What four consecutive integers have a sum of 66?
Q2. What three consecutive even integers have a sum of 36
Q3. What three consecutive odd integers have a sum of 57
Q4. What three consecutive integers have a sum of 33
Answers
1.X + (X+1) + (X+2) + (X+3) = 66
4X + 1 + 2 + 3 = 66
4X + 6 = 66
4X = 60
X = 15
2.in image
3.in image
4. 1st integer: x
2nd integer: x+1
3rd integer: x+2
That is necessary because they are consecutive integers. Since the sum is 33, we need to create an equation.
x+x+1+x+2=33.
Simplify:
3x+3=33.
Opposite operations:
3x=-3+33.
To get the 3 close to the 33, we needed to make it negative, which is the opposite operation of the positive 3.
So,
3x=30.
Divide by 3:
x=10.
The first integer, x, equals 10.
To go with the guide that we already created,
1st integer: x=10
2nd integer: x+1=11
3rd integer:x+2=12.
Therefore, the three consecutive integers are 10, 11, and 12.
To check that, add them up. They all equal 33 and they are consecutive, which means this is the right answer!
Step-by-step explanation:
(a) X + (X+1) + (X+2) + (X+3) = 66
Combine like terms by summing the X values together and the constants:
4X + 1 + 2 + 3 = 66
4X + 6 = 66
Subtract 6 from both sides:
4X = 60
Divide both sides by 4:
X = 15
This means the first number in the sequence of consecutive numbers is 15. So the four consecutive integers that sum to 66 are then: 15, 16, 17, and 18.
(b) Let 2n = the first even number, where n is an integer.
Let 2n + 2 = the second even number, and ...
Let 2n + 4 = the third even number.
With the above three equalities in mind, the statement that, "The sum of three consecutive even numbers is 36" can be translated mathematically into the following equation to be solved for n:
2n + (2n + 2) + (2n + 4) = 36
2n + 2n + 2 + 2n + 4 = 36
Collecting like-terms on the left, we have:
6n + 6 = 36
6n + 6 - 6 = 36 - 6
6n + 0 = 30
6n = 30
(6n)/6 = 30/6
(6/6)n = 5
(1)n = 5
n = 5
Therefore, ...
2n = 2(5) = 10
2n + 2 = 10 + 2 = 12
2n + 4 = 10 + 4 = 14
CHECK:
2n + (2n + 2) + (2n + 4) = 36
2(5) + (2(5) + 2) + (2(5) + 4) = 36
10 + 12 + 14 = 36
36 = 36
Therefore, the three consecutive even numbers are: 10, 12, and 14.
(c) Let the first and the smallest integer be
x
−
2
Then... the second integer will be
x
Then.... the third integer will be
x
+
2
Now... their sum is
57
x
−
2
+
x
+
x
+
2
=
57
You get
3
x
=
57
x
=
57
3
x
=
19
( that's the second integer)
First integer
x
−
2
=
19
−
2
x
−
2
=
17
Third integer
x
+
2
=
19
+
2
x
+
2
=
21
(d) We are told that the sum of three consecutive integers is
−
33
. Let's call the lowest of these three integers
x
. Since the numbers are consecutive, it must be the case that the next smallest integer is
x
+
1
and the greatest integer is
x
+
2
.
So we can rewrite the problem as the algebraic statement
x
+
(
x
+
1
)
+
(
x
+
2
)
=
−
33
. The rest is algebra.
x
+
(
x
+
1
)
+
(
x
+
2
)
=
−
33
x
+
x
+
1
+
x
+
2
=
−
33
3
x
+
3
=
−
33
3
x
=
−
36
x
=
−
12
Our lowest integers is
−
12
. It follows that our next two integers are
−
11
and
−
10
. We confirm that
−
12
−
11
−
10
=
−
23
−
10
=
−
33
.