Find three consecutive integers such that the sum of twice the second integer and 4 times the third integer is two less than seven times the first integer.

12,13 and 14

Step-by-step explanation:

Let the three consecutive integers be n, n+1 and n+2

ATQ,

The sum of twice the second integer and 4 times the third integer is two less than seven times the first integer.

[2(n+1)+4(n+2)] = 7n-2

or

2n+2+4n+8=7n-2

Taking like terms together,

6n-7n = -2-8-2

n = 12

So,

First integer = 12

Second integer = 12+1 = 13

Third integer = 12+2 = 14

Hence, three consecutive integers are 12,13 and 14.

1) 2/5

2) zero=0/1

3) zero=0/2

4) -4/5

5) undefined=3/0

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