Three consecutive integers have a sum of 78. Find the integers

Let the numbers be x, x+1 and x+2 .

Equation :-

=> x+(x+1)+(x+2)=78

=> x+x+1+x+2=78

=> 3x+3=78

Subtract 3 from both sides..

3x=75

Divide both sides by 3 .

x=25

The three consecutive integers are

x , x+1 and x+2..In the place of x replace 25

∴ 25 , 26 and 27 are the three consecutive integers.

Given : Three consecutive integers have a sum of 78. Find the integers

Solution : Let's assume that the variable be x

The first one be x itself. Now, as it's consecutive the number should be increased by 1. So, let the second one be x + 1. As we know again it will be increased by 1. Hence, the third one be x + 1 + 1 i.e. x + 2

So, we got all the three now let's proceed

The result is 78

Taking 3 as common on the RHS of the equation

1st number = x = 25

2nd number = x + 1 = 26

3rd number = x + 2 = 27

b. would be the answer : )

step-by-step explanation:

step 1: add -16x to both sides.

16x−4y+−16x=64+−16x  

−4y=−16x+64

step 2: divide both sides by -4.

−4y /-4 = -16x + 64/-4

answer:   y=4x−16

hope i'm right

the answer is 21%.