Solving Linear Equations with Fractional Coefficients

Overview of Linear Equations with Fractional Coefficients

  • Linear equations in one variable are algebraic statements of equality where the unknown quantity is raised strictly to the first power.
  • When rational numbers appear as coefficients or constants, the equations take the linear form:   ax+b=ca x + b = c   where aa, bb, and cc are rational numbers and a≠0a \neq 0.
  • Solving equations containing fractions requires maintaining equivalence across the equality sign by systematically applying inverse arithmetic operations (Addition, Subtraction, Multiplication, and Division Properties of Equality).

Linear equations with fractional coefficients

Systematic Methods for Solving Fractional Equations

  • Clearing Denominators Using the Least Common Denominator (LCD):
    • Determine the least common multiple of all integer denominators present across every term on both sides of the equation.
    • Multiply every individual term on both sides of the equation by the LCDLCD.
    • Cancel each denominator against the common multiplier to produce an equivalent equation consisting entirely of integer coefficients.
  • Direct Combination of Rational Terms:
    • If fractional terms containing the variable share an identical denominator, combine their numerators directly across the common denominator.
    • Isolate the variable term on one side of the equation using standard inverse operations.
    • Multiply both sides by the reciprocal of the fractional coefficient to isolate the variable.
  • Verification by Direct Substitution:
    • Substitute the solved value back into the original, unaltered algebraic expression.
    • Evaluate the left-hand side (LHSLHS) and right-hand side (RHSRHS) independently to verify arithmetic equality.

Detailed Solution: Problem 30

  • Original Algebraic Statement:b13−3b13=813\frac{b}{13} - \frac{3b}{13} = \frac{8}{13}
  • Structural Analysis:
    • The equation contains a single unknown variable, bb.
    • Every term on both the left-hand side and right-hand side shares a common denominator of 1313.
  • Solution via Direct Combination:
    • Group the numerators over the common denominator:     b−3b13=813\frac{b - 3b}{13} = \frac{8}{13}
    • Combine like terms in the numerator (1b−3b=−2b1b - 3b = -2b):     −2b13=813\frac{-2b}{13} = \frac{8}{13}
    • Multiply both sides by 1313 to eliminate the denominator:     −2b=8-2b = 8
    • Divide both sides by the coefficient −2-2:     b=8−2b = \frac{8}{-2}b=−4b = -4
  • Solution via Clearing Denominators (LCDLCD Method):
    • The least common denominator of all terms is 1313.
    • Multiply every term across the equation by 1313:     13×(b13)−13×(3b13)=13×(813)13 \times \left(\frac{b}{13}\right) - 13 \times \left(\frac{3b}{13}\right) = 13 \times \left(\frac{8}{13}\right)
    • Simplify each term:     b−3b=8b - 3b = 8
    • Combine like terms:     −2b=8-2b = 8
    • Divide by −2-2:     b=−4b = -4
  • Step-by-Step Verification:
    • Substitute b=−4b = -4 into the left-hand side of the original equation:     LHS=−413−3(−4)13LHS = \frac{-4}{13} - \frac{3(-4)}{13}LHS=−413−−1213LHS = \frac{-4}{13} - \frac{-12}{13}LHS=−4−(−12)13LHS = \frac{-4 - (-12)}{13}LHS=−4+1213LHS = \frac{-4 + 12}{13}LHS=813LHS = \frac{8}{13}
    • Compare with the right-hand side:     RHS=813RHS = \frac{8}{13}
    • Because LHS=RHSLHS = RHS, the solution b=−4b = -4 is verified.

Detailed Solution: Problem 33

  • Original Algebraic Statement:23+3m5=3115\frac{2}{3} + \frac{3m}{5} = \frac{31}{15}
  • Structural Analysis:
    • The equation contains the variable mm.
    • The denominators present in the equation are 33, 55, and 1515.
  • Determination of the Least Common Denominator (LCDLCD):
    • Prime factorizations:     3=33 = 35=55 = 515=3×515 = 3 \times 5
    • The least common multiple is:     LCD(3,5,15)=15\text{LCD}(3, 5, 15) = 15
  • Solution via Clearing Denominators:
    • Multiply each term across the equation by the LCD=15LCD = 15:     15×(23)+15×(3m5)=15×(3115)15 \times \left(\frac{2}{3}\right) + 15 \times \left(\frac{3m}{5}\right) = 15 \times \left(\frac{31}{15}\right)
    • Simplify each product:     5×2+3×3m=1×315 \times 2 + 3 \times 3m = 1 \times 3110+9m=3110 + 9m = 31
    • Isolate the variable term by subtracting 1010 from both sides:     9m=31−109m = 31 - 109m=219m = 21
    • Isolate mm by dividing both sides by 99:     m=219m = \frac{21}{9}
    • Simplify the fraction to lowest terms by dividing the numerator and denominator by their greatest common divisor, 33:     m=73m = \frac{7}{3}
  • Alternative Solution via Direct Fraction Arithmetic:
    • Subtract 23\frac{2}{3} from both sides:     3m5=3115−23\frac{3m}{5} = \frac{31}{15} - \frac{2}{3}
    • Convert 23\frac{2}{3} into an equivalent fraction with denominator 1515:     23=2×53×5=1015\frac{2}{3} = \frac{2 \times 5}{3 \times 5} = \frac{10}{15}
    • Subtract the rational numbers:     3m5=3115−1015=2115\frac{3m}{5} = \frac{31}{15} - \frac{10}{15} = \frac{21}{15}
    • Reduce 2115\frac{21}{15} by dividing numerator and denominator by 33:     3m5=75\frac{3m}{5} = \frac{7}{5}
    • Multiply both sides by 55:     3m=73m = 7
    • Divide both sides by 33:     m=73m = \frac{7}{3}
  • Step-by-Step Verification:
    • Substitute m=73m = \frac{7}{3} into the left-hand side of the original equation:     LHS=23+3(73)5LHS = \frac{2}{3} + \frac{3\left(\frac{7}{3}\right)}{5}LHS=23+75LHS = \frac{2}{3} + \frac{7}{5}
    • Express both fractions with the common denominator of 1515:     LHS=2×515+7×315LHS = \frac{2 \times 5}{15} + \frac{7 \times 3}{15}LHS=1015+2115LHS = \frac{10}{15} + \frac{21}{15}LHS=3115LHS = \frac{31}{15}
    • Compare with the right-hand side:     RHS=3115RHS = \frac{31}{15}
    • Because LHS=RHSLHS = RHS, the solution m=73m = \frac{7}{3} is confirmed.

