SOLUTION: Find four consecutive even integers such that the sum of twice the fourth integer and half the second integer is 68.

SOLUTION: Find four consecutive even integers such that the sum of twice the fourth integer and half the second integer is 68.

Algebra.Com

Question 158470: Find four consecutive even integers such that the sum of twice the fourth integer and half the second integer is 68.
Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
X, X+2, X+4, X+6 ARE THE 4 EVEN INTEGERS.
2(X+6)+(X+2)/2=68
2X+12+(X+2)/2=68
2X+(X+2)/2=68-12
(2*2X+X+2)/2=56
(4X+X+2)/2=56
5X+2=2*56
5X=112-2
5X=110
X=110/5
X=22 THE FIRST INTEGER.
PROOF:
2(22+6)+(22+2)/2=68
2*28+24/2=68
56+12=68
68=68

RELATED QUESTIONS
find four consecutive even integers such that the sum of the second integer and the... (answered by JulietG)

Find five consecutive integers such that the sum of the first ,second ,and fifth integers (answered by Alan3354)

find the 4 consecutive even integers such that the sum of the first and the fourth even... (answered by vleith)

please help me solve this problem... find four consecutive even integers such that twice... (answered by MathTherapy)

Find four consecutive even integers such that twice the third integer subtracted from... (answered by josgarithmetic)

1.find two consecutive odd integers if the larger is four less than one third the... (answered by josgarithmetic)

can you find the second of four consecutive even integers such that the product of the... (answered by stanbon)

Find two consecutive even integers such that the sum of the first integer and twice the... (answered by ewatrrr)

Find four consecutive even integers such that the twice the first added to three times... (answered by Alan3354)