Question

lets check darials solution: first, substitute -2 for x.
\\(\frac{1}{3}x + 6 = x + 10\\)
\\(\frac{1}{3}\left( -2 \
ight) + 6 = -2 + 10\\)
\\(5 = 8\\) false
is \\(x = -2\\) a solution of the equation?
no
where did daniel make a mistake?
step 1 \quad step 2 \quad step 3
daniels work
\\(\frac{1}{3}x + 6 = x + 10\\)
step 1 \quad \\(-\frac{1}{3}x\\) \quad \\(=-\frac{1}{3}x\\)
\\(6 = \frac{2}{3}x + 10\\)
step 2 \quad \\(-10\\) \quad \\(=-10\\)
\\(-4 = \frac{2}{3}x\\)
step 3 \quad \\(-4 + 2 = \frac{2}{3}x + 2\\)
\\(-2 = x\\)

Explanation:

Step1: Analyze Step 1 validity

Starting equation: $\frac{1}{2}x + 6 = x + 10$
Subtract $\frac{1}{2}x$ from both sides:
$\frac{1}{2}x + 6 - \frac{1}{2}x = x + 10 - \frac{1}{2}x$
$6 = \frac{1}{2}x + 10$ (Step 1 is correct)

Step2: Analyze Step 2 validity

Subtract 10 from both sides:
$6 - 10 = \frac{1}{2}x + 10 - 10$
$-4 = \frac{1}{2}x$ (Step 2 is correct)

Step3: Analyze Step 3 validity

Solve for $x$: multiply both sides by 2 (not divide by 2):
$-4 \times 2 = \frac{1}{2}x \times 2$
$-8 = x$
Daniel incorrectly divided by 2 instead of multiplying.

Answer:

1. Is $x=-2$ a solution of the equation?

No

1. Where did Daniel make a mistake?

Step 3