Aline passes through point (-2, 5) and has a slope of 2/3 points a(x, 3) and b(-2, y) lie on the line

Y = mx + b
slope(m) = 2/3
(-2,5)...x = -2 and y = 5
now we sub and find b, the y int
5 = 2/3(-2) + b
5 = -4/3 + b
5 + 4/3 = b
15/3 + 4/3 = b
19/3 = b

so ur equation of the line is : y = 2/3x + 19/3...but lets put it in standard form
y = 2/3x + 19/3
-2/3x + y = 19/3
2x - 3y = -19this is the equation of the line (standard form)

so A (x,3)this is saying y = 3...so we sub in 3 for y and solve for x
2x - 3y = -19
2x - 3(3) = -19
2x - 9 = -19
2x = -19 + 9
2x = -10
x = -10/2
x = -5so point A is (-5,3)

point B (-2,y)...this is saying x = -2...so sub in -2 for x and solve for y
2x - 3y = -19
2(-2) - 3y = -19
-4 - 3y = -19
-3y = -19 + 4
-3y = - 15
y = -15/-3
y = 5so point B is (-2,5)

What are 5e numbers?

That is what i got (fgx) hope this