Math, asked by harshali5019, 10 months ago

Solve 5x– 3 < 7, when (i) x is an integer (ii) x is a real number

Answers

Answered by amitnrw
1

Given :  5x– 3 < 7

To find : x (i) x is an integer (ii) x is a real number

Solution:

5x– 3 < 7

lets add 3 on both sides

=> 5x - 3 + 3 < 7 + 3

=> 5x < 10

Dividing by 5 on both sides

=> x <  2

(i) x is an integer

=> x  =    -∞ , ........... -2  , - 1, 0 , 1  

x = { - ∞,......... -2  , - 1, 0 , 1 }

(ii) x is a real number

=> x  ∈ (-∞ , 2)

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Answered by AnIntrovert
35

(i) Given that, 5x – 3 < 7

Now by adding 3 both side we get,

5x – 3 + 3 < 7 + 3

Above inequality becomes

5x < 10

Again by dividing both sides by 5 we get,

5x/5 < 10/5

x < 2

When x is an integer then

It is clear that that the integer number less than 2 are…, -2, -1, 0, 1.

Thus, solution of 5x – 3 < 7 is …,-2, -1, 0, 1, when x is an integer.

Therefore the solution set is {…, -2, -1, 0, 1}

(ii) Given that, 5x – 3 < 7

Now by adding 3 both side we get,

5x – 3 + 3 < 7 + 3

Above inequality becomes

5x < 10

Again by dividing both sides by 5 we get,

5x/5 < 10/5

x < 2

When x is a real number then

It is clear that the solutions of 5x – 3 < 7 will be given by x < 2 which states that all the real numbers that are less than 2.

Hence the solution set is x ∈ (-∞, 2)

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