In the coordinate plane, region J is defined by all the points (x,y)

Leo8
If you extend the line b=2a+4, it will cross the line 4y-6x=32. Thus Option A is not sufficient.

(1) b/2 = a + 2
Since the point (a,b) can take sufficient values as per this equation.
Lets test 2 points and see if it satisfies the inequality 4y−6x<32?
If b=6,a=1, and we substitute these values in the expression, it will return a TRUE value.
But, if b=20,a=8, substituting these values will return a FALSE value for the expression(4x-6y<32)
Hence, insufficient.

(2) 12b=18a+60
This expression can be simplified as 2b = 3a + 10
Multiplied by 2 and taking the a to the left hand side, we will get 4b - 6a = 20
This is of the form 4y - 6x < 32
We know that 20 is always lesser than 32. Hence, sufficient(Option B)

Oops, modified the solution. Thanks for pointing out the error!

It should be 2b=3a+10 .... neverthless the answer should be B.

i opt for B In the same line of reasoning

Imo B
Statement can have values that can be outside of the region
But statement 2 is limited and will have values within the region as it equals 10 which less than 16

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We don't necessarily have to diagram this problem so much as apply algebra

Statement 1

b/2 = a +2
b= 2a +4

44 = 2(20) + 4; 44(2) - 20(3) = 28 > 16 yet - -
2 = 2(-1) + 4; 2(2) - (-1)(3) = 7 < 16

Insuff


Statement 2

Can be algebraically simplified to

2b-3a = 10

Suff


Thus

"B"

Here is my approach.

Line f: 4y - 6x = 32

Statment 1: b/2 = a+2 --> 4b = 8a + 16 --> 4b - 8a = 16
--> Point (a,b) lies on line g: 4y - 8x = 16
From the equations of Line f and Line g, we can conclude that these 2 lines intersect. --> Point (a,b) can be inside or outside region J
--> Insufficient.

Statement 2: 12b=18a+60 --> 4b = 6a + 20 --> 4b - 6a = 20
--> Point (a,b) lies on line h: 4y - 6x = 20
From the equations of Line f and Line h, we can conclude that these 2 lines are parallel, and line h is located within Region J (because 20 < 32)
--> Sufficient.

Answer B.

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