Examkin

Question

Difficulty: Very hardLinear Equations in One Variable
Consider the linear equation in one variable xx:
2x−33−x−54=x+96\frac{2x - 3}{3} - \frac{x - 5}{4} = \frac{x + 9}{6}
If xx is the solution to this equation, what is the value of 4x−14x - 1?
  1. A
    5
  2. B
    7
  3. C
    11
  4. 19Answer
  5. E
    59

Answer

19
To solve the linear equation, first multiply all terms by the least common denominator, 12, yielding 4(2x - 3) - 3(x - 5) = 2(x + 9). Expanding both sides yields 8x - 12 - 3x + 15 = 2x + 18. Combining like terms on the left side yields 5x + 3 = 2x + 18. Subtracting 2x and 3 from both sides gives 3x = 15, so x = 5. Substituting x = 5 into the target expression 4x - 1 gives 4(5) - 1 = 19.

Step-by-Step Solution

1
Clear the fractional denominators by multiplying every term by the least common multiple (LCM) of 3, 4, and 6, which is 12.
12 \cdot \left(\frac{2x - 3}{3}\right) - 12 \cdot \left(\frac{x - 5}{4}\right) = 12 \cdot \left(\frac{x + 9}{6}\right) \implies 4(2x - 3) - 3(x - 5) = 2(x + 9)
Multiplying every term by 12 eliminates all denominators without changing the equation's solution set.
2
Distribute the factors across the binomial terms inside parentheses.
8x - 12 - 3x + 15 = 2x + 18
Distributing -3 across (x - 5) yields -3x + 15 because the product of two negative numbers is positive.
3
Combine like terms on the left side of the equation.
5x + 3 = 2x + 18
Combining 8x - 3x gives 5x, and -12 + 15 gives +3.
4
Isolate x by subtracting 2x and 3 from both sides of the equation.
3x = 15 \implies x = 5
Subtracting 2x from both sides yields 3x + 3 = 18, then subtracting 3 yields 3x = 15, giving x = 5.
5
Substitute x = 5 into the expression 4x - 1 to find the final value.
4(5) - 1 = 20 - 1 = 19
The question asks for the evaluation of 4x - 1 rather than the value of x alone.

Key Concept

Solving linear equations in one variable involving fractional expressions and evaluating algebraic expressions.
Estimated Time:2m 0s
Rate this question

Topics

Linear Equations in One Variable

More questions from this topic