Chapter 6: Linear Inequalities

Let the length of shortest side = ‘x’ cm

According to the question,

The longest side of a triangle is twice the shortest side

⇒ Length of largest side = 2x

Also, the third side is 2 cm longer than the shortest side

⇒ Length of third side = (x + 2) cm

Perimeter of triangle = sum of three sides

= x + 2x + x + 2

= 4x + 2 cm

Now, we know that,

Perimeter is more than 166 cm

⇒ 4x + 2 ≥ 166

⇒ 4x ≥ 164

⇒ x ≥ 41

Hence, minimum length of the shortest side should be = 41 cm.

According to the question,

T = 30 + 25(x – 3), 3 ≤ x ≤ 15; where, T = temperature and x = depth inside the earth

The Temperature should be between 155°C and 205°C,

So, we get,

⇒ 155 < T < 205

⇒ 155 < 30 + 25(x – 3) < 205

⇒ 155 < 30 + 25x – 75 < 205

⇒ 155 < 25x – 45 < 205

Adding 45 to each term, we get

⇒ 200 < 25x < 250

Dividing each term by 25, we get

⇒ 8 < x < 10

Hence, temperature varies from 155° C to 205° C at a depth of 8 km to 10 km.

Long Answer Type

According to the question,

For the above fraction be greater than 0, either both the denominator and numerator should be greater than 0 or both should be less than 0.

⇒ 6 – 33x > 0 and 7x – 1 > 0

⇒ 33x < 6 and 7x > 1

⇒ x < 2/11 and x > 1/7

⇒ 1/7 < x < 2/11 …(i)

Or

⇒ 6 – 33x < 0 and 7x – 1 < 0

⇒ 33x > 6 and 7x < 1

⇒ x > 2/11 and x < 1/7

⇒ 2/11< x < 1/7 …(which is not possible since 1/7 > 2/11)

Also,

For the above fraction be greater than 0, either both the denominator and numerator should be greater than 0 or both should be less than 0.

⇒ 23 – x > 0 and x – 8 > 0

⇒ x < 23 and x > 8

⇒ 8 < x < 23 …(ii)

Or

23 – x < 0 and x – 8 < 0

⇒ x > 23 and x < 8

⇒ 23 < x < 8 …(which is not possible, as 23 > 8]

Therefore, from equations (i) and (ii), we infer that there is no solution satisfying both inequalities.

Hence, the given system has no solution.

According to the question,

Considering 3x + 2y = 48,

The shaded region and the origin are both on the same side of the graph of the line, and (0, 0) satisfy the constraint 3x + 2y ≤ 48.

Considering x + y = 20,

The shaded region and the origin both are on the same side of the graph of the line, and (0, 0) satisfy the constraint x + y ≤ 20.

We also know that,

Shaded region is in the first quadrant i.e. x ≥ 0 and y ≥ 0,

Hence, the linear inequalities are

3x + 2y ≤ 48

x + y ≤ 20

x ≥ 0

y ≥ 0

According to the question,

Considering x + y = 8,

The shaded region and the origin are both on the same side of the graph of the line, and (0, 0) satisfy the constraint x + y ≤ 8.

Considering x + y = 4,

The origin is on the opposite side of the shaded region, and (0, 0), hence, doesn’t satisfy the constraint x + y ≥ 4; therefore required constraint is x + y ≥ 4

We see that,

The shaded region is in the first quadrant i.e. x ≥ 0 and y ≥ 0,

Also, shades region is below the line y = 5 and left to the line x = 5

⇒ y ≤ 5 and x ≥ 5

Hence, the linear inequalities are

x + y ≤ 8

x + y ≥ 4

x ≥ 0

y ≥ 0

x ≤ 5

y ≤ 5

According to the question,

x + 2y ≤ 3

Line: x + 2y = 3

Also, (0, 0) satisfies the x + 2y ≤ 3, hence region is towards the origin

3x + 4y ≤ 12

Line: 3x + 4y = 12

Also, (0, 0) satisfies the 3x + 4y ≤ 3, hence region is towards the origin

x ≥ 0 ⇒ region is to the right of the y-axis

And, y ≥ 1 ⇒ region is up above the line x = 1,

Therefore graph can be plotted as,

Hence, we can conclude from the graph that the above system has no common region as a solution.