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

Since your integers are consecutive and even, they are 2 apart. So you can call the integers x, x+2, and x + 4.

Now plug those numbers into the information your are given.

7x (seven times the first integer)

-4 (you have to subtract 4 since it is 4 more than the other number)

x+2+x+4 (the sum of the second and third integer)

7x - 4 = x + 2 + x + 4

Combining like terms on the right: 7x - 4 = 2x + 6

Subtract 2x from both sides: 5x - 4 = 6.

Add 4 to both sides: 5x = 10

Divide both sides by 5: x = 2

If x = 2, x + 2 = 4 and x + 4 = 6. Your three numbers are 2, 4, and 6.

Answer 2

Let 2n-2, 2n and 2n+2 be the integers.

7(2n-2) = (4n+2)+4

14n-14 = 4n+6

10n=20

n=2 so that the integers are 2, 4 and 6

Answer 3

The integers are n, n + 2, and n + 4

7n = (n + 2) + (n + 4) + 7

Can you solve for n and answer?