Question

Find three consecutive even integers. Such that 7 times of first integer is 4 more than the sum of second and third integers

Answer 1

1st even integer = 2

2nd even integer = 4

3rd even integer = 6

Explanation:

Let the consecutive even integers be:

1st = 2(x)

2nd= 2(x +1) = 2x + 2

3rd = 2(x + 2) = 2x + 4

According to Given conditions:

7(2x) = 4 + 2x +2 + 2x + 4

By Simplifying:

14x = 10 + 4x

Subtracting 4x from both sides

14x - 4x = 10 + 4x -4x

10x = 10

Dividing both sides by 10 we get:

x = 1

Now putting value of x in supposed integers:

1st even integer = 2(1) = 2

2nd even integer = 2(1)+2 = 4

3rd even integer = 2(1) + 4 = 6

I hope it will help you!