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Difficulty: Very hardLinear Equations in One Variable
In the equation below, aa and bb are constants.
a(x−1)−2(3x+b)3=4(x−5)−a(b−x)6\frac{a(x - 1) - 2(3x + b)}{3} = \frac{4(x - 5) - a(b - x)}{6}
If the equation has infinitely many solutions for xx, what is the value of a−ba - b?
  1. A
    9
  2. B
    10
  3. 15Answer
  4. D
    17

Answer

The value of a−ba - b is 15.
The correct answer is 15. Multiplying the entire equation by 6 to clear the denominators, distributing the terms, and grouping them yields (2a−12)x−(2a+4b)=(a+4)x−(ab+20)(2a - 12)x - (2a + 4b) = (a + 4)x - (ab + 20). For the equation to have infinitely many solutions, the coefficients of xx on both sides must be equal (2a−12=a+42a - 12 = a + 4), and the constant terms must also be equal (2a+4b=ab+202a + 4b = ab + 20). Solving this system of equations yields a=16a = 16 and b=1b = 1. The difference a−ba - b is 16−1=1516 - 1 = 15.

Step-by-Step Solution

1
Eliminate the denominators by multiplying both sides of the equation by 6.
2[a(x−1)−2(3x+b)]=4(x−5)−a(b−x)2[a(x - 1) - 2(3x + b)] = 4(x - 5) - a(b - x)
Multiplying both sides by the least common multiple of 3 and 6 simplifies the equation by removing the fractions.
2
Expand both sides of the equation using the distributive property.
2[ax−a−6x−2b]=4x−20−ab+ax2[ax - a - 6x - 2b] = 4x - 20 - ab + ax which simplifies to 2ax−2a−12x−4b=ax+4x−ab−202ax - 2a - 12x - 4b = ax + 4x - ab - 20
Distributing the constants outside the parentheses allows us to group like terms.
3
Group the xx-terms and constant terms on each side of the equation.
(2a−12)x−(2a+4b)=(a+4)x−(ab+20)(2a - 12)x - (2a + 4b) = (a + 4)x - (ab + 20)
Structuring the equation in the form Ax+B=Cx+DAx + B = Cx + D makes it easier to compare the coefficients.
4
Set up a system of equations by equating the coefficients of xx and the constant terms from both sides.
2a−12=a+42a - 12 = a + 4 and 2a+4b=ab+202a + 4b = ab + 20
For a linear equation to have infinitely many solutions, the coefficient of xx on both sides must be equal, and the constant terms on both sides must also be equal.
5
Solve the first equation for aa.
a=16a = 16
Subtracting aa and adding 12 to both sides isolates the variable aa.
6
Substitute a=16a = 16 into the second equation and solve for bb.
2(16)+4b=16b+20  ⟹  32+4b=16b+20  ⟹  12=12b  ⟹  b=12(16) + 4b = 16b + 20 \implies 32 + 4b = 16b + 20 \implies 12 = 12b \implies b = 1
Substituting the known value of aa leaves a single linear equation in terms of bb.
7
Calculate the value of a−ba - b.
16−1=1516 - 1 = 15
This is the final value requested by the question.

Key Concept

Solving linear equations in one variable with infinitely many solutions by equating coefficients on both sides of the equation.
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Topics

Linear Equations in One Variable

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