# 1.) The sum of three consecutive integers is 78. Find the integers. 2.) Find two consecutive odd integers such that twice the first increased by three times the second equals 111.

2
by airamaelaurio

## Answers

2014-11-18T14:43:54+08:00
Question 1:
x=1st integer
x+1=2nd integer
x+2=3rd integer
then,

x + x+1+x+2 = 78
then

3x+3=78
3x=75
x=75/3
x=25

Substituting,

x = 25 (1st integer)
x+1= 26 (2nd integer)
x+2= 27 (3rd integer)

Check:

25+26+27 = 78

Done.

Question 2:

Let x be 1st odd
Let x +2 be 2nd consecutive odd number

then,

2x + 3(x+2) = 111
2x+3x+6 = 111
5x+6=111
5x=105
x= 105/5
x=21

Substituting, we'll have,

1st odd (x) = 21
2nd odd (x+2) = 23

checking,

(21 x 2) + ( 23 x 3) = 111
42 + 69 = 111
111=111

Done, have fun! :p

brilliant... we have the same answer... LOL
2014-11-18T15:02:36+08:00
1. Let an integer be a.

So we can write it as.

a + (a + 1) + (a + 2) = 78

*Notice I added 1 and 2 to a to represents the other two consecutive integers

lets add

3a + 3 = 78

Transpose 3

3a = 78 - 3

Subtract

3a = 75

Divide both by 3

3a/3 = 75/3

Therefore

a = 25

So now we can determine the 3 consecutive integers

The integers are 25, 26, and 27

2. Let the integer be a again

We can write it as

Twice the first is (2a)

Increased by is (2a +)

Three times the second is (2a + 3(a + 2))

*Notice that I added 2 to 3a since we are looking for and odd consecutive integer.

Equals to 111 is

(2a) + 3(a + 2) = 111

Let's solve by multiplying 3 to a + 2

2a + 3a + 6 = 111

Now add like terms

5a + 6 = 111

Transpose 6

5a = 111 - 6

Subtract

5a = 105

Divide both by 5

5a/5 = 105/5

Therefore

a = 21

1st odd is 21

2nd odd is 23

Let a be 21 and b is 23

The new equation will be like this

2a + 3b = 111

Substitute a and b value

2(21) + 3(23) = 111

Multiply

42 + 69 = 111

Add

111 = 111 ◘ Correct

I hope it helps