The tables represent two linear functions in a system

table one
x -6, -3, 0, 3
y= -22, -10, 2, 14

table 2
x = -6, -3, 0, 3
y= -30, -21, -12, -3
what is the solution to this system?
a) [-13/3 , -25]
b) [-14/3, -54]
c) (-13, 50)
d) (-14, 54)

D. (-14, -54)
 WORKINGS
 According to Table 1: y = 4x + 2 … (Equation I)
According to Table 2: y = 3x – 12 … (Equation II)
 Substitute the value of y in equation I (4x + 2) for y in Equation II to solve for x
y = 3x – 12
4x + 2 = 3x – 12
4x – 3x = -12 – 2
x = -14
 Substitute -14 for x in Equation I to solve for y
y = 4x + 2
y = 4(-14) + 2
y = -56 + 2
y = -54
 Therefore, x= -14; y = -54

The answer for you question is b