Binary operations

Step 1: (2*x=2^2+x=4+x). Step 2: (4+x=9) gives (x=5). Step 3: Carefully identify which element is squared.

Step 1: (x*3=x+3^2=x+9). Step 2: (x+9=15) gives (x=6). Step 3: First square the second element, then solve the equation.

Step 1: (x*4=|x-4|=3). Step 2: This gives (x=1) or (x=7), and (1) is in the options. Step 3: Absolute value equations usually give two possibilities.

Step 1: Put (a=2) and (b=6) in the rule. Step 2: (2*6=\frac{2\cdot 6}{2+6}=\frac{12}{8}=\frac{3}{2}). Step 3: Do not forget to simplify the fraction.

Step 1: (1*2=\frac{1+2}{1+1\cdot 2}). Step 2: This becomes (\frac{3}{3}=1). Step 3: Keep numerator and denominator correct in fraction-based operations.

Step 1: (b*a=\frac{b+a}{1+ba}). Step 2: Since (b+a=a+b) and (ba=ab), the value is the same. Step 3: For commutativity, reverse the whole rule and compare.

Step 1: The rule chooses the greater of the two elements. Step 2: Between (2) and (4), the greater element is (4). Step 3: In such rules, compare the elements instead of calculating.

Step 1: The rule chooses the smaller of the two elements. Step 2: Between (3) and (1), the smaller element is (1). Step 3: In a minimum-based operation, the answer is the selected element.

Step 1: The sum of two even integers is again an even integer. Step 2: So (a+b) does not go outside the set of even integers. Step 3: For special sets, check the property with examples and general reasoning.

Step 1: The sum of two odd integers is an even integer. Step 2: For example, (3+5=8), which is not odd. Step 3: For closure, the result must stay in the same set.

Step 1: Two multiples of (3) can be written as (3m) and (3n). Step 2: Their sum is (3m+3n=3(m+n)), again a multiple of (3). Step 3: For multiples, use general form reasoning.

Step 1: Take two multiples (5m) and (5n). Step 2: (5m-5n=5(m-n)), which is again a multiple of (5). Step 3: The multiple property can remain true even under subtraction.

Step 1: (a*b=a+b-1) and (b*a=b+a-1). Step 2: Since (a+b=b+a), both are equal. Step 3: Addition-based rules with the same constant often remain commutative.

Step 1: ((a*b)*c=(a+b-1)+c-1=a+b+c-2). Step 2: (a*(b*c)=a+(b+c-1)-1=a+b+c-2). Step 3: If both forms match, associativity is true.

Step 1: First, the identity is (1) because (a+1-1=a). Step 2: (4*x=1) gives (4+x-1=1), so (x=-2). Step 3: Confirm the identity before finding inverse.

Step 1: If (5) is the identity, then (a*5=a). Step 2: (a+5-k=a) gives (k=5). Step 3: For an unknown constant, the identity condition gives the fastest route.

Step 1: Write (a*(-4)=a). Step 2: (a-4+k=a) gives (k=4). Step 3: In sign-based questions, write the equation line by line.

Step 1: For commutativity, (ka+b=kb+a) must hold for all (a,b). Step 2: This generally works when (k=1), giving (a+b) on both sides. Step 3: For unknown coefficients, apply the condition for all (a,b).

Step 1: (2*3=2+3k) and (3*2=3+2k). Step 2: Equating them gives (2+3k=3+2k), so (k=1). Step 3: Use two unequal numbers to find the parameter in a commutativity condition.

Step 1: When an element does not change any (a) under the operation, it is the identity element. Step 2: Here (a*0=a) and (0*a=a) also holds. Step 3: To identify identity, check whether the result stays as the original element.