$3 x-1 \geq 5$ and $x+2>-1$ $3 x-1 \geq 5$ $3 x-1+1>5+1$ $3 x \geq 6$ Dividing both the sides by 3 we get, $3 x / 3 \geq 6 / 3$ $x \geq 2$ $\therefore \mathrm{X} \in(2, \infty) \ldots(1)$ Now,...

$2 x+5 \leq 0$ and $x-3 \leq 0$ Let us consider the first inequality. $2 x+5 \leq 0$ $2 x+5-5 \leq 0-5$ $2 x \leq-5$ Dividing both the sides by 2 we get, 2x/2 ≤ –5/2 x ≤ – 5/2 ∴ x ∈ (–∞, -5/2]… (1)...

$3 x-6>0$ and $2 x-5>0$ $3 x-6>0$ $3 x-6+6>0+6$ $3 x>6$ Dividing both the sides by 3 we get, $3 x / 3>6 / 3$ $x>2$ $\therefore \mathrm{x} \in(2, \infty) \ldots(1)$ Now, $2...

$x-2>0$ and $3 x<18$ $x-2>0$ $x-2+2>0+2$ $x>2$ $\therefore \mathrm{X} \in(2, \infty) \ldots(1)$ Now, $3 x<18$ Dividing both the sides by 3 we get, $3 x / 3<18 / 3$...

x + 3 > 0 and 2x < 14 Let us consider the first inequality. \[\begin{array}{*{35}{l}} x\text{ }+\text{ }3\text{ }>\text{ }0  \\ x\text{ }+\text{ }3\text{ }-\text{ }3\text{ }>\text{...

\[\begin{array}{*{35}{l}} \left( x\text{ }-\text{ }1 \right)/3\text{ }+\text{ }4\text{ }<\text{ }\left( x\text{ }-\text{ }5 \right)/5\text{ }-\text{ }2  \\ Subtract\text{ }both\text{ }sides\text{...

\[\begin{array}{*{35}{l}} 2\text{ }\left( 3\text{ }-\text{ }x \right)\text{ }\ge \text{ }x/5\text{ }+\text{ }4  \\ 6\text{ }-\text{ }2x\text{ }\ge \text{ }x/5\text{ }+\text{ }4  \\ 6\text{ }-\text{...

\[\begin{array}{*{35}{l}} x\text{ }+\text{ }5\text{ }>\text{ }4x\text{ }-\text{ }10  \\ x\text{ }+\text{ }5\text{ }-\text{ }5\text{ }>\text{ }4x\text{ }-\text{ }10\text{ }-\text{ }5  \\...

(iii) \[\begin{array}{*{35}{l}} ~x~\in ~N  \\ ~2\text{ }<\text{ }5/2\text{ }<\text{ }3  \\ \end{array}\] So when, when x is a natural number, the maximum possible value of x is 2. We know that...

(iii) x ∈ N As natural numbers start from 1 this implies it can never be negative, when x is a natural number, the solution of the given inequation is ∅.

(iii) x ∈ N When, 4 < 25/6 < 5 So when, x is a natural number, the maximum possible value of x is 4. Since,the natural numbers start from 1, the solution of the given inequation is {1, 2, 3,...

Solution: Let us take C1 = 30o C and C2 = 35o We know that the expression for relating celsius and faranheit degree is given by: F = 9/5C + 32 Making use of the formula aove, we can write: $...

Solution: Let us take F1 = 86o F and F2 = 95o We know that the Faranhiet and Celsius relation is given by: F = 9/5C + 32 F1 = 9/5 C1 + 32 F1 – 32 = 9/5 C1 C1 = 5/9 (F1 – 32) = 5/9 (86 – 32) C1  =...

Solution: Given: Rohit obtained 65 and 70 marks in two tests. Let the third test's marks be x. So let's discover a minimum x for which the combined average of all three papers is at least 65 points....

Solution: Let 'x' be the lesser of the two even positive integers that follow. Then x + 2 is the other even integer. Both even integers are higher than 5 and their aggregate is less than 23,...

Solution: Let 'x' be the lesser of the two odd natural numbers that follow. x + 2 is the other odd number. According to the question, the sum of the natural numbers is less than 40, and they are...

Solution: Let ‘x' be the lesser of the two odd positive integers that follow.  Then, x + 2 is the other odd integer. According to the question,...

Solution: It is given to us that : (|x + 2| – x) / x < 2 We can rewrite the above equation as: |x + 2|/x – x/x < 2 |x + 2|/x – 1 < 2 Upon adding 1 to both sides, we get |x + 2|/x – 1 + 1...

Solution: We already know that when we take the reciprocal of any inequality, we must also change the inequality. Also, |x| – 3 ≠ 0 This implies that : |x| > 3 or |x| < 3 Now, For |x| < 3...

Solution: It is given that: |x – 2| / (x – 2) > 0 The above equation clearly states that x≠2 so two cases arise: Case1: x–2>0 It implies that : x>2 In this case, we have:  |x–2| = x – 2...

Solution: It is given that: | (3x – 4)/2 | ≤ 5/12 We can write the above equation as | 3x/2 – 4/2 | ≤ 5/12 | 3x/2 – 2 | ≤ 5/12 Consider ‘r’ to be a positive real number and let ‘a’ be a fixed real...

 Solution: |4 – x | + 1 < 3 Upon subtracting 1 from each side, we get: |4 – x| + 1 – 1 < 3 – 1 |4 – x| < 2 Consider ‘r’ to be a positive real number and let ‘a’ be a fixed real number....

Solution: Consider ‘r’ to be a positive real number and let ‘a’ be a fixed real number. Then, we can write: |x + a| > r ⟺ x > r – a or x < – (a + r) Here, a = 1/3 and r = 8/3 x > 8/3 –...