To calculate: The solution set of −2|3y−10|=4=−6.

Question

Jennifer Hill · Accepted Answer

Formula Used:
From the definition of absolute value,
$| u | = k$ is equivalent to $u = k$ or $u = - k$
Calculation:
First isolate the absolute value. Subtract 4 from all two parts of the inequality.
$- 2 | 3 y - 10 | + 4 - 4 = - 6 - 4$
$- 2 | 3 y - 10 | = - 10$
Now divide —2 from all two parts of the inequality.
$\frac{- 2 | 3 y - 10 |}{- 2} = \frac{- 10}{- 2}$
$| 3 y - 10 | = 5$
The inequality is in the form , where . Write the equivalent compound inequality,
$3 y - 10 = 5 or 3 y - 10 = - 5$
$3 y = 5 + 10 or 3 y = - 5 + 10$

$3 y = 15 or 3 y = 5$
$y = 5 or y = \frac{5}{3}$
Check at $y = 5$
$- 2 | 3 (5) - 10 | + 4 - 6$
$- 2 | 15 - 10 | + 4 - 6$
$- 2 | 5 | + 4 - 6$
$- 6 = - 6$
The left side value is equal to the right-side value. Thus, the value y = Sis verified.
Check at $y = \frac{5}{3}$
$- 2 | 3 (5) - 10 | + 4 - 6$
$- 2 | 15 - 10 | + 4 - 6$
$- 2 | 5 | + 4 - 6$
$- 6 = - 6$
The left side value is equal to the right-side value. Thus, the value $y = 3$ is verified.
Check at $y = \frac{5}{3}$
$- 2 ∣ 3 (5) - 10 ∣ + 4 \begin{matrix} \end{matrix} = - 6$
$- 2 ∣ 15 - 10 ∣ + 4 \begin{matrix} \end{matrix} = - 6$
$- 2 ∣ 5 ∣ + 4 \begin{matrix} \end{matrix} = - 6$
$- 6 = - 6$
The left side value is equal to the right-side value. Thus, the value $y = 3$ is
verified.
Therefore, the solution set of $- 2 ∣ 3 y - 10 ∣ + 4 = - 6 i s {\frac{5}{3}, 5}$ .

To calculate: The solution set of -2\mid3y-10\mid=4=-6.

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