find integer number x such that 4x-3 divisible by x-2

Hươu 🦒🦒🦒Let's solve this problem step-by-step.

We are given that 4x - 3 is divisible by x - 2. In other words, we need to find an integer x such that:

(4x - 3) / (x - 2) is an integer.

We can rewrite 4x - 3 as follows:

4x - 3 = 4(x - 2) + 5

Now, let's divide (4x - 3) by (x - 2):

(4x - 3) / (x - 2) = [4(x - 2) + 5] / (x - 2) = 4(x - 2) / (x - 2) + 5 / (x - 2) = 4 + 5 / (x - 2)

For (4x - 3) / (x - 2) to be an integer, 5 / (x - 2) must be an integer. This means that (x - 2) must be a divisor of 5.

The divisors of 5 are: -5, -1, 1, 5.

Therefore, we have the following cases:

x - 2 = -5 => x = -3

x - 2 = -1 => x = 1

x - 2 = 1 => x = 3

x - 2 = 5 => x = 7

Thus, the integer values of x that satisfy the condition are:

x = -3, 1, 3, 7

Verification:

If x = -3: 4(-3) - 3 = -15, -3 - 2 = -5, -15 / -5 = 3 (integer)

If x = 1: 4(1) - 3 = 1, 1 - 2 = -1, 1 / -1 = -1 (integer)

If x = 3: 4(3) - 3 = 9, 3 - 2 = 1, 9 / 1 = 9 (integer)

If x = 7: 4(7) - 3 = 25, 7 - 2 = 5, 25 / 5 = 5 (integer)

Final Answer: The final answer is −3,1,3,7

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