Final answer:
To solve the inequality (7^(x+y))/7 > 1, we subtract the exponents based on the quotient rule to get 7^(x+y-1). Since 7 raised to any positive number is greater than 1, the inequality implies x+y-1 > 0, which simplifies to x+y > 1.
Step-by-step explanation:
The question appears to involve the simplification of an inequality with exponential expressions. To solve the inequality (7^(x+y))/7 > 1, we can start by observing that the expression on the left side can be simplified since the bases are the same.
First, we employ the rules of exponents, specifically the quotient rule, which states that when dividing like bases, we subtract the exponents. In our case, the 7 in the denominator can be considered as 7^1. So, the expression simplifies to 7^((x+y)-1).
Next, we see that for the inequality to hold true, the exponent (x+y-1) must be greater than 0, because 7 raised to any positive exponent is greater than 1. This leads us to x+y-1 > 0.
To isolate the variables, we add 1 to both sides of the inequality to obtain x+y > 1. So, the solution to the inequality is the set of all x and y that satisfy this condition.
The complete question is: equality: ⁽⁷ˣ⁺ʸ⁾/⁽⁷⁾>¹ provided and will update whenever a value is updated. The regions will be a is: