Question 1164026
There are three consecutive even integers
a, (a+2), (a+4)
 such that the sum of twice the smallest, one less than the middle, and four more than the largest is -15.
2a + (a+2-1) + (a+4+4) = -15
2a + a + 1 + a + 8 = -15
4a + 9 = -15
4a = - 15 - 9
4a = -24
a = -24/4
a = -6
 What are the three integers?
-6, -4, -2
:
There are two consecutive integers
a. (a+1)
 such that four times the smaller plus three less than twice the larger is 71.
4a + 2(a+1) -3 = 71
4a + 2a + 2 -3 = 71
6a - 1 = 71
6a = 71 + 1
a = 72/6
a = 12
 What are the two integers?
12, 13
:
You should check these in the given statements to be sure they are correct!