Find two consecutive even integers such that the sum of the larger and twice the smaller is 62. PLEASE HELP I DON'T GET IT!/:
let n = smaller, then since they are "consecutive even integers", n + 2 = larger.
(n + 2) + 2n = 62
Solve for n.
No problem, I can help you with that!
Let's use algebra to solve the problem.
Step 1: Let's represent the smaller even integer as 'x'. Since it is an even integer, the next even integer can be represented as 'x + 2' (since the difference between consecutive even integers is always 2).
Step 2: According to the problem, the sum of the larger even integer (x + 2) and twice the smaller integer (2x) is 62. We can write this as an equation:
(x + 2) + 2x = 62
Step 3: Now, let's solve this equation to find the value of 'x'.
Combine like terms:
3x + 2 = 62
Subtract 2 from both sides of the equation:
3x = 60
Step 4: Divide both sides of the equation by 3 to solve for 'x':
x = 20
So the smaller even integer is 20.
Step 5: Now that we know the value of 'x', we can find the next even integer by adding 2 to it:
x + 2 = 20 + 2 = 22
So the larger even integer is 22.
Therefore, the two consecutive even integers are 20 and 22.
To find two consecutive even integers, we can use the following steps:
Step 1: Define the variables:
Let's assume the first even integer as 'x', and the next consecutive even integer as 'x + 2'.
Step 2: Set up the equation:
The sum of the larger (x + 2) and twice the smaller (2x) is 62.
So, we can write the equation as: (x + 2) + 2x = 62.
Step 3: Solve the equation:
Combine like terms: 3x + 2 = 62.
Subtract 2 from both sides of the equation:
3x = 60.
Divide both sides by 3:
x = 20.
Step 4: Find the other even integer:
Substitute the value of x back into one of the integers:
x + 2 = 20 + 2 = 22.
Therefore, the two consecutive even integers are 20 and 22.