Question

tia makes an error solving the equation \\(\frac{1}{3}(10x - 15) = 5x + 20\\).
part a.
select the line where tia’s error first appears.
\\(\frac{1}{3}(10x - 15) = 5x + 20\\)
\\(10x - 15 = 3(5x + 20)\\)
\\(10x - 15 = 15x + 20\\)
\\(-15 = 5x + 20\\)
\\(-35 = 5x\\)
\\(-7 = x\\)
part b.
what is the correct solution to \\(\frac{1}{3}(10x - 15) = 5x + 20\\)?

Explanation:

Step1: Multiply both sides by 3

To eliminate the fraction, multiply each term in the equation $\frac{1}{3}(10x - 15)=5x + 20$ by 3.
$$3\times\frac{1}{3}(10x - 15)=3\times(5x + 20)$$
Simplifying, we get $10x - 15 = 15x + 60$.

Step2: Subtract 10x from both sides

Subtract $10x$ from each side to get:
$$10x - 10x - 15=15x - 10x + 60$$
Simplifying, we have $- 15 = 5x + 60$.

Step3: Subtract 60 from both sides

Subtract $60$ from each side:
$$-15 - 60=5x + 60 - 60$$
Simplifying, we get $-75 = 5x$.

Step4: Divide both sides by 5

Divide each side by $5$:
$$\frac{-75}{5}=\frac{5x}{5}$$
Simplifying, we find $x = - 15$. Wait, let's check again. Wait, in step 1, when we multiply $5x + 20$ by 3, it's $15x + 60$, correct. Then step 2: $10x - 15 = 15x + 60$; subtract $10x$: $-15 = 5x + 60$; subtract $60$: $-75 = 5x$; divide by 5: $x = -15$. Wait, but maybe I made a mistake earlier. Wait, let's start over.

Correct steps:

Given equation: $\frac{1}{3}(10x - 15)=5x + 20$

Multiply both sides by 3:

$10x - 15 = 3(5x + 20)$

$10x - 15 = 15x + 60$ (distribute the 3)

Subtract $10x$ from both sides:

$10x - 10x - 15 = 15x - 10x + 60$

$-15 = 5x + 60$

Subtract $60$ from both sides:

$-15 - 60 = 5x + 60 - 60$

$-75 = 5x$

Divide both sides by 5:

$x=\frac{-75}{5}=-15$

Wait, but maybe the original problem in Part B was misread. Wait, the equation is $\frac{1}{3}(10x - 15)=5x + 20$. Let's check again.

Wait, maybe I made a mistake in the constant term. Let's re-express the multiplication:

$3\times(5x + 20)=15x + 60$, correct. Then $10x - 15 = 15x + 60$. Then, subtract $10x$: $-15 = 5x + 60$. Subtract $60$: $-75 = 5x$. Then $x = -15$. But maybe the user's Part B was written as $\frac{1}{3}(10x - 15)=5x + 20$, so the correct solution is $x=-15$? Wait, no, maybe I messed up the multiplication. Wait, $3\times(5x + 20)$ is $15x + 60$, yes. Then $10x - 15 = 15x + 60$. Then, moving terms: $10x - 15x = 60 + 15$; $-5x = 75$; $x = -15$. Yes, that's correct.

Wait, but maybe the original problem in the image had a typo, but according to the given Part B: "What is the correct solution to $\frac{1}{3}(10x - 15)=5x + 20$?", the solution is $x = -15$. Wait, but let's check with the original steps. Wait, maybe I made a mistake in the first multiplication. Let's do it again:

$\frac{1}{3}(10x - 15)=5x + 20$

Multiply both sides by 3:

$10x - 15 = 3\times5x + 3\times20$

$10x - 15 = 15x + 60$

Subtract $10x$:

$-15 = 5x + 60$

Subtract $60$:

$-75 = 5x$

Divide by 5:

$x = -15$

Yes, that's correct.

Answer:

$x = - 15$