Explanation:
The system of Linear equation is x + 2y ≤ 4 and 2x - y > 0
Explanation:
Given: Point ( 2 , 1 )
To find: System of Linear Inequalities which contain given point in
its solution set
There are 2 lines passing through point ( 2 , 1 )
So first we find the equation of both lines
Equation of line using Two-Point form is given by,
(y-y_1)=\frac{y_2-y_1}{x_2-x_1}(x-x_1)(y−y1)=x2−x1y2−y1(x−x1)
Equation of line 1 passing through ( 0 , 2 ) & ( 4 , 0 )
(y-2)=\frac{0-2}{4-0}(x-0)(y−2)=4−00−2(x−0)
(y-2)=\frac{-2}{4}x(y−2)=4−2x
(y-2)=\frac{-x}{2}(y−2)=2−x
y=\frac{-x}{2}+2y=2−x+2
y+\frac{x}{2}=2y+2x=2
x+2y=4x+2y=4 ....... (1)
Equation line 2 passing through ( 0 , 0 ) & ( 2 , 1 )
(y-0)=\frac{0-2}{0-1}(x-0)(y−0)=0−10−2(x−0)
y=\frac{-2}{-1}xy=−1−2x
y=2xy=2x
2x-y=02x−y=0 ....... (2)
Now from given graph is clear line 1 contain the given point and area shaded by it is toward origin then we have less than equal sign,
⇒ x + 2y ≤ 4
Line 2 is dotted line. So, it does not contain the given point and area shaded by it is toward 4th quadrant then we have greater than sign in it.
⇒ 2x - y > 0
Therefore, The system of Linear equation is x + 2y ≤ 4 and 2x - y > 0