Detailed Solution: Problem 36

  • Original Algebraic Statement:14+4x5=1120\frac{1}{4} + \frac{4x}{5} = \frac{11}{20}
  • Structural Analysis:
    • The equation contains the variable xx.
    • The denominators present in the equation are 44, 55, and 2020.
  • Determination of the Least Common Denominator (LCDLCD):
    • Prime factorizations:     4=224 = 2^25=55 = 520=22×520 = 2^2 \times 5
    • The least common multiple is:     LCD(4,5,20)=20\text{LCD}(4, 5, 20) = 20
  • Solution via Clearing Denominators:
    • Multiply every term in the equation by the LCD=20LCD = 20:     20×(14)+20×(4x5)=20×(1120)20 \times \left(\frac{1}{4}\right) + 20 \times \left(\frac{4x}{5}\right) = 20 \times \left(\frac{11}{20}\right)
    • Simplify each product:     5×1+4×4x=1×115 \times 1 + 4 \times 4x = 1 \times 115+16x=115 + 16x = 11
    • Isolate the variable term by subtracting 55 from both sides:     16x=11−516x = 11 - 516x=616x = 6
    • Isolate xx by dividing both sides by 1616:     x=616x = \frac{6}{16}
    • Simplify the fraction to lowest terms by dividing the numerator and denominator by their greatest common divisor, 22:     x=38x = \frac{3}{8}
  • Alternative Solution via Fraction Arithmetic:
    • Subtract 14\frac{1}{4} from both sides:     4x5=1120−14\frac{4x}{5} = \frac{11}{20} - \frac{1}{4}
    • Convert 14\frac{1}{4} to an equivalent fraction with denominator 2020:     14=1×54×5=520\frac{1}{4} = \frac{1 \times 5}{4 \times 5} = \frac{5}{20}
    • Subtract the fractions:     4x5=1120−520=620\frac{4x}{5} = \frac{11}{20} - \frac{5}{20} = \frac{6}{20}
    • Reduce 620\frac{6}{20} to lowest terms:     4x5=310\frac{4x}{5} = \frac{3}{10}
    • Multiply both sides by the reciprocal of 45\frac{4}{5}, which is 54\frac{5}{4}:     x=310×54x = \frac{3}{10} \times \frac{5}{4}x=1540x = \frac{15}{40}
    • Divide numerator and denominator by 55:     x=38x = \frac{3}{8}
  • Step-by-Step Verification:
    • Substitute x=38x = \frac{3}{8} into the left-hand side of the original equation:     LHS=14+4(38)5LHS = \frac{1}{4} + \frac{4\left(\frac{3}{8}\right)}{5}LHS=14+1285LHS = \frac{1}{4} + \frac{\frac{12}{8}}{5}LHS=14+325LHS = \frac{1}{4} + \frac{\frac{3}{2}}{5}LHS=14+310LHS = \frac{1}{4} + \frac{3}{10}
    • Express both terms with the common denominator of 2020:     LHS=520+620LHS = \frac{5}{20} + \frac{6}{20}LHS=1120LHS = \frac{11}{20}
    • Compare with the right-hand side:     RHS=1120RHS = \frac{11}{20}
    • Because LHS=RHSLHS = RHS, the solution x=38x = \frac{3}{8} is confirmed.

Common Algebraic Pitfalls and Best Practices

  • Incomplete Multiplication Across Terms:
    • When multiplying an equation by the LCDLCD, every term on both sides must be multiplied by that factor. Overlooking constant terms or terms that do not have fractional coefficients results in an incorrect equation.
  • Proper Sign Distribution:
    • When a minus sign precedes a fraction, such as in b13−3b13\frac{b}{13} - \frac{3b}{13}, the subtraction applies to the entire numerator. Maintaining correct sign operations prevents widespread sign errors.
  • Representation of Solutions:
    • Fractional solutions must be reduced to their lowest terms using the greatest common divisor (GCDGCD) of the numerator and denominator.
    • Improper fractions, such as m=73m = \frac{7}{3}, are standard and mathematically exact forms; converting to repeating or rounded decimals introduces precision loss.