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Linear Equations in One Variable - Solving Equations with Linear Expressions on one Side and Numbers on the other

Grade 8CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A linear equation in one variable is an algebraic equation where the highest power of the variable is 11, typically written in the form ax+b=cax + b = c. Visually, if you represent the solution on a number line, it corresponds to a single unique point.

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The Equals sign (==) indicates that the value of the expression on the Left Hand Side (LHS) is exactly equal to the value on the Right Hand Side (RHS). Think of this as a balanced weighing scale where both sides carry the same weight.

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The Balance Method involves performing the same mathematical operation (addition, subtraction, multiplication, or division) on both sides of the equation to keep the 'scale' balanced while isolating the variable.

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Transposition is a shortcut for the balance method where a term is moved from one side of the '=' sign to the other with its sign changed. Visually, imagine a number 'jumping' over the equals sign and transforming: plus becomes minus, and minus becomes plus.

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To solve an equation where a linear expression is on one side and a number is on the other, the goal is to isolate the variable (e.g., xx) by removing constants through inverse operations in the reverse order of BODMAS.

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Inverse operations are pairs of operations that 'undo' each other. Addition is the inverse of subtraction, and multiplication is the inverse of division. For example, if a variable is divided by a number, you multiply the RHS by that same number to solve it.

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Verification of the result is the final step where the calculated value of the variable is substituted back into the original LHS. If the resulting LHS equals the RHS, the solution is verified as correct.

📐Formulae

Standard Form: ax+b=cax + b = c

Addition Transposition: x+a=b  ⟹  x=b−ax + a = b \implies x = b - a

Subtraction Transposition: x−a=b  ⟹  x=b+ax - a = b \implies x = b + a

Multiplication Transposition: ax=b  ⟹  x=baax = b \implies x = \frac{b}{a} (where a≠0a \neq 0)

Division Transposition: xa=b  ⟹  x=b×a\frac{x}{a} = b \implies x = b \times a

General Solution for ax+b=cax + b = c: x=c−bax = \frac{c - b}{a}

💡Examples

Problem 1:

Solve the equation: 2x−5=112x - 5 = 11

Solution:

Step 1: Transpose −5-5 from LHS to RHS. It becomes +5+5. 2x=11+52x = 11 + 5 2x=162x = 16 Step 2: Transpose the coefficient 22 (which is multiplied by xx) to the RHS. It becomes a divisor. x=162x = \frac{16}{2} x=8x = 8 Verification: Substitute x=8x = 8 in LHS: 2(8)−5=16−5=112(8) - 5 = 16 - 5 = 11. Since LHS = RHS, the solution is correct.

Explanation:

We first isolate the term containing the variable by moving the constant through addition, and then isolate the variable itself by dividing by its coefficient.

Problem 2:

Solve the equation: x3+52=−32\frac{x}{3} + \frac{5}{2} = -\frac{3}{2}

Solution:

Step 1: Transpose +52+\frac{5}{2} to the RHS. x3=−32−52\frac{x}{3} = -\frac{3}{2} - \frac{5}{2} Step 2: Simplify the RHS since the denominators are the same. x3=−3−52\frac{x}{3} = \frac{-3 - 5}{2} x3=−82\frac{x}{3} = \frac{-8}{2} x3=−4\frac{x}{3} = -4 Step 3: Transpose the divisor 33 to the RHS by multiplying. x=−4×3x = -4 \times 3 x=−12x = -12

Explanation:

First, we eliminate the constant fraction by subtraction. Then, we clear the denominator of the variable by multiplying the entire RHS by that denominator.