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
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
Solution:
F = 9/5 c + 32
F < 0
⟹ 9/5 C + 32
9c + 160 < 0
9c < – 160
c < –160/9
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
x ≤ – 24/7
From (1) and (2) we get
– 34/7 ≤ x ≤ – 24/7
Solution:
- √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
- √3x – 1 – 2√8 – 2x = 0
Squaring both sides
3x – 1 = 4(8 – 2x)
3x – 1 = 32 – 8x
11x = 33
x = 3
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
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}
