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Sagot :
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
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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