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Linear Equations in One Variable - Reducing Equations to Simpler Form

Grade 8CBSE

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

πŸ”‘Concepts

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Reducing Equations to Simpler Form: Many linear equations contain fractions or parentheses that make them look complex. The primary goal is to simplify both the Left Hand Side (LHS) and the Right Hand Side (RHS) before solving. Imagine a balance scale where you first organize the weights on each side into single piles of 'variables' and 'numbers' before trying to find the value of one variable.

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Clearing Denominators using LCM: When an equation has fractions, you can eliminate them by multiplying every term on both sides by the Least Common Multiple (LCM) of all the denominators. Visually, this transforms an equation with multiple levels (fractions) into a single horizontal line of integers, which is much easier to manage.

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The Distributive Property: If an equation contains terms like a(x+b)a(x + b), you must multiply the outside term aa by every term inside the parentheses to get ax+abax + ab. Think of this as 'distributing' a factor equally to everything inside a container to break the container open.

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Transposition Method: This involves moving terms from one side of the equality sign to the other. When a term moves across the == sign, its sign changes: addition becomes subtraction (++ becomes βˆ’-) and subtraction becomes addition (βˆ’- becomes ++). In your mind, see the term 'jumping' over the equal sign and flipping its operational symbol.

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Cross-Multiplication: For equations in the form ax+bcx+d=pq\frac{ax + b}{cx + d} = \frac{p}{q}, we use cross-multiplication where the numerator of the first side is multiplied by the denominator of the second side, and vice versa. This creates an 'X' shape across the equals sign, resulting in the equation q(ax+b)=p(cx+d)q(ax + b) = p(cx + d).

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Variables on Both Sides: If the variable xx appears on both sides, use transposition to collect all terms containing xx on one side (usually the LHS) and all constant numbers on the other side (RHS). This creates a clear path to isolate the variable.

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Verification of the Result: After finding the value of the variable, it is important to check the answer by substituting it back into the original equation. If the calculated LHSLHS equals the RHSRHS, the solution is correct. This is like a final audit to ensure the balance scale is perfectly level.

πŸ“Formulae

Standard form of a linear equation: ax+b=0ax + b = 0

Distributive Law: a(b+c)=ab+aca(b + c) = ab + ac

Transposition Rule: x+a=bβ€…β€ŠβŸΉβ€…β€Šx=bβˆ’ax + a = b \implies x = b - a

Multiplication/Division Rule: ax=bβ€…β€ŠβŸΉβ€…β€Šx=baax = b \implies x = \frac{b}{a}

Cross Multiplication: ab=cdβ€…β€ŠβŸΉβ€…β€Šad=bc\frac{a}{b} = \frac{c}{d} \implies ad = bc

LCM Method for fractions: LCM(d1,d2,...)Γ—eachΒ termLCM(d_1, d_2, ...) \times \text{each term}

πŸ’‘Examples

Problem 1:

Solve the equation: 6x+13+1=xβˆ’36\frac{6x + 1}{3} + 1 = \frac{x - 3}{6}

Solution:

Step 1: Find the LCM of the denominators 33 and 66, which is 66. Step 2: Multiply every term by 66 to clear the fractions: 6(6x+13)+6(1)=6(xβˆ’36)6 \left( \frac{6x + 1}{3} \right) + 6(1) = 6 \left( \frac{x - 3}{6} \right) Step 3: Simplify the terms: 2(6x+1)+6=xβˆ’32(6x + 1) + 6 = x - 3 Step 4: Expand the brackets: 12x+2+6=xβˆ’312x + 2 + 6 = x - 3 12x+8=xβˆ’312x + 8 = x - 3 Step 5: Transpose xx to LHS and 88 to RHS: 12xβˆ’x=βˆ’3βˆ’812x - x = -3 - 8 11x=βˆ’1111x = -11 Step 6: Solve for xx: x=βˆ’1111=βˆ’1x = \frac{-11}{11} = -1

Explanation:

To simplify, we first cleared the denominators using the LCM of 33 and 66. This converted the fractional equation into a linear one. We then used the distributive property and transposition to isolate xx.

Problem 2:

Solve for zz: zβˆ’53=zβˆ’35\frac{z - 5}{3} = \frac{z - 3}{5}

Solution:

Step 1: Use cross-multiplication: 5(zβˆ’5)=3(zβˆ’3)5(z - 5) = 3(z - 3) Step 2: Expand both sides using the distributive property: 5zβˆ’25=3zβˆ’95z - 25 = 3z - 9 Step 3: Transpose 3z3z to the LHS and βˆ’25-25 to the RHS: 5zβˆ’3z=βˆ’9+255z - 3z = -9 + 25 Step 4: Simplify both sides: 2z=162z = 16 Step 5: Solve for zz: z=162z = \frac{16}{2} z=8z = 8

Explanation:

Since the equation is a ratio on both sides, cross-multiplication is the fastest way to reduce it to a simpler form. After expanding the brackets, we grouped the variable terms and constant terms to find the solution.