given the equation:
y = 5/4x - 3
to complete the table for the values of x and y using intergers
fristly,
when x = -4
y = 5/4(-4) - 3
y = -5 - 3
y = -8
when x = 0
y = 5/4(0) - 3
y = 0 - 3
y = -3
when = 4
y = 5/4(4) - 3
y = 5 - 3
y = 2
when x = 8
y = 5/4(8) - 3
y = 5(2) - 3
y = 10 - 3
y = 7
so completing the table:
y = 5/4x - 3
x y
-4 -8
0 -3
4 2
8 7