Find three consecutive even integers such that the sum of the smallest integer and twice the median integer is 20 more than the largest integer.
x + 2(x+2) = x+4 + 20
Now just solve for x, the smallest
shush
Let's call the three consecutive even integers x, x+2, and x+4.
According to the problem, the sum of the smallest integer (x) and twice the median integer (2(x+2)) is 20 more than the largest integer (x+4).
x + 2(x+2) = (x+4) + 20
Simplifying this equation step-by-step:
x + 2x + 4 = x + 24
Combining like terms:
3x + 4 = x + 24
Subtracting x from both sides:
2x + 4 = 24
Subtracting 4 from both sides:
2x = 20
Dividing both sides by 2:
x = 10
Now we have the value of the smallest integer. We can find the other two integers by adding 2 and 4 to x:
x+2 = 10+2 = 12
x+4 = 10+4 = 14
So the three consecutive even integers are 10, 12, and 14.
To find three consecutive even integers, we can use algebraic expressions. Let's assume the smallest even integer is "x". Since we need three consecutive even integers, the next two even integers will be "x + 2" and "x + 4".
According to the problem, the sum of the smallest integer and twice the median integer (2nd number) is 20 more than the largest integer. So, we can set up the following equation:
x + 2( x + 2) = (x + 4) + 20
Now let's solve this equation step by step:
x + 2( x + 2) = (x + 4) + 20
Distribute 2 to x and 2:
x + 2x + 4 = x + 4 + 20
Combine like terms:
3x + 4 = x + 24
Subtract x from both sides:
3x - x + 4 = x - x + 24
2x + 4 = 24
Subtract 4 from both sides:
2x + 4 - 4 = 24 - 4
2x = 20
Divide both sides by 2:
(2x)/2 = 20/2
x = 10
So, the smallest even integer is 10. The next two consecutive even integers are:
1st Consecutive even integer: x = 10
2nd Consecutive even integer: x + 2 = 10 + 2 = 12
3rd Consecutive even integer: x + 4 = 10 + 4 = 14
Therefore, the three consecutive even integers are 10, 12, and 14.