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Linear Equations in One Variable - Solving Equations with Variables on both Sides

Grade 8CBSE

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

🔑Concepts

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A linear equation in one variable with variables on both sides is an equality where the unknown (like xx) appears on both the Left-Hand Side (LHS) and the Right-Hand Side (RHS). Visually, think of this as a balance scale where both pans contain unknown weights and some fixed weights.

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The primary goal is to isolate the variable on one side of the equation (usually the LHS) and the constant numbers on the other side (usually the RHS) to reach a form like x=kx = k.

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Transposition is the most common method used; it involves moving terms from one side to another. When a term moves across the '=' sign, its sign changes: a positive term becomes negative (+→−+ \rightarrow -) and a negative term becomes positive (−→+- \rightarrow +).

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If the equation contains terms within parentheses, use the Distributive Property, a(b+c)=ab+aca(b + c) = ab + ac, to simplify the expression before attempting to move terms across the sides.

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The Balance Rule states that any operation (addition, subtraction, multiplication, or division) performed on the LHS must also be performed on the RHS to maintain equality. Visually, if you remove 2kg2kg from the left pan of a balanced scale, you must remove 2kg2kg from the right pan to keep it level.

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Grouping Like Terms involves collecting all terms containing the variable on one side and all constant numbers on the other side. For example, in 5x+3=2x+95x + 3 = 2x + 9, we group 5x5x and 2x2x together and 33 and 99 together.

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Verification of the result is a critical final step. Substitute the calculated value of the variable back into the original equation's LHS and RHS. If LHS=RHSLHS = RHS, the solution is correct.

📐Formulae

General Form: ax+b=cx+dax + b = cx + d

Distributive Property: a(x+b)=ax+aba(x + b) = ax + ab

Transposition of Addition: If x+a=bx + a = b, then x=b−ax = b - a

Transposition of Subtraction: If x−a=bx - a = b, then x=b+ax = b + a

Transposition of Multiplication: If ax=bax = b, then x=bax = \frac{b}{a}

Transposition of Division: If xa=b\frac{x}{a} = b, then x=a×bx = a \times b

💡Examples

Problem 1:

Solve for xx: 8x+4=3(x−1)+78x + 4 = 3(x - 1) + 7

Solution:

Step 1: Open the brackets on the RHS using the distributive property: 8x+4=3x−3+78x + 4 = 3x - 3 + 7

Step 2: Simplify the constants on the RHS: 8x+4=3x+48x + 4 = 3x + 4

Step 3: Transpose 3x3x to the LHS (it becomes −3x-3x): 8x−3x+4=48x - 3x + 4 = 4 5x+4=45x + 4 = 4

Step 4: Transpose +4+4 to the RHS (it becomes −4-4): 5x=4−45x = 4 - 4 5x=05x = 0

Step 5: Divide both sides by 55: x=05x = \frac{0}{5} x=0x = 0

Explanation:

We first simplified the right side by distributing the 33 into the parentheses. Then, we moved all xx terms to the left and all constants to the right to isolate the variable.

Problem 2:

Solve the equation: x2+5=x3+7\frac{x}{2} + 5 = \frac{x}{3} + 7

Solution:

Step 1: Transpose the variable term x3\frac{x}{3} to the LHS: x2−x3+5=7\frac{x}{2} - \frac{x}{3} + 5 = 7

Step 2: Transpose the constant 55 to the RHS: x2−x3=7−5\frac{x}{2} - \frac{x}{3} = 7 - 5 x2−x3=2\frac{x}{2} - \frac{x}{3} = 2

Step 3: Find a common denominator for the LHS (which is 66): 3x−2x6=2\frac{3x - 2x}{6} = 2 x6=2\frac{x}{6} = 2

Step 4: Multiply both sides by 66 to solve for xx: x=2×6x = 2 \times 6 x=12x = 12

Explanation:

This problem involves fractions. By transposing terms first, we grouped variables on one side. Finding the Least Common Multiple (LCM) of the denominators allowed us to combine the fractions into a single variable term.