SOLUTION: There are three consecutive even integers such that the sum of twice the smallest, one less than the middle, and four more than the largest is -15. What are the three integers? and

Question 1164026: There are three consecutive even integers such that the sum of twice the smallest, one less than the middle, and four more than the largest is -15. What are the three integers? and There are two consecutive integers such that four times the smaller plus three less than twice the larger is 71. What are the two integers?
Found 2 solutions by greenestamps, ankor@dixie-net.com:
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

first problem...

In my experience, the algebra is often easier if, instead of calling the three consecutive integers x, x+2, and x+4, we call then x-2, x, and x+2. So that's how I will set up the problem.

Twice the smallest (x-2), plus 1 less then the middle (x), plus 4 more than the largest (x+2), equals -15:

Solve using basic algebra....

Second problem...

Let the two integers be x and x+1.

The sum of 4 times the smaller (x) and 3 less then twice the larger (x+1) is 71:

Again solve using basic algebra....

Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
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!
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