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put this solution on YOUR website! Find four consecutive multiples of five such that twice the sum of the two biggest exceeds twice the smallest by ten.
5k = smallest multiple of 5
5(k+1) = next to smallest multiple of 5
5(k+2) = next to biggest multiple of 5
5(k+3) = biggest multiple of 5
>>twice the sum of the two biggest...
sum of the two biggest = 5(k+2) + 5(k+3)
twice the sum of the two biggest = 2[5(k+2) + 5(k+3)]
...exceeds twice the smallest by ten.<<
twice the smallest = 2(5k)
2[5(k+2) + 5(k+3)] = 2(5k) + 10
Can you solve that for k? If not post again asking how.
Solution: k = -4
smallest multiple of 5 = 5k = 5(-4) = -20
next to smallest multiple of 5 = 5(k+1) = 5(-4+1) = 5(-3) = -15
next to biggest multiple of 5 = 5(k+2) = 5(-4+2) = 5(-2) = -10
biggest multiple of 5 = 5(k+3) = 5(-4+3) = 5(-1) = -5
So they are -20, -15, -10, -5.
Edwin