Question

There are 3 consecutive year integers. If 6 times the first is increased by the third, the result is 51. Find the smallest integer.

Answer 1

Hi there !!
Here's your answer

Let the smallest consecutive integer be x

Since these integers are consecutive,

2nd integer = x + 1

3rd integer = x + 2

Given,
If 6 times the first is increased by the third, the result is 51

6 times the 1st integer increased by the third
= 6(x) + x + 2

= 6x + x + 2 = 7x + 2

The result is 52

So,
7x + 2 = 51

7x = 51 - 2

7x = 49

$x = \frac{49}{7}$
x = 7

Thus,

the smallest integer is 7