Question 694986

find 2 consecutive integers such that the larger is nine more than twice the smaller


Let the smaller integer be S
Then larger = S + 1


Therefore, S + 1 = 2S + 9


S - 2S = 9 - 1


- S = 8


S, or smaller number = {{{8/-1}}}, or {{{highlight_green(- 8)}}}


Larger integer = - 8 + 1, or {{{highlight_green(- 7)}}}


You can do the check!!


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