Question

4 carl solved this equation $\frac{1}{2}(1 - 10x) + 3x = x + 1 + \frac{1}{2}$, but his answer is wrong. carl’s work is shown. carl’s work $\frac{1}{2}(1 - 10x) + 3x = x + 1 + \frac{1}{2}$ $\frac{1}{2} - 2x + 3x = x + 1 + \frac{1}{2}$ $\frac{1}{2} + x = \frac{3}{2}x + 1$ $\frac{1}{2} = \frac{1}{2}x$ $4 = x$ describe and correct the error carl made.

Explanation:

Step1: Identify the error

In the step where Carl simplifies the right - hand side of the equation \(x + 1+5\), he made a mistake. The correct simplification of \(x + 1+5\) is \(x+(1 + 5)=x + 6\), but Carl wrote \(\frac{3}{2}x+1\) which is incorrect. Also, when he distributed \(\frac{1}{3}\) to \(- 6x\), \(\frac{1}{3}\times(-6x)=-2x\) (which is correct), but the initial equation seems to have a typo, assuming the original equation is \(\frac{1}{3}(9 - 6x)+3x=x + 1+5\) (correcting the likely typo in the original problem statement). Let's re - solve the equation correctly.

Step2: Correctly simplify the equation

Start with the original equation (assuming it is \(\frac{1}{3}(9 - 6x)+3x=x + 1+5\)):
- Distribute \(\frac{1}{3}\) in \(\frac{1}{3}(9 - 6x)\): \(\frac{1}{3}\times9-\frac{1}{3}\times6x+3x=x + 1+5\)
- Calculate \(\frac{1}{3}\times9 = 3\) and \(\frac{1}{3}\times6x = 2x\), so the left - hand side becomes \(3-2x + 3x\), and the right - hand side becomes \(x+6\).
- Combine like terms on the left - hand side: \(3 + x=x + 6\).

Step3: Solve for \(x\)

Subtract \(x\) from both sides of the equation \(3 + x=x + 6\):
\(3+x - x=x + 6-x\)
\(3=6\), which is a contradiction. But if we assume the original equation was \(\frac{1}{3}(9 - 6x)+3x=\frac{1}{2}x+1 + 5\) (to match Carl's wrong step), let's solve it correctly.
- Distribute \(\frac{1}{3}\): \(3-2x + 3x=\frac{1}{2}x+6\)
- Combine like terms on the left: \(3 + x=\frac{1}{2}x+6\)
- Subtract \(\frac{1}{2}x\) from both sides: \(3 + x-\frac{1}{2}x=\frac{1}{2}x+6-\frac{1}{2}x\)
- We get \(3+\frac{1}{2}x = 6\)
- Subtract 3 from both sides: \(\frac{1}{2}x=6 - 3=3\)
- Multiply both sides by 2: \(x = 6\)

But going back to Carl's work, the main error is in combining the constant terms on the right - hand side. He should have \(x+6\) instead of \(\frac{3}{2}x + 1\). Let's do the correct steps for the equation as per the original (with the typo fixed):

1. Original equation (corrected): \(\frac{1}{3}(9 - 6x)+3x=x + 1+5\)
- Distribute: \(3-2x + 3x=x + 6\)
- Combine like terms: \(3 + x=x + 6\)
- Subtract \(x\) from both sides: \(3=x + 6-x\)
- We get \(3 = 6\), which means the equation has no solution. But if we take the equation as Carl mis - wrote the right - hand side, say the original equation was \(\frac{1}{3}(9 - 6x)+3x=\frac{3}{2}x+1 + 5\) (to match his wrong step):
- Distribute: \(3-2x + 3x=\frac{3}{2}x+6\)
- Combine like terms: \(3 + x=\frac{3}{2}x+6\)
- Subtract \(x\) from both sides: \(3=\frac{3}{2}x - x+6\)
- Simplify \(\frac{3}{2}x-x=\frac{1}{2}x\), so \(3=\frac{1}{2}x+6\)
- Subtract 6 from both sides: \(\frac{1}{2}x=3 - 6=-3\)
- Multiply both sides by 2: \(x=-6\)

The error Carl made was in combining the constant terms on the right - hand side of the equation. He incorrectly simplified \(x + 1+5\) as \(\frac{3}{2}x+1\) instead of \(x + 6\) (or a similar correct combination depending on the actual original equation). To correct it, we should properly combine the like terms on both sides of the equation. For example, if we take the equation \(\frac{1}{3}(9 - 6x)+3x=x + 1+5\):

- Step 1: Distribute \(\frac{1}{3}\) in \(\frac{1}{3}(9 - 6x)\): \(\frac{1}{3}\times9-\frac{1}{3}\times6x+3x=x + 1+5\) gives \(3-2x + 3x=x + 6\)
- Step 2: Combine like terms on the left: \(3 + x=x + 6\)
- Step 3: Subtract \(x\) from both sides: \(3=6\), which shows the equation is inconsistent (no solution). If we assume the equation was \(\frac{1}{3}(9 - 6x)+3x=\frac{3}{2}x+1 + 5\) (to follow Carl's wrong step structure):
- Step 1: Distribute: \(3-2x + 3x=\frac{3}{2}x+6\)
- Step 2: Combine like terms: \(3 + x=\frac{3}{2}x+6\)
- Step 3: Subtract \(x\) from both sides: \(3=\frac{1}{2}x+6\)
- Step 4: Subtract 6: \(\frac{1}{2}x=3 - 6=-3\)
- Step 5: Multiply by 2: \(x=-6\)

Answer:

The error Carl made was in combining the constant terms on the right - hand side of the equation (he incorrectly simplified \(x + 1+5\)). To correct the solution:

1. Start with the equation \(\frac{1}{3}(9 - 6x)+3x=x + 1+5\) (assuming the original equation):

• Distribute: \(3-2x + 3x=x + 6\)
• Combine like terms: \(3 + x=x + 6\)
• Subtract \(x\) from both sides: \(3 = 6\), so the equation has no solution.

If we assume the equation was \(\frac{1}{3}(9 - 6x)+3x=\frac{3}{2}x+1 + 5\) (to match Carl's wrong step structure):

• Distribute: \(3-2x + 3x=\frac{3}{2}x+6\)
• Combine like terms: \(3 + x=\frac{3}{2}x+6\)
• Subtract \(x\): \(3=\frac{1}{2}x+6\)
• Subtract 6: \(\frac{1}{2}x=-3\)
• Multiply by 2: \(x = - 6\)