Attached solution with work.
Round each fraction to help you estimate the solution for the following equation:
1 answer:
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Answer:
y =
Step-by-step explanation:
Given
6 | 5y - 1 | - 1 = 29 ( add 1 to both sides )
6 | 5y - 1 | = 30 ( divide both sides by 6 )
| 5y - 1 | = 5
The absolute value function always returns a positive value. However, the expression inside can be positive or negative, thus
5y - 1 = 5 ( add 1 to both sides )
5y = 6 ( divide both sides by 5 )
y =
OR
- (5y - 1) = 5, that is
- 5y + 5 = 5 ( subtract 5 from both sides )
5y = 0 ⇒ y = 0
As a check substitute these values into the left side and if equal to the right side then they are the solutions.
x = 0 : 6|0 - 1 | - 1 = 6 | - 1| - 1 = 6(1) - 1 = 6 - 1 = 5 ≠ 29
x = : 6 | 6 - 1 | - 1 = 6(5) - 1 = 30 - 1 = 29
Thus x = is a solution, x = 0 is extraneous