Please I really need help with this question and if you do answer please add the steps :)

Question

Write a system of linear inequalities so the points (1, 2) and (4, −3) are solutions of the system, but the point (−2, 8) is not a solution of the system.

Answer 1

the co ordinate points given to make the inequation true are (1,2) and (4,-3)

finding the slope of the equation with the help of these 2 co ordinate points

slope m = (y2 - y1) / (x2 - x1)

m = (2 -(-3))/(1-4)

= 5/-3 = -5/3

slope m = -5/3

now point slope formula to get an equation of line in y = mx+b or (mx+c) form

(y-y1) = m(x-x1)

y- 2 = (-5/3)(x - 1)

3(y-2) = -5(x-1)

3y - 6 = -5x + 5

= 5x+3y =11 ( standard form of equation ax+by = c)

or 3y = -5x + 11

y = (-5/3)x + 11/3 ( y = mx + c form)

check for the 2 points if they make the equation true i.e for (1,2) and (4,-3)

5(1) + 3(2) = 11 = 5+6 = 11 true

5(4) + 3(-3) = 11 = 20 - 9 = 11 true

so the equation is true for given two co ordinates

now this same have to be an inequation and should be false for the co ordinate (-2,8)

substitue values and check

5(-2) + 3(8) = -10 + 24 = 14 != 11

so the inequation sign for the above equation should be <= sign not a >= sign ( so that it can make it true for the first two given co ordinates and false for the third co ordinate according to the requirement)

so our inequation is 5x + 3y <= 11

hope it is clear and helpful! can ask still any doubts.