Linear Equations in One Variable Class 8: Step-by-Step Solutions

Imagine you know that three identical chocolate bars and a Rs. 10 note together cost Rs. 55. How much does each chocolate bar cost? You'd set up the equation $3x + 10 = 55$ and solve for $x$. That's a linear equation in one variable!

In NCERT Class 8 Math (Chapter 2: Linear Equations in One Variable), you level up from the simple equations you learned in Class 7. Now you'll handle equations with variables on both sides, equations involving fractions, and more challenging word problems. These skills are absolutely essential for algebra in Classes 9 and 10.

A linear equation in one variable is an equation of the form:

The word "linear" means the variable $x$ appears only to the first power (no $x^2$, $x^3$, or $\sqrt{x}$). The word "one variable" means there's only one unknown.

Examples of linear equations:
- $2x + 5 = 13$
- $7y - 3 = 4y + 9$
- $\frac{x+1}{3} = \frac{2x-1}{5}$

Not linear equations:
- $x^2 + 3x = 7$ (has $x^2$, so it's quadratic)
- $2x + 3y = 10$ (has two variables)

The solution of a linear equation is the value of the variable that makes both sides equal.

Think of an equation as a weighing balance. Both sides must always be equal. You can do anything to one side, as long as you do the exact same thing to the other side.

The key operations:
- Add the same number to both sides.
- Subtract the same number from both sides.
- Multiply both sides by the same non-zero number.
- Divide both sides by the same non-zero number.

Transposing is a shortcut: when you move a term from one side to the other, you change its sign.

Always verify your answer by substituting back: $2(4) + 5 = 8 + 5 = 13$. Correct!

This is the big upgrade in Class 8. When the variable appears on both sides, your goal is to collect all variable terms on one side and all constant terms on the other.

Example 1: Solve $5x - 3 = 3x + 7$.

Verify: LHS $= 5(5) - 3 = 22$. RHS $= 3(5) + 7 = 22$. Correct!

Example 2: Solve $8x + 4 = 3(x - 1) + 7$.

First, expand the bracket:

Verify: LHS $= 8(0) + 4 = 4$. RHS $= 3(0-1) + 7 = -3 + 7 = 4$. Correct!

Example 3: Solve $2(3x - 1) = 10 - (2x - 5)$.

Fractions as answers are perfectly normal! Don't assume you've made a mistake just because the answer isn't a whole number.

Many CBSE questions involve fractions and require simplification before solving. Here's the systematic approach:

Example 1: Solve $\frac{x+1}{3} = \frac{2x-1}{5}$.

Step 1: Cross-multiply to eliminate fractions.

Step 2: Expand.

Step 3: Solve.

Verify: LHS $= \frac{8+1}{3} = \frac{9}{3} = 3$. RHS $= \frac{2(8)-1}{5} = \frac{15}{5} = 3$. Correct!

Example 2: Solve $\frac{3x}{4} - \frac{x-2}{3} = \frac{5}{6}$.

Step 1: Find LCM of $4, 3, 6$, which is $12$. Multiply every term by $12$.

Step 2: Expand and solve.

The trick of multiplying by the LCM to clear all fractions at once is a powerful technique. Use it whenever you see multiple fractions in an equation!

Word problems are where linear equations truly shine. The key is translating the problem into an equation, then solving. Here's a systematic approach:

1. Read the problem carefully. Identify what's unknown.
2. Let the unknown be $x$ (or any variable).
3. Translate each sentence into a mathematical expression.
4. Set up the equation.
5. Solve and verify the answer makes sense.

Example 1: Age Problem
The sum of a father's and son's ages is $50$. Five years ago, the father's age was $4$ times the son's age. Find their present ages.

Let the son's present age $= x$. Then father's age $= 50 - x$.

Five years ago: son was $(x - 5)$, father was $(50 - x - 5) = (45 - x)$.

Son is $13$ years old, father is $37$ years old. Check: $5$ years ago, son was $8$ and father was $32 = 4 \times 8$. Correct!

Example 2: Number Problem
The sum of two consecutive odd numbers is $56$. Find them.

Let the numbers be $x$ and $x + 2$.

The numbers are $27$ and $29$. Check: $27 + 29 = 56$. Correct!

Example 3: Perimeter Problem
The length of a rectangle is $5$ cm more than twice its breadth. If the perimeter is $46$ cm, find the dimensions.

Let breadth $= x$. Then length $= 2x + 5$.

Breadth $= 6$ cm, length $= 17$ cm. Check: $2(6 + 17) = 2(23) = 46$. Correct!

Linear equations aren't just exam material. They model real situations:

• Shopping: If apples cost Rs. $x$ per kg and you buy $3$ kg plus a bag for Rs. $10$, and the total is Rs. $130$, then $3x + 10 = 130$ gives $x = 40$.
  - Speed-Distance: If a car travels at $60$ km/h and reaches a city $240$ km away, the time taken is $t = \frac{240}{60} = 4$ hours.
  - Mixtures: A chemist mixing solutions of different concentrations uses linear equations to find the right proportions.
  - Profit & Loss: Finding the cost price when selling price and profit percentage are known.

Every time you see a relationship between quantities that can be expressed as a first-degree equation, you're using linear equations!

Here's your essential checklist:

• A linear equation in one variable has the form $ax + b = 0$ with $a \neq 0$.
  - Transpose terms by changing their sign when moving across the equals sign.
  - For equations with variables on both sides, collect variables on one side and constants on the other.
  - Clear fractions by multiplying through by the LCM of all denominators.
  - Word problems follow a clear process: read, assign variable, translate, solve, verify.
  - Always check your answer by substituting back.

Ready to solve some equations? Jump into SparkEd's Linear Equations practice and sharpen your skills with adaptive questions that match your level!