4)+3=20+3=23\), RHS = \(2(4)+15=8+15=23\) ✓ - **Answer:** \(x = 4\) *Example 3: Solve: \(3(x - 4) + 5 = 2(x + 1) + 1\) Solution: - Expand brackets: \(3x - 12 + 5 = 2x + 2 + 1\) - Simplify: \(3x - 7 = 2x + 3\) - Subtract \(2x\) from both sides: \(3x - 2x - 7 = 2x - 2x + 3\) - \(x - 7 = 3\) - Add 7 to both sides: \(x - 7 + 7 = 3 + 7\) - \(x = 10\) - Check: LHS = \(3(10-4)+5=3(6)+5=18+5=23\), RHS = \(2(10+1)+1=2(11)+1=22+1=23\) ✓ - **Answer:** \(x = 10\)

Quick recap

Collect variables on one side, numbers on the other.
Expand parentheses before solving.
A linear equation has variable with highest power 1
Solving means finding the value that makes the equation true
Perform same operation on both sides to maintain balance
Use transposition to move terms (sign changes when moving across equals sign)
Always check your answer by substituting back into original equation
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✓ Quick check

Solve: 5x = 2x + 12.

3x = 12, so x = 4.

Solve: 2(x + 1) = 10.

2x + 2 = 10, so 2x = 8 and x = 4.

Equations with Fractions

What are Fractional Equations? Fractional equations contain fractions with the variable in the numerator, denominator, or both. To solve them, we eliminate denominators by multiplying both sides by the Least Common Multiple (LCM) of all denominators.

Steps to Solve Fractional Equations:

Find the LCM of all denominators
Multiply both sides of the equation by the LCM
Simplify (cancel denominators)
Solve the resulting linear equation
Check that the solution doesn't make any denominator zero

How to Approach Word Problems:

Step	Action
1	Read the problem carefully
2	Identify what is being asked (the unknown)
3	Choose a variable to represent the unknown
4	Translate the words into a mathematical equation
5	Solve the equation
6	Answer the question (with appropriate units)

Common Word Problem Types:

Age problems: Compare ages now and in the future
Number problems: Find unknown numbers based on given relationships
Money problems: Total cost, change, etc.
Consecutive integers: \(n\), \(n+1\), \(n+2\)
Perimeter/Geometry: Use formulas to set up equations

Example 1: Solve x/2 + x/3 = 5.

Multiply by 6: 3x + 2x = 30, so x = 6.

Example 2: Solve x/4 = 3.

x = 3 × 4 = 12.

Example 3: Solve: \(\frac{x+2}{3} = \frac{x-1}{2}\)

- LCM of 3 and 2 is 6 - Multiply both sides by 6: \(6 \times \frac{x+2}{3} = 6 \times \frac{x-1}{2}\) - Simplify: \(2(x+2) = 3(x-1)\) - Expand: \(2x + 4 = 3x - 3\) - Subtract \(2x\) from both sides: \(4 = x - 3\) - Add 3 to both sides: \(x = 7\) - Check: LHS = \((7+2)/3 = 9/3=3\), RHS = \((7-1)/2 = 6/2=3\) ✓ - **Answer:** \(x = 7\) *Example 2 (Word Problem - Number): The sum of three consecutive integers is 72. Find the integers. Solution: - Let the first integer be \(n\) - Then the three consecutive integers are \(n\), \(n+1\), \(n+2\) - Sum: \(n + (n+1) + (n+2) = 72\) - \(3n + 3 = 72\) - \(3n = 69\) - \(n = 23\) - Integers: \(23, 24, 25\) - Check: \(23+24+25=72\) ✓ - **Answer:** \(23, 24, 25\) *Example 3 (Word Problem - Age): A father is three times as old as his son. In 10 years, he will be twice as old as his son. Find their present ages. Solution: - Let son's present age = \(x\) years - Father's present age = \(3x\) years - In 10 years: son = \(x + 10\), father = \(3x + 10\) - Equation: \(3x + 10 = 2(x + 10)\) - \(3x + 10 = 2x + 20\) - \(3x - 2x = 20 - 10\) - \(x = 10\) (son's age) - Father's age = \(3 \times 10 = 30\) - Check: In 10 years: son=20, father=40 (twice) ✓ - **Answer:** Son is 10 years, Father is 30 years

Quick recap

Clear fractions by multiplying through by the LCM.
Then solve the simpler equation.
To solve fractional equations, multiply by LCM of denominators
Always check that solution doesn't make any denominator zero
Word problems: read carefully, define variable, translate to equation
Consecutive integers: use \(n\), \(n+1\), \(n+2\), etc.
Age problems: express future/past ages by adding/subtracting years
Always verify your answer in the original word problem
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✓ Quick check

Solve: x/3 + 2 = 5.

x/3 = 3, so x = 9.

