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Find three consecutive integers such that twice the greatest integer is 3 less than 3 times the least integer

Mar 13, 2020

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Find three consecutive integers such that twice the greatest integer is 3 less than 3 times the least integer

Three consecutive integers: $$(n-1),\ n,\ (n+1)$$

Twice the greatest integer is 3 less than 3 times the least integer:

$$\begin{array}{|rcll|} \hline 2(n+1) &=& 3(n-1)-3 \\ 2n+2 &=& 3n-3-3 \\ 2n+2 &=& 3n-6 \\ 2n+8 &=& 3n \\ 8 &=& 3n-2n \\ 8 &=& n \\ \mathbf{n} &=& \mathbf{8} \\ \hline \end{array}$$

The three consecutive integers are: $$\mathbf{7,\ 8,\ 9}$$

Mar 13, 2020