Examkin

Question

Difficulty: MediumLinear Equations in One Variable

In the equation below, aa is a constant.

4(x−2)−a(3−2x)=8x−54(x - 2) - a(3 - 2x) = 8x - 5

If the equation has no solution for xx, what is the value of aa?

  1. A
    -2
  2. 2Answer
  3. C
    4
  4. D
    6

Answer

2
For the linear equation to have no solution, the coefficients of the xx terms on both sides of the equation must be equal, while the constant terms must be different. First, expand the left side of the equation: 4(x−2)−a(3−2x)=4x−8−3a+2ax4(x - 2) - a(3 - 2x) = 4x - 8 - 3a + 2ax. Grouping the xx terms and constant terms gives (4+2a)x−(8+3a)=8x−5(4 + 2a)x - (8 + 3a) = 8x - 5. Setting the coefficient of xx on the left side equal to the coefficient of xx on the right side gives 4+2a=84 + 2a = 8. Solving for aa yields 2a=42a = 4, which simplifies to a=2a = 2. Substituting a=2a = 2 back into the equation yields 8x−14=8x−58x - 14 = 8x - 5, or −14=−5-14 = -5, which is a false statement with no solution. Therefore, the value of aa is 22.

Step-by-Step Solution

1
Expand and simplify the left side of the equation.
The left side expands to 4x−8−3a+2ax4x - 8 - 3a + 2ax. Grouping the terms by the variable xx gives (4+2a)x−(8+3a)(4 + 2a)x - (8 + 3a).
To find when the linear equation has no solution, we need to rewrite it in the standard form px+q=rx+spx + q = rx + s by distributing terms and grouping like terms.
2
Equate the coefficients of the xx terms from both sides of the equation.
4+2a=84 + 2a = 8
A linear equation of the form px+q=rx+spx + q = rx + s has no solution if the coefficients of the variable are equal (p=rp = r) but the constants are not (q≠sq \neq s).
3
Solve for the constant aa.
2a=4  ⟹  a=22a = 4 \implies a = 2
Subtracting 44 from both sides of 4+2a=84 + 2a = 8 isolates the variable term, and dividing by 22 gives the value of aa.
4
Verify that the constants are different when a=2a = 2.
Substituting a=2a = 2 back into the constant terms yields a left-side constant of −(8+3(2))=−14-(8 + 3(2)) = -14 and a right-side constant of −5-5. Since −14≠−5-14 \neq -5, the equation has no solution.
If the constant terms were also equal, the equation would have infinitely many solutions instead of no solution.

Key Concept

Linear Equations in One Variable (No Solution Case)
Rate this question

Topics

Linear Equations in One Variable

More questions from this topic