# 9th Class Math Chapter 7 Linear Equations & Inequalities Short Questions Answer

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## 9th Class Math Chapter 7 Linear Equations & Inequalities Short Questions Answer

State the trixhotomy and transitive properties of inequalities.

Solution:

LAW OF TRICHOTOMY

For any a,b ϵ R, one and only of the statements is true.

A < b or a = b or a > b

LAW OF TRANSITY

Let a,b,c ϵ R

(a) If a > b and b > c then a > c

(b) If a < b and b < c then a < c

The formula relating degrees Fahrenheit to degrees Celsius is F = 9/5 C + 32, for what value of C is F<0?

Solution:

F = 9/5 c + 32

F < 0

9/5 C + 32

9c + 160 < 0

9c < – 160

c < –160/9

Seven times, the sum of an integer and 12 is at least 50 and at most 60. Write and solve the inequality that expresses this relationship.

Solution:

Let x be the integer. So according to the question

50 ≤ 7(x + 12) ≤ 60

This is equivalent to two inequalities

50 ≤ 7(x + 2)

7(x + 2) ≤ 60

The first inequality gives

50 ≤ 7 (x + 2)

50 ≤ 7 x + 84

50 – 84 ≤ 7 x

– 34 ≤ 7 x

– 34/7 ≤ x

The second inequality gives

7(x + 12) ≤ 60

7x + 84 ≤ 60

7x ≤ 60 – 84

From (1) and (2) we get

– 34/7 ≤ x ≤ – 24/7

Solve each of the following and check for extraneous solution if any: √2t + 4 = √t – 1, and √3x – 1 – 2√8 – 2x = 0

Solution:

1. √2t + 4 = √t – 1

Squaring both sides

2t + 4 = t – 1

2t t = – 1 – 4

t = – 5

On checking

√–10+4 = √–5–1

√–6 = √–6

Since √–6 = imaginary

1.    √3x – 1 – 2√8 – 2x = 0

Squaring both sides

3x – 1 = 4(8 – 2x)

3x – 1 = 32 – 8x

11x = 33

x = 3

Solve for x. |3x + 14| – 2 = 5x, and 1/3|x – 3| = ½ |x + 2|

Solution

|3x + 14| – 2 = 5x

This equivalent to

3x + 14 = ±(5x + 2)

3x + 14 = 5x + 2

3x – 5x = 2 – 14

– 2x = – 12

x = 6

On checking we see that x = 6 satisfies the given equation but x = – 12 does not satisfy the given equation. So, the solution set is {6}

1/3|x – 3| = ½ |x + 2|

Multiplying both sides by 6 we get

2|x – 3| = ½ |x + 2|

Which is equivalent to

2(x – 3) = 3(x + 2)

2x – 6 = 3x + 6

2x – 3x = 6 + 6

x = – 12

On checking x = – 12

Solve the following inequality – 1/3 x + 5 ≤ 1 and – 3 < 1–2x/5 < 1

Solution:

– 1/3 x + 5 ≤ 1

Multiplying by 3

x + 15 ≤ 3

x ≤ 3 – 15

x ≤ 3 – 15

x ≥ 12

– 3 < 1–2x/5 < 1

– 3 < 1–2x/5

– 15 < 1 + 15

2x < 1 + 15

2x < 16

x < 8

1 – 2x < 5

– 2x < 5 – 1

– 2x > 4

x > – 2

8 > x > – 2

– 2 < x < 8

Solution set is {x|8 > x > – 2}

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