Simplification and Advanced Applications

Theory: When equations include fractions with different denominators, we multiply both sides by the LCM of all denominators to eliminate the fractions and simplify the equation.

Example:

Question: Solve \(\frac{1}{2}x – \frac{1}{5} = \frac{3}{4}\)

Solution: LCM of 2, 5, 4 = 20
Multiply both sides by 20:
\(\Rightarrow 20\left(\frac{1}{2}x – \frac{1}{5}\right) = 20 \cdot \frac{3}{4} \Rightarrow 10x – 4 = 15 \Rightarrow 10x = 19 \Rightarrow x = \frac{19}{10}\)

Theory: Multiplying both sides of an equation by the LCM removes all fractions and simplifies to an integer-based equation.

Example:

Question: Solve \(\frac{3x}{5} – \frac{2x}{7} = \frac{1}{2}\)

Solution: LCM = 70
  \(70\left(\frac{3x}{5} – \frac{2x}{7}\right) = 70 \cdot \frac{1}{2} \Rightarrow 42x – 20x = 35 \Rightarrow 22x = 35 \Rightarrow x = \frac{35}{22}\)

Theory: Use distributive property to eliminate brackets and convert the equation into a simpler linear form.

Example:

Question: Solve \(3(t – 3) = 5(2t + 1)\)

Solution: 3t – 9 = 10t + 5
\(\Rightarrow -9 – 5 = 10t – 3t \Rightarrow -14 = 7t \Rightarrow t = -2\)

Theory: After removing brackets, similar terms on the same side should be combined for clarity and ease of solving.

Example:

Question: Solve \(15(y – 4) – 2(y – 9) + 5(y + 6) = 0\)

Solution: \(15y – 60 – 2y + 18 + 5y + 30 = 0 \Rightarrow 18y – 12 = 0 \Rightarrow y = \frac{12}{18} = \frac{2}{3}\)

Theory: Clear fractions by multiplying through with the LCM, then bring all variables to one side and constants to the other.

Example:

Question: Solve \(\frac{2x}{3} + 1 = \frac{7x}{15} + 9\)

Solution: LCM = 15
15\(\left(\frac{2x}{3} + 1\right) = 15\left(\frac{7x}{15} + 9\right) \Rightarrow 10x + 15 = 7x + 135 \Rightarrow x = 40\)

Theory: This technique simplifies fractional equations into standard linear form, making them easier to solve.

Example:

Question: Solve \(\frac{1}{3}x + \frac{1}{6} = \frac{2}{3}\)

Solution: LCM = 6
6\(\left(\frac{1}{3}x + \frac{1}{6}\right) = 6\cdot\frac{2}{3} \Rightarrow 2x + 1 = 4 \Rightarrow x = \frac{3}{2}\)

Theory: Equations involving decimals can be simplified by multiplying all terms by powers of 10 to remove decimals.

Example:

Question: Solve \(0.25(4f – 3) = 0.05(10f – 9)\)

Solution: LHS = \(1.0f−0.75, RHS = 0.5f−0.45\)
\(f−0.75=0.5f−0.45⇒0.5f=0.3⇒f=0.6\)

Theory: These equations require bracket removal, simplification, LCM application, and variable isolation across multiple steps.

Example:

Question: Solve \(3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17\)

Solution: LHS = \(15z−21−18z+22=−3z+1\)
RHS = \(32z−52−17=32z−69\)
\(-3z + 1 = 32z – 69 \Rightarrow -31z = -70 \Rightarrow z = \frac{70}{31}\)

Theory: Transposition still works for fractional and decimal terms, moving terms with inverse operations.

Example:

Question: Solve \(x + \frac{14}{5} = \frac{3x}{2}\)

Solution: Multiply by 10 to remove denominators: \(10x + 28 = 15x \Rightarrow x = \frac{28}{5}\)

Theory: Such equations combine all levels of complexity: brackets, fractions, and variables on both sides. Solve by applying distributive laws and LCM.

Example:

Question: Solve \(\frac{6x + 1}{3} + 2 = \frac{x – 3}{6}\)

Multiply both sides by 6: \(2(6x + 1) + 12 = x – 3 \Rightarrow 12x + 2 + 12 = x – 3 \Rightarrow 11x = -17 \Rightarrow x = -\frac{17}{11}\)

Theory: Involves a combination of simplification, multiplication, transposition, and fraction handling.

Example:

Question: Solve \(5x – 2(2x – 7) = 2(3x – 1) + \frac{7}{2}\)

Solution: LHS: \(5x−4x+14=x+14\)
RHS: \(6x – 2 + \frac{7}{2}\) = \(6x + \frac{3}{2}\)
\(x + 14 = 6x + \frac{3}{2} \Rightarrow 14 – \frac{3}{2} = 5x \Rightarrow x = \frac{5}{2}\)