Critiquing Reasoning about Solution Types

Micah solves a linear equation and concludes that [tex]$x = 0$[/tex] is the solution. His work is shown below.

[tex]
\begin{aligned}
\frac{5}{6}(1-3x) &= 4\left(-\frac{5x}{8} + 2\right) \\
\frac{\frac{5}{6} - \frac{5x}{2}}{+\frac{5x}{2}} &= \frac{-\frac{5x}{2} + 8}{+\frac{5x}{2}} \\
0 &= x
\end{aligned}
[/tex]

Which statement is true about Micah's solution?

A. Micah's solution is wrong. There are no values that make the statement true.
B. Micah's solution is correct, and the value of [tex]$x$[/tex] that makes the statement true is 0.
C. Micah should have divided by [tex]$\frac{5}{2}$[/tex].
D. Micah should have subtracted [tex]$\frac{5}{2}$[/tex].



Answer :

Let's critically analyze Micah's steps and determine if his conclusion is correct and what the correct form of the solution should be.

1. The original equation to solve is:
[tex]\[ \frac{5}{6}(1 - 3x) = 4\left(-\frac{5x}{8} + 2\right) \][/tex]

2. To eliminate the fractions, we clear the denominators by multiplying every term by the least common multiple (LCM) of 6 and 8. The LCM of 6 and 8 is 24:
[tex]\[ 24 \cdot \frac{5}{6}(1 - 3x) = 24 \cdot 4\left(-\frac{5x}{8} + 2\right) \][/tex]

3. Multiply and simplify both sides:
[tex]\[ 4 \cdot 5(1 - 3x) = 96\left(-\frac{5x}{8} + 2\right) \][/tex]
[tex]\[ 20(1 - 3x) = 96\left(-\frac{5x}{8} + 2\right) \][/tex]

4. Simplify inside the parentheses:
[tex]\[ 20 - 60x = 96\left(-\frac{5x}{8}\right) + 192 \][/tex]

5. Distribute and simplify the right-hand side:
[tex]\[ 20 - 60x = -60x + 192 \][/tex]

6. Now, observe that when we simplify the equation further, we notice:
[tex]\[ (20 - 60x) + 60x = (-60x + 192) + 60x \][/tex]
[tex]\[ 20 = 192 \][/tex]

7. This simplifies to a contradiction:
[tex]\[ 20 = 192 \][/tex]

Since the simplification leads to a contradiction, it indicates there is no value of \( x \) that satisfies the given equation. Therefore, Micah's solution is incorrect.

Given this detailed analysis, the correct statement about Micah's solution is:
Micah's solution is wrong. There are no values of [tex]\( x \)[/tex] that make the statement true.