Linear Inequalities

42. The given system of inequality is

x+y≤ 6 - (1)

x+y≥ 4- (2)

So the corresponding equations are

x+y=6

and x + y = 4

Putting (x, y)= (0,0) in equality (1) and (2),

0+0 ≤ 6 and 0 + 0 ≥ 4

0 ≤ 6 is true.  => 0 ≥ 4 is false.

So, solution of plane of inequality (1) includes the origin and inequality (2) does not includes the origin.

? The reqd solution of the given system of inequality is the shaded region.

41. The given system of inequality is

2x – y> 1 - (1)

x – 2y< 1- (1)

So the corresponding equations are

2x – y=1

and x – 2y= –1

Putting (x, y)= (0,0) in (1) and (2) to cheek the inequality

2 * 0 – 0 > 1

0 > 1 which is not true.

and 0 – 2 * 0< 1

0< 1 which is not true.

So, the solution of plane of inequality (1)and (2) does not include the plane with point (0,0) or origin.

? The reqd. solution of the given system of inequality is the shaded region.

40. The given system of inequalities is

x + y ≥ 4.- (1)

2x – y< 0.- (2)

The corresponding equations are x+y=4 and 2x – y=0.

and

Put (x, y)= (1,1) in (1) and (2).

So, 1+1 ≥ 4

2 ≥ 4 which is not true.

and 2 * 1 – 1<0

1<0 which is not true.

So solution of plane of inequality (1) and (2) does not include the plane with point (1,1).

? The reqd. solution of the given system of inequality is the shaded portion.

38. 

2. The given system of inequalities are

3x+2y≤ 12- (1)

x≥ 1- (2)

y≥ 2- (3)

We draws the graphs of the lines 3x+2y=12 using points and as 3 * 0 + 2 * 0 ≤ 12

The solution is plane which includes the origin (0, 0).

0 ≤ 12

$\begin{array}{l}x\\ y\end{array}|\begin{array}{l}0\\ 6\end{array}|\begin{array}{l}4\\ 0\end{array}|$

and x = 1 and y = 2.

The inequality (1), (2) and (3) represents the region between these three lines including the points on the respective lines. So, every point on the shaded region in first quadrant represents a solution of the given system of inequalities.

36. For the given inequality, x> –3.

The equation of the line is x= –3.

This line divides the xy-plane into planer I and II. We take a point (0,0) to check the correctness of the inequality.

So, 0> –3 which is true.

So, the solution of the region is I which includes the origin.

The dotted line indicates that any point on the line does not satisfy the inequality.

35. For the given equation inequality y< 2,  the equation of line is y= 2

This line devides the xy-plane into two planer I and II. We take a point (0,0) to check the correctness of the inequality.

So, 0< 2

0< 2 which is false.

So, the solution of the region is II which does not include the origin.

The dotted line indicates that any point on the line does not satisfy the inequality.

34. For the inequality 3y – 5x<30, the equation of line is 3y 5x=30. We consider the table below to plot 3y 5x =30.

$\begin{array}{l}x\\ y\end{array}|\begin{array}{l}0\\ 10\end{array}|\begin{array}{l}-6\\ 0\end{array}|$

This line divides the xy-plane into two planer I and II. We select a point (0,0) and check the correctness the inequality.

3 * 0 – 5 * 0 < 30

0 < 30 which is true.

So, the solution region is I which includes the origin. The dotted line indicates that any point on the line will not satisfy the given inequality.

33.For the inequality –3x+2y≥ –6 the equation of line is – 3x+2y=6.

We consider the table below to plot – 3x+2y= –6.

$\begin{array}{l}x\\ y\end{array}|\begin{array}{l}0\\ -3\end{array}|\begin{array}{l}2\\ 0\end{array}|$

This line divides the xy-plane into two planer I and II. We select a point (0,0) and check the correctness the inequality,

–3 * 0+2 * 0 ≥ –6

0 ≥ –6 which is true.

So, the solution region is I which includes the origin. The continuous line also indicates that any point on the line also satisfy the given inequality.