Question 571856: Find four consecutive even integers such that 4 times the sum of the first and fourth is 8 greater than 12 times the third. I have no idea how to do this problem. Someone explain it to me please?
Answer by mathsmiles(68)   (Show Source):
You can put this solution on YOUR website! So we're looking for 4 numbers. Let's make the first number X. (This is algebra, right?)
Since they're asking for even integers, the next number is X+2. Make sense?
The 3rd number is X+4.
The 4th number is X+6.
Next they say if we add the first and 4th and then multiply by 4 ...
4[(X) + (X+6)]
The results will be 8 more than 12 times the 3rd
So the other side of the equation is 12(X+4) but we need to add 8 to equal the first side since we know the first side is 8 more than this.
8+12(X+4)
Putting these together:
4[X + (X + 6)] = 8 + 12(X + 4) Now combine like terms on the left:
4(2X + 6) = 8 + 12(X + 4) Now multiplying the paren on the right
4(2X + 6) = 8 + 12X + 48 Combining terms on the right:
4(2X + 6) = 12X + 56 Multiplying the remaining paren on the left:
8X + 24 = 12X + 56 Subtract 8X from both sides to bring X to one side alone:
24 = 4X + 56 Subtract 56 from both sides:
-32 = 4X Divide by 4 on both sides:
X = -8
Going back to the problem, we're supposed to find 4 consecutive even integers so the first is -8, the next would be -6, -4, -2.
Let's check:
4 times the sum of the 1st and 4th:
sum of the 1st and 4th is -8 + -2 = -10
4 times that is -40
8 greater than 12 times the third:
12 times the 3rd is 12 x -4 = -48
adding the 8 we get -48 + 8 = -40
-40 = -40 Correct!
Answer: -8, -6, -4, -2
Hope this helps so you can do more problems like this with more confidence. Good luck!!
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