SOLUTION: Find four consecutive even integers such that if the sum of the first and third is multiplied by 6 the result is 8 more than 10 times the 4th. can you please show me the formula a

Question 5983: Find four consecutive even integers such that if the sum of the first and third is multiplied by 6 the result is 8 more than 10 times the 4th.
can you please show me the formula also because i have looked all through my text book and it doesnt explain it in detail and skips steps.
Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
Let the first even integer be x.
The second consecutive even integer will be (x + 2)
The third one will be (x + 4)
And the fourth one will be (x + 6)
Now write the necessary equations.
The sum of the 1st and 3rd multiplied by 6 is:
6[x + (x + 4)]
The result is (=): 8 more (+8) than 10 times the fourth 10(x + 6) or:
10(x + 6) + 8
Put it all together and solve for x:
6[x + (x + 4)] = 10(x + 6) + 8 Simplify.
6[2x + 4] = 10x + 60 + 8
12x + 24 = 10x + 68 Subtract 10x from both sides.
2x + 24 = 68 Subtract 24 from both sides.
2x = 44 Divide both sides by 2.
x = 22 This is the first even integer.
x + 2 = 24 This is the second consecutive even integer.
x + 4 = 26 This is the third consecutive even integer.
x + 6 = 28 This is the fourth consecutive even integer.
Check:
6[x + (x + 4)] = 6[22 + (22 + 4)] = 6[22 + 26] = 6[48] = 288
10(x + 6) + 8 = 10(22 + 6) + 8 = 10(28) + 8 = 280 + 8 = 288
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