Solve: (x + 1)/2 = 4.

x + 1 = 8, so x = 7.

Word Problems

What are Simultaneous Equations? Simultaneous equations are two or more equations that share the same variables. They are solved together (simultaneously) to find values that satisfy all equations at the same time.

Example: \(x + y = 10\) \(x - y = 4\)

Why do we need them? Sometimes one equation with two variables has infinitely many solutions. Adding another equation gives a unique solution.

Methods to Solve Simultaneous Equations:

Method	How it works	Best for
Elimination Method	Add/subtract equations to eliminate one variable	When coefficients are friendly
Substitution Method	Solve one equation for one variable, substitute into other	When one variable is easily isolated

Elimination Method Steps:

Align equations so like terms are stacked
Multiply one or both equations to make coefficients of one variable opposites
Add or subtract to eliminate that variable
Solve for the remaining variable
Substitute back to find the other variable

Substitution Method Steps:

Solve one equation for one variable (e.g., \(x = something\))
Substitute this expression into the other equation
Solve for the remaining variable
Substitute back to find the other variable

Example 1: Three consecutive integers add up to 27. Find them.

n + (n+1) + (n+2) = 27, so 3n + 3 = 27, n = 8 → 8, 9, 10.

Example 2: A father is 30 years older than his son and three times as old. Find the son's age.

x + 30 = 3x, so 2x = 30 and x = 15.

Example 3: Solve: \(2x + y = 7\) and \(x - y = 2\)

- Align equations: \(2x + y = 7\) \(x - y = 2\) - Add the equations: \((2x + x) + (y - y) = 7 + 2\) - \(3x = 9 \implies x = 3\) - Substitute \(x = 3\) into \(x - y = 2\): \(3 - y = 2\) - \(-y = -1 \implies y = 1\) - Check in first: \(2(3) + 1 = 6 + 1 = 7\) ✓ - **Answer:** \(x = 3\), \(y = 1\) *Example 2 (Substitution Method): Solve: \(x = 2y + 3\) and \(3x - 2y = 5\) Solution: - First equation gives \(x = 2y + 3\) - Substitute into second: \(3(2y + 3) - 2y = 5\) - \(6y + 9 - 2y = 5\) - \(4y + 9 = 5\) - \(4y = -4 \implies y = -1\) - Substitute back: \(x = 2(-1) + 3 = -2 + 3 = 1\) - Check: \(3(1) - 2(-1) = 3 + 2 = 5\) ✓ - **Answer:** \(x = 1\), \(y = -1\) *Example 3 (Real-Life Application): The sum of two numbers is 25 and their difference is 7. Find the numbers. Solution: - Let the numbers be \(x\) and \(y\) (with \(x > y\)) - Equation 1: \(x + y = 25\) - Equation 2: \(x - y = 7\) - Add equations: \((x+y) + (x-y) = 25 + 7\) - \(2x = 32 \implies x = 16\) - Substitute: \(16 + y = 25 \implies y = 9\) - Check: \(16 - 9 = 7\) ✓ - **Answer:** The numbers are 16 and 9

Quick recap

Translate words into an equation, then solve.
Consecutive integers: n, n+1, n+2.
Simultaneous equations share variables and are solved together
Elimination method: add/subtract equations to remove one variable
Substitution method: solve for one variable, substitute into other
Both methods give the same solution; choose based on convenience
Solution is an ordered pair \((x, y)\) that satisfies both equations
Graphically, the solution is the intersection point of two lines
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✓ Quick check

Two consecutive numbers add up to 15. The smaller one is ___ ?

n + (n+1) = 15, so 2n + 1 = 15 and n = 7.

When 5 is added to a number, the result equals twice the number. The number is ___ ?

x + 5 = 2x, so x = 5.