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Find two consecutive integers such that 6 times the smaller integer is 26 less than 7 times the larger integer.

Sagot :

Answer:

19 and 20

Step-by-step explanation:

let 'x' = first integer (smaller of the two)

let 'x+1' = next consecutive integer

6x = 7(x+1) - 26

6x = 7x + 7 - 26

6x = 7x - 19

-x = -19

x = 19

x+1 = 20

Check:

19(6) = 7(19+1) - 26

114 = 140 - 26

114 = 114

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