Simplifying a system of linear inequalities Given two inequalities y &#x2265;<!-- ≥ -->

Leland Morrow

Leland Morrow

Answered question

2022-06-15

Simplifying a system of linear inequalities
Given two inequalities
y m 1 ( x x 1 ) + y 1
and
y m 2 ( x x 2 ) + y 2
Is there anyway to solve for the space that they both exist in?

Answer & Explanation

britspears523jp

britspears523jp

Beginner2022-06-16Added 28 answers

Consider two regions: R 1 = { ( x , y ) : y a 1 x + b 1 } and R 2 = { ( x , y ) : y a 2 x + b 2 }. You are looking for R = R 1 R 2 .
So, if a 1 x + b 1 < a 2 x + b 2 , then { ( x , y ) : y a 2 x + b 2 } lies in R; otherwise, { ( x , y ) : y a 1 x + b 1 } lies in R
First, you need to compute the intersection point of two lines y = a 1 x + b 1 = a 2 x + b 2 ( ( 4 / 3 , 2 / 3 ) in the figure). Afterwards, determine the range of x when a 1 x + b 1 < a 2 x + b 2 . In the example, x < 4 / 3 for a 1 = 1, b 1 = 4, a 2 = 0.5 and b 2 = 2. Then
y { 0.5 x + 2  if  x < 4 / 3 ; x + 4  if  x 4 / 3 .

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