The graph of a linear equation passes through three of these points: (0, 5), (2, 2), (3, 1), and (4, -1). the point that the graph does not pass through is

Label the given points as follows:
A (0, 5)
B (2, 2)
C (3, 1)
D (4, -1)

The straight line has a constant slope. Therefore it should have the same value when any two of the four points are used to calculate the slope.

Try A and B:
Slope = (2 - 5)/(2 - 0) = -3/2

Try A and C:
Slope = (1 - 5)/(3 - 0) = -4/3

Try A and D.
Slope = (-1 - 5)/(4 - 0) = -3/2

Try B and D.
Slope = (-1 - 2)/(4 - 2) = -3/2

Clearly, C does not lie on the straight line.

 The point that the graph does not pass through is (3,1). 

First subtract 10 from both side
y^2+6y-16=0
(y+8)(y-2)=0
-8 and 2

2xy/9 is the answer to this problem