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Difficulty: Very hardSolving Linear Equations

For what value of the constant aa does the linear equation a(x−2)3−3x−12=−5x+76\frac{a(x - 2)}{3} - \frac{3x - 1}{2} = -\frac{5x + 7}{6} have no real solution for xx?

Answer: 2

Answer

The constant aa must be equal to 22 for the equation to have no solution.
A linear equation in the form Ax+B=Cx+DAx + B = Cx + D has no solution if the coefficients of the variable are equal (A=CA = C) but the constant terms are different (B≠DB \neq D). Multiplying the given equation by the least common denominator, 6, clears the fractions and yields 2a(x−2)−3(3x−1)=−(5x+7)2a(x - 2) - 3(3x - 1) = -(5x + 7). Expanding both sides results in 2ax−4a−9x+3=−5x−72ax - 4a - 9x + 3 = -5x - 7, which simplifies to (2a−9)x+(3−4a)=−5x−7(2a - 9)x + (3 - 4a) = -5x - 7. Equating the coefficients of xx gives 2a−9=−52a - 9 = -5, which solves to a=2a = 2. Substituting a=2a = 2 back into the constants yields a left-side constant of −5-5 and a right-side constant of −7-7. Since −5≠−7-5 \neq -7, the variable terms cancel out while leaving an inequality, meaning the equation has no solution when a=2a = 2.

Step-by-Step Solution

1
Clear the denominators by multiplying the entire equation by the least common denominator, which is 6.
2a(x−2)−3(3x−1)=−(5x+7)2a(x - 2) - 3(3x - 1) = -(5x + 7)
Multiplying by the LCD simplifies the rational expressions into polynomial terms.
2
Distribute and expand the terms on both sides of the equation.
2ax−4a−9x+3=−5x−72ax - 4a - 9x + 3 = -5x - 7
Expanding the terms allows us to group variable terms and constant terms together.
3
Group the xx terms and constant terms on the left side.
(2a−9)x+(3−4a)=−5x−7(2a - 9)x + (3 - 4a) = -5x - 7
Structuring the equation in the standard form Ax+B=Cx+DAx + B = Cx + D makes it easier to compare coefficients.
4
Set the coefficients of xx on both sides equal to each other.
2a−9=−52a - 9 = -5
For a linear equation to have no solution, the variable terms must cancel out, meaning their coefficients must be identical.
5
Solve for the parameter aa and verify the constant terms are unequal.
2a=4  ⟹  a=22a = 4 \implies a = 2. Constant check: 3−4(2)=−53 - 4(2) = -5, and −5≠−7-5 \neq -7.
If the constant terms were equal, the equation would have infinitely many solutions instead of no solution.

Key Concept

Identifying the parameter value that results in a linear equation having no solution by equating variable coefficients and ensuring constant terms are unequal.
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Topics

Solving Linear Equations

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