Answer: To find the least positive value of x that satisfies the given congruence, we can use the trial and error method or algebraic manipulation.
Using the trial and error method, we can start by plugging in values of x and checking if the congruence is satisfied.
For x = 1, (x + 3) = 4, so 89 ≡ 0 (mod 4), which is not a multiple of 3.
For x = 2, (x + 3) = 5, so 89 ≡ 1 (mod 4), which is not a multiple of 3.
For x = 3, (x + 3) = 6, so 89 ≡ 2 (mod 4), which is not a multiple of 3.
For x = 4, (x + 3) = 7, so 89 ≡ 3 (mod 4), which is not a multiple of 3.
For x = 5, (x + 3) = 8, so 89 ≡ 0 (mod 4), which is a multiple of 3.
Therefore, the least positive value of x that satisfies the congruence is x = 5.
Alternatively, we can use algebraic manipulation to solve the congruence. We have:
89 ≡ (x + 3) (mod 4)
=> 89 ≡ x + 3 (mod 4)
=> 86 ≡ x (mod 4) (subtracting 3 from both sides)
Now we need to find the least positive value of x that satisfies this congruence and is a multiple of 3.
The solutions for this congruence are x = 2 (mod 4) and x = 6 (mod 4).
Plugging in x = 2, we get 89 ≡ 5 (mod 4), which is not a multiple of 3.
Plugging in x = 6, we get 89 ≡ 3 (mod 4), which is not a multiple of 3.
Therefore, the least positive value of x that satisfies the congruence and is a multiple of 3 is x = 10 (which is equivalent to x = 2 (mod 4) and x = 6 (mod 4)), but this is not the answer to the original question since x must be positive.
Plugging in x = 14, we get 89 ≡ 1 (mod 4), which is not a multiple of 3.
Plugging in x = 18, we get 89 ≡ 3 (mod 4), which is not a multiple of 3.
Finally, plugging in x = 22, we get 89 ≡ 1 (mod 4), which is not a multiple of 3.
Plugging in x = 26, we get 89 ≡ 3 (mod 4), which is not a multiple of 3.
Since we want x to be positive, we can stop here and conclude that the least positive value of x that satisfies the given congruence is x = 5.
Explanation: