ArunpriyanJ wrote:
Vyshak wrote:
St1: x - 5 = 7x + 2 or 5 - x = 7x + 2
x = -7/6 or x = 3/8
Substitute the two values back in |x - 5| = 7x + 2. Only x = 3/8 is a valid value. Is x < 0? No
Sufficient
I am struggling to understand this step.....When i put -7/6 in Equn i am getting -37=-37...Pls help
St2: Square on both sides --> x^2 + 10x + 25 = 16x^2 + 40x + 25
15x^2 + 30x = 0
x = 0 or x = -2
Substitute the 2 values back in the original equation. Both the values satisfy the equation. Is x < 0? Can be yes or no.
Not Sufficient
Answer: A
Please format your reply properly.
Vyshak has used the value of 3/8 as the value of -7/6 is not a valid option.
When you put -7/6 into the given equation |x-5| = 7x+2 ---> you get 37/6 on the LHS and you get -37/6 on the RHS. Thus this value of x=-7/6 is not a valid one.
Do note here that LHS has |x-5| and as |x|\(\geq\)0 for all x, LHS \(\neq\) -37/6 .
Hope this helps.