Jerry solved this equation: [tex]3\left(x-\frac{1}{4}\right)=\frac{13}{6}[/tex]

1. [tex]3x-\frac{3}{4}=\frac{13}{6}[/tex]

2. [tex]3x-\frac{3}{4}+\frac{3}{4}=\frac{13}{6}+\frac{3}{4}[/tex]

3. [tex]3x=\frac{26}{12}+\frac{9}{12}[/tex]

4. [tex]3x=\frac{35}{12}[/tex]

5. [tex]\left(\frac{3}{1}\right) \frac{3}{1} x=\frac{35}{12}\left(\frac{3}{1}\right)[/tex]

6. [tex]x=\frac{105}{12}[/tex]

In which step did Jerry make an error?

A. In step 2, he should have subtracted [tex]\frac{3}{4}[/tex] from both sides.
B. In step 3, he should have found an LCD of 10.
C. In step 4, he should have subtracted 9 from 26.
D. In step 5, he should have multiplied both sides by [tex]\frac{1}{3}[/tex].



Answer :

Let's analyze each step Jerry took in solving the equation \( 3\left(x - \frac{1}{4}\right) = \frac{13}{6} \):

1. Starting with the given equation:
[tex]\[ 3\left(x - \frac{1}{4}\right) = \frac{13}{6} \][/tex]
Jerry correctly distributed the 3:
[tex]\[ 3x - \frac{3}{4} = \frac{13}{6} \][/tex]

2. To isolate \( x \), Jerry added \( \frac{3}{4} \) to both sides. This step looks like:
[tex]\[ 3x - \frac{3}{4} + \frac{3}{4} = \frac{13}{6} + \frac{3}{4} \][/tex]
This is correct since adding \( \frac{3}{4} \) balances the equation.

3. Next, Jerry combined the fractions on the right side:
[tex]\[ 3x = \frac{13}{6} + \frac{3}{4} \][/tex]
To add these fractions, Jerry needed a common denominator. The least common denominator (LCD) of 6 and 4 is 12:
[tex]\[ \frac{13}{6} = \frac{26}{12} \][/tex]
[tex]\[ \frac{3}{4} = \frac{9}{12} \][/tex]
Adding these gives:
[tex]\[ 3x = \frac{26}{12} + \frac{9}{12} = \frac{35}{12} \][/tex]

4. Therefore, step 4 (simplifying to \( 3x = \frac{35}{12} \)) is correct:
[tex]\[ 3x = \frac{35}{12} \][/tex]

5. However, in step 5, Jerry mistakenly multiplied by \( \frac{3}{1} \) instead of dividing both sides by 3 (or multiplying by \( \frac{1}{3} \)):
[tex]\[ 3x = \frac{35}{12} \][/tex]
To solve for \( x \), we should divide both sides by 3:
[tex]\[ x = \frac{\frac{35}{12}}{3} = \frac{35}{12} \cdot \frac{1}{3} = \frac{35}{36} \][/tex]

Based on this detailed analysis, Jerry made the error in step 5. Thus, the error occurs in step 5, where Jerry should have multiplied both sides by [tex]\( \frac{1}{3} \)[/tex].