Question

Solve 3x+8>2, when
(i) x is an integer
(ii) x is a real number
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Answer 1

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$\Large\fbox{\color{purple}{QUESTION}}$

Solve 3x+8>2, when

(i) x is an integer

(ii) x is a real number

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➡️Given Linear inequality: 3x+8>2

➡️The given inequality can also be written as:

The given inequality can also be written as:3x+8 -8 > 2 -8 …(1)

➡️In the above inequality, -8 is multiplied on both the sides, as it does not change the definition of the given expression.

➡️Now, simplify the expression (1)

➡️⇒ 3x > -6

➡️Now, both the sides, divide it by 3

➡️⇒ 3x/3 > -6/3

➡️⇒ x > -2

✴➡️(i) x is an integer

➡️Hence, the integers greater than -2 are

-1,0,1,2,…etc

➡️Thus, when x is an integer, the solutions of the given inequality are -1,0,1,2,…

➡️Hence, the solution set for the given linear inequality is {-1,0,1,2,…}

✴➡️(ii) x is a real number

➡️If x is a real number, the solutions of the given inequality are all the real numbers, which

If x is a real number, the solutions of the given inequality are all the real numbers, whichare greater than 2.

➡️Therefore, in the case of x is a real number, the solution set is (-2, ∞)

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Answer 2

Step-by-step explanation:

The given inequality is 3x + 8 > 2. (i) The integers greater than –2 are –1, 0, 1, 2,… Thus, when x is an integer, the solutions of the given inequality are –1, 0, 1, 2 … Hence, in this case, the solution set is {–1, 0, 1, 2, …}. (ii) When x is a real number, the solutions of the given inequality are all the real numbers, which are greater than –2. Thus, in this case

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