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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.

Sagot :

Alvedeguzidn
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
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