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# RD Chapter 15- Linear Inequations Ex-15.6 Interview Questions Answers

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Question 1 :
Solve the following systems of linear inequations graphically.
(i) 2x + 3y ≤ 6, 3x + 2y ≤ 6, x ≥ 0, y ≥ 0
(ii) 2x + 3y ≤ 6, x + 4y ≤ 4, x ≥ 0, y ≥ 0
(iii) x – y ≤ 1, x + 2y ≤ 8, 2x + y ≥ 2, x ≥ 0, y ≥ 0
(iv) x + y ≥ 1, 7x + 9y ≤ 63, x ≤ 6, y ≤ 5, x ≥ 0, y ≥ 0
(v) 2x + 3y ≤ 35, y ≥ 3, x ≥ 2, x ≥ 0, y ≥ 0

(i) 2x + 3y ≤ 6, 3x + 2y ≤ 6, x ≥ 0, y ≥ 0
We shall plot the graph of the equation and shade the side containing solutions of the inequality,
You can choose any value but find the two mandatory values which are at x = 0 and y = 0, i.e., x and y–intercepts always,
2x + 3y ≤ 6
So when,
 x 0 1 3 y 2 1.33 0
3x + 2y ≤ 6
So when,
 x 0 1 2 y 3 1.5 0

x ≥ 0, y ≥ 0 (ii) 2x + 3y ≤ 6, x + 4y ≤ 4, x ≥ 0, y ≥ 0

We shall plot the graph of the equation and shade the side containing solutions of the inequality,

You can choose any value but find the two mandatory values which are at x = 0 and y = 0, i.e., x and y–intercepts always,

2x + 3y ≤ 6

So when,

 x 0 1 3 y 2 1.33 0

x + 4y ≤ 4

So when,

 x 0 2 4 y 1 0.5 0

x ≥ 0, y ≥ 0 (iii) x – y ≤ 1, x + 2y ≤ 8,2x + y ≥ 2, x ≥ 0, y ≥ 0

We shall plot thegraph of the equation and shade the side containing solutions of theinequality,

You can choose anyvalue but find the two mandatory values which are at x = 0 and y = 0, i.e., xand y–intercepts always,

x – y ≤ 1

So when,

 x 0 2 1 y -1 1 0

x + 2y≤ 8

So when,

 x 0 4 8 y 4 2 0

2x + y ≥ 2

So when,

 x 0 2 1 y 2 -2 0

x ≥ 0, y ≥ 0 (iv) x + y ≥ 1, 7x + 9y ≤63, x ≤ 6, y ≤ 5, x ≥ 0, y ≥ 0

We shall plot thegraph of the equation and shade the side containing solutions of theinequality,

You can choose anyvalue but find the two mandatory values which are at x = 0 and y = 0, i.e., xand y–intercepts always,

x + y ≥ 1

So when,

 x 0 2 1 y 1 -1 0

7x + 9y ≤ 63

So when,

 x 0 5 9 y 7 3.11 0

x ≤ 6, y ≤ 5 and x ≥0, y ≥ 0 (v) 2x + 3y ≤ 35, y ≥ 3, x≥ 2, x ≥ 0, y ≥ 0

We shall plot thegraph of the equation and shade the side containing solutions of theinequality,

You can choose anyvalue but find the two mandatory values which are at x = 0 and y = 0, i.e., xand y–intercepts always,

2x + 3y ≤ 35

So when,

 x 0 5 17.5 y 11.667 8.33 0

y ≥ 3, x ≥ 2, x ≥ 0, y≥ 0 Question 2 :
Show that the solution set of the following linear inequations is empty set:
(i) x – 2y ≥ 0, 2x – y ≤ –2, x ≥ 0, y ≥ 0
(ii) x + 2y ≤ 3, 3x + 4y ≥ 12, y ≥ 1, x ≥ 0, y ≥ 0

(i) x – 2y ≥ 0, 2x – y ≤–2, x ≥ 0, y ≥ 0

We shall plot thegraph of the equation and shade the side containing solutions of the inequality,

You can choose anyvalue but find the two mandatory values which are at x = 0 and y = 0, i.e., xand y–intercepts always,

x – 2y ≥ 0

So when,

 x 0 2 4 y 0 1 2

2x – y ≤ –2

So when,

 x 0 1 -1 y 2 4 0

x ≥ 0, y ≥ 0 The lines do notintersect each other for x ≥ 0, y ≥ 0

Hence, there is nosolution for the given inequations.

(ii) x + 2y ≤ 3, 3x +4y ≥ 12, y ≥ 1, x ≥ 0, y ≥ 0

We shall plot thegraph of the equation and shade the side containing solutions of theinequality,

You can choose anyvalue but find the two mandatory values which are at x = 0 and y = 0, i.e., xand y–intercepts always,

x + 2y ≤ 3

So when,

 x 0 1 3 y 1.5 1 0

3x + 4y ≥ 12

So when,

 x 0 2 4 y 3 1.5 0

y ≥ 1, x ≥ 0, y ≥ 0 Question 3 :

Find the linear inequations for which the shaded area in Fig. 15.41 isthe solution set. Draw the diagram of the solution set of the linearinequations. Here, we shall applythe concept of a common solution area to find the signs of inequality by usingtheir given equations and the given common solution area (shaded part).

We know that,

If a line is in theform ax + by = c and c is positive constant. (In case of negative c, therule becomes opposite), so there are two cases which are,

If a line is above theorigin:

(i) If the shaded areais below the line then ax + by < c

(ii) If the shadedarea is above the line then ax + by > c

If a line is below theorigin then the rule becomes opposite.

So, according to therules Question 4 : Find the linear inequations for which the solution set is the shaded region given in Fig. 15.42. 