Answer :
Final answer:
The problem involves setting up an equation with the smaller even integer as x, leading to the larger consecutive integer being x + 2, and the relation that x + 2 = (1/2)x + 11. Solving this algebraic equation reveals that the two consecutive even integers in question are 18 and 20.
Explanation:
To solve the problem where the greater of two consecutive even integers is eleven more than half the lesser, we must first set up an algebraic expression to represent the scenario.
Let the smaller even integer be represented as x, then the next consecutive even integer, being larger, would be x + 2.
The problem states that the larger integer is eleven more than half of the smaller integer, so we can write the equation x + 2 = (1/2)x + 11.
To find the value of x, we can manipulate the equation:
- Multiply both sides of the equation by 2 to eliminate the fraction: 2(x + 2) = x + 22.
- Simplify the left side: 2x + 4 = x + 22.
- Subtract x from both sides: x + 4 = 22.
- Subtract 4 from both sides to solve for x: x = 18.
Therefore, the smaller even integer is 18, and the larger even integer, being 18 + 2, is 20. This satisfies the condition as 20 is indeed eleven more than half of 18 (which is 9).
