Find three consecutive integers such that the sum of the first and the second is 4 more than the third.

1st integer = 5

2nd integer  = 6

3rd integer  = 7

Step-by-step explanation:

Let:

1st integer = x

2nd integer = x + 1

3rd integer = x + 2

Now according to given conditions:

x + x + 1 = 4+ x +2

Adding like terms

2x +1 = 6 + x

Taking terms with x on left side and others on right

2x - x = 6 -1

x = 5

Proof:

x + x + 1 = 4+ x +2

Putting x = 5

5 + 5 + 1 = 4 +5 +2

11 = 11

Hence proved!

So,

1st integer = 5

2nd integer = 5 + 1 = 6

3rd integer = 5+ 2  = 7

I hope it will help you!

a - base

b - exponent

so,

about 0.75 hours

step-by-step explanation: