Question

what are three consecutive integers whose sum is 63 ​

Answer 1

Answer:

Let the three consecutive integers be x, x+1, x+2.

x + x+1 + x+2 = 63

3x + 3 = 63

3x = 63 - 3

3x = 60

x =$\frac{60}{3}$

x = 20

Because 60 ÷ 3 = 20.

Answer 2

Answer: let the three consecutive integers be a, a+1, a+2

a+ (a+1) + (a+2) = 63

or 3a+ 3 = 63

or 3a= 63 - 3

or a= 60/3

therefore, a= 20, a+1= 21, a+2= 22

thus the integers are 20, 21 and 22.

Step-by-step explanation: just so you know or is written to begin the next step.

By taking the three consecutive integers as a, a+1, a+2 and reading the question we get the equation: a+ (a+1) + (a+2) = 63

so in a+ (a+1) + (a+2) = 63 we cant add the numbers and variables as we don't know the value of a so we add the numbers and variables differently. We need to separate the variables and numbers in order to do that...

a+a+a+1+2= 63

So a, a and a add up to be 3a and 1+2=3

So we get 3a+3=63

we take 3 and transfer it from left side of '=' to right, and as we all know when we change sides the sign changes so 3 becomes -3 (minus 3).

So we get 3a= 63-3 [then we subtract 3 from 63 that gives us 60]

3a=60 [then we take 3 from 3a and bring it to right hand side so 3a= 60 becomes a= 60 divided by 3]

a= 60/3 [ 20 multiplied by 3= 60 so 60/3=20]

a= 20

now that we have the value of a we can find out the integers: in the beginning we wrote let the three consecutive integers be a, a+1, a+2 so

1st integer that is a=20

2nd integer that is a+1=20=1 that is equal to 21

3rd integer that is a+2=20+2 that is equal to 22

Done....