# A little help here?..algebra 2?

Question:The Sum of two numbers is 68. Six times the smaller number is 8 less than half the larger number. What are the numbers?
6x=small number and (1/2)x=larger number
What i have so far is (1/2)x-8+6x=68..but then my answer comes out to be=(120/13) and i dont think thats right =/

Find three consecutive integers such that the sum of the first and third is 6 less than twice the second.
x= first, (2x+1)=second, (x+2)=third
and i came up with 2x+1-6=2x+2 and i ended up with -5=2..and thats not right..

what am i doing wrong?

1. Denote the smaller number by x and the larger number by y. Six times the smaller number is 8 less than half the larger number means that you need to add 8 to (6x), in order to obtain y/2:
y/2 = 6x + 8
y = 12x + 16.
The Sum of two numbers is 68:
x + y
= x + 12x + 16 = 68
13x = 52
x=4
y = 68 - x = 64

2. You're doing everything correct except for the second integer is not (2x+1), but (x+1), if the first is x.

But nevertheless, this problem has no solution. For any three consecutive integers the sum of the first and third is twice the second. For example:
1,2,3 : 1 + 3 = 4 = 2(2)
6,7,8: 6 + 8 = 14 = 2(7)
-11,-10,-9: -11 + (-9) = -20 = 2(-10)

In general:
x= first, (x+1)=second, (x+2)=third
first + third
= x + x + 2
= 2x + 2
= 2(x+1)
= 2second
LET THE NO.=X
SO, OTHER IS 68-X
6X=(68-X)/2-8

OR 12X=68-X-16 OR 13X=62 OR X=4
OTHER NO.=64

SECOND PROBLEM HAS NO SOLUTION!

This article contents is post by this website user, EduQnA.com doesn't promise its accuracy.