Questions on Word Problems: Problems with consecutive odd even integers answered by real tutors!

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lets call the 2 numbers a and b
:
sum of the numbers is a+b which equals 24----------------->a+b=24...eq 1
:
the product of the numbers is ab which equals 143.-------->ab=143...eq 2
:
lets take eq 1 and write it in terms of a--->a=24-b
:
take a's value from eq 1 and plug it into eq 2
:
(24-b)b=143---distribute--->
:

so b can be 11 or 13 if b is 11 then a is 13 if b is 13 the a is 11
:our two numbers are

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Quadratic equation (in our case ) has the following solutons: For these solutions to exist, the discriminant should not be a negative number. First, we need to compute the discriminant : . Discriminant d=4 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 13, 11. Here's your graph:

:

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we are looking for 2 consecutive odd integers that are also positive. Lets call them a and a+2.
distribute and combine
:

:
factor
:

:
a=5 and -2 but we can throw out the negative so a=5
:
integers are and

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taking the formula n+1,n+3,n+5....where n=1,2,3,4,........to get even integers we have
2,4,6,8,.......48 this becomes an arithmetic series ie
2+4+6+8+......+48 where first term(a)=2,common diffrence(d)=2
we first find the number of terms in this seies
using the formula where L is the last term.
48=2+(n-1)2
n=24 therefore there are 24 terms.
Now we find the the sum using the formula
where S=sum n=number of terms
S=[24/2(2*2+(24-1)2]
=12(4+46)
=600
therefore the sum is 600

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Let x & x+1 be the 2 integers.
2(x+1)^2=3x+57
2(x^2+2x+1)=3x+57
2x^2+4x+2=3x+57
2x^2+4x-3x-57+2=0
2x^2+x-55=0
(2x+11)(x-5)=0
2x=11=0
2x=-11
x=-11/2
x=-5.5 ans. -5.5+1=-4.5 ans.
x-5=0
x=5 ans. 5+1=6 ans.
Proof:
2*6^2=3*5+57
2*36=15+57
72=72

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The sum of 3 consecutive odd numbers is 39
how would you figure it out?
----------------
n is the 1st number
n+2 is the 2nd
n+4 is the 3rd
n + n+2 + n+4 = 39
3n+6 = 39
3n = 33
n = 11
so it's 11, 13, 15
That's one way.

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Let x, x+2 & x+4 be the three integers.
3x=(x+4)+2
3x=x+6
3x-x=6
2x=6
x=6/2
x=3 ans.
Proof:
3*3=(3+4)+2
9=7+2
9=9

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To solve for 'i', isolate the unknown to one side of the equation:
12i+18=9i
Subtracting 9i from both sides:
3i+18=0
Subtracting 18 from both sides:
3i = -18
Dividing both sides by 3:
i = -6

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find 4 consecutive intergers such that the sum of the first three exceeds the 4th by 18. I can not figure out how to write out the problem to solve it.
x+x+2+x+4=x+6-18
is that correct
------------------
It says integers, not even integers. And, the sum exceeds the 4th, so add 18 to the 4th, not subtract.
------------
So x + x+1 + x+2 = x+3 + 18
3x+3 = x+21
2x = 18
x = 9 (the 1st)
Check:
9+10+11 = 30
12 + 18 = 30

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your on the right track but this is 4 consecutive integers not 4 consecutive odd or even integers.....so x,x+1,x+2,x+3
x+x+1+x+2=x+3+18
3x+3=x+21
2x+18
so integers are 9,10,11,12

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find 4 consecutive intergers such that the sum of the first three exceeds the 4th by 18.
.
Let x = first of 4 consecutive integers
then
x+1 = second of 4 consecutive integers
x+2 = third of 4 consecutive integers
x+3 = fourth of 4 consecutive integers
.
Then from "sum of the first three exceeds the 4th by 18" we get:
x+(x+1)+(x+2) = (x+3)+18
3x+3 = x+21
2x+3 = 21
2x = 18
x = 9
.
Answer:
9, 10, 11, and 12

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lets call our consecutive even integers: x, x+2, x+4, x+6
(x+6)=1/2(x+(x+2)+(x+4))+2------multiply all terms by 2
2x+12=x+x+2+x+4+4
:
2x+12=3x+10
:
so integers are 2,4,6,and 8

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call the four integers x,x+2,x+4,x+6
x+(x+2)=(x+6)+57
2x+2=x+63
so our integers are 61,63,65,67

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Consecutive Even Integers --->
Continuing,
,EQN 1:
Expand & Transpose
, factor out, perfect square

2 values ----->
Therefore:
, 1st even number
, 2nd even number
Check Via EQN 1:


Thank you,
Jojo

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Please help me solve this problem: The greater of two consecutive even integers is twelve less than twice the smaller.Find the consecutive even integers.
.
Let x = smaller of two consecutive even integers
then
x+2 = larger of two consecutive even integers
.
From: "The greater of two consecutive even integers is twelve less than twice the smaller." we get:
x+2 = 2x-12
2 = x-12
14 = x (smaller integer)
.
larger integer:
x+2 = 14+2 = 16 (larger integer)
.
Solution: 14 and 16

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consecutive odd numbers follow the pattern 2x+1, 2x+3, 2x+5, etc


So the equation is


Start with the given equation.


Combine like terms on the left side.


Subtract from both sides.


Combine like terms on the right side.


Divide both sides by to isolate .


Reduce.


Go back to the first number


Plug in


Multiply


Add. So the first number is 11

-------------------

Go back to the second number


Plug in


Multiply


Add. So the second number is 13

-------------------



Go back to the third number


Plug in


Multiply


Add. So the third number is 15




----------------------------------------------------------------------

Answer:

So the consecutive odd integers are 11, 13 and 15


Check

11+13+15 = 39
39 = 39

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sum of three consecutive integers is -14.Find the numbers?
n , n+1=? , n+2=?
----------------
n + n+1 + n+2 = -14
3n+3 = -14
3n = -17
There is no integer solution. The sum of 3 consecutive integers will always be a multiple of 3.

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call the 2 consecutive integers x and x+2(remember consecutive odds or evens are 2 apart)
4x=3(x+2)-9
4x=3x+6-9
small integer
larger integer

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lets call the two integers a and b
a+b>12 eq 1
a=2b-10 eq 2
lets take a's value from eq 2 and plug it into eq 1.
2b-10+b=12--->--->

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the sum of two integers is greater than 12, one integer is ten less than twice the other. what are the least values of the integers?
.
Let x = one integer
and y = other integer
.
x+y > 12
x = 2y-10
.
2y-10 + y > 12
3y-10 > 12
3y > 22
y > 22/3
.
The smallest integer value of y is 8.
.
to find x:
x = 2y-10
x = 2(8)-10
x = 16-10
x = 6
.
The two integers: 6 and 8

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The sum of 2 consecutive intergers is 118. Define a variable for the smaller interger? and What must you add to an even interger to get the next greater even interger? Write an expression for the second interger?
.
Let x = smaller even integer
then
x+2 = next consecutive even integer
.
x + x+2 =118
2x + 2 = 118
2x = 116
x = 58
x+2 = 60
.
The integers are 58 and 60

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the sum of the squares of six consecutive integers is 1111. what are the integers and what is the problem solving strategy?

Let consecutive the integers be , , , , , and 

Then we have the equation:







Collect terms:




 
Divide through by 6:



Factor:



Use the zero factor property:



So there are two solutions:

If ,  then the consecutive integers are

, , , , , and 

or

, , , , , and 

or

, , , , , and 

and if ,  then the consecutive integers are

, , , , , and 

or

, , , , , and 

or

, , , , , and 

Edwin

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Consecutive integers follow the pattern: x, x+1, x+2, x+3...

Since the "sum of the squares of six consecutive integers is 1111", this means that


Now FOIL the expressions to get




and combine like terms to get





I'll let you solve the equation.

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lets call the integers x,x+2,and x+4.
(x+(x+2)+(x+4)-18=x---->3x-12=x---->2x=12
so integers are 6,8,10
hope it helps!!!

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Odd integers follow the pattern , , etc


So if the "sum of two odd integers whose sum is 48" then


Combine like terms on the left side.


Subtract from both sides.


Combine like terms on the right side.


Divide both sides by to isolate .


Reduce.


----------------------------------------------------------------------

Answer:

So the answer is which means that the two numbers are and

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lets call our integers x,x+2,and x+4
so x+(x+2)+(x+4)=256-10
----> so
the integers are 80,82,84

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let these numbers be 2x 2x+2 and 2x+4.
two times the sum of the first and the second is equal to three times the third means that:
2*(2x + 2x+2)=3*(2x+4)
2*(4x+2)=6x+12
8x+4=6x+12
2x=8
x=4
so the three consecutive even integers are 8, 10 and 12.

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X(X+2)=11(X+X+2)-106
X^2+2X=11(2X+2)-106
X^2+2X=22X+22-106
X^2+2X-22X-22+106=0
X^2-20X+84=0
(X-14)(X-6)=0
X-14=0
X=14 ANSWER.
X-6=0
X=6 ANSWER.
PROOFS:
14*16=11(14+16)-106
224=11*30-106
224=330-106
224=224
---------------
6*8=11(6+8)-106
48=11*14-106
48=154-106
48=48

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so lets call our intergers x+1,x+3,x+5
lets distribute the right and left sides and simplify simplifying further it appears that there are NO unique solutions

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Two consecutive ODD Integers ----
Condition:
, EQN 1
Continuing,

---> , divide whole eqn by 2


Remember ---
By PYTH.THEOREM:



---->
2 VALUES:




USE HIGHLIGHTED ONE:
-------------> 1st odd intgr.
------> 2nd odd intgr.
Check via EQN 1,



Thank you,
Jojo

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Let n be the first integer, (n+2) would be the next consecutive odd integer.




You can factor this quadratic equation.

.
.
.
First solution:


The two integers are then -13 and -11.
.
.
.
Second solution:


The two integers are then 11 and 13.

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The first group is one person here is a list of
group numbers and number of people in group. Any
group has 2 more than the previous group
-------------
1 --- 1
2 --- 3
3 --- 5
4 --- 7
5 --- 9
6 --- 11
-------------
Now here is a list of group numbers and total of people so far
-------------
1 --- 1
2 --- 1 + 3 = 4
3 --- 5 + 5 = 9
4 --- 9 + 7 = 16
5 --- 16 + 9 = 25
6 --- 25 + 11 = 36
--------------
Each total is the square of the count number of the group
---------------




etc.
The question then becomes "What group count number squared is under
2250 when the next group count number squared goes over 2250?
gives me That's too much
gives me still too much
gives me
That's under , and
So, people of the 48th group bring the total
up to . So, 48 groups have to arrive at school
(There can be
people in the 48th group)
------------------
For a formula, if = the total number in the school
and = the group count