The lesser of two consecutive even integers is 10 more than one-half the greater. Find the integers.
Let's assume the first even integer as 'x', which means the next even integer would be 'x + 2' since they are consecutive even integers.
According to the given information, the lesser integer (x) is 10 more than one-half the greater integer ((1/2)*(x + 2)). We can express this information as the following equation:
x = (1/2)*(x + 2) + 10
To solve this equation, we will start by eliminating the fraction by multiplying both sides by 2:
2x = x + 2 + 20
Simplifying further, we have:
2x = x + 22
We solve for x by subtracting x from both sides:
x = 22
So the first even integer is 22. Substituting this value back into the expression for the second even integer:
x + 2 = 22 + 2 = 24
Therefore, the two consecutive even integers are 22 and 24.
To solve this problem, let's assume that the smaller even integer is x, and the larger even integer is x + 2 (since they are consecutive even integers).
According to the given information, the smaller integer is 10 more than one-half of the larger integer. Translating this into an equation, we have:
x = (1/2)(x + 2) + 10
Let's solve for x now:
Multiply both sides of the equation by 2 to eliminate the fraction:
2x = x + 2 + 20
Simplify the equation:
2x = x + 22
Subtract x from both sides:
2x - x = 22
Simplify further:
x = 22
Therefore, the smaller even integer is x = 22, and the larger even integer is x + 2 = 22 + 2 = 24.
So, the two consecutive even integers are 22 and 24.
One integer is n and the other is n+2.
1/2(n+2)=n-10
solve for n (the lesser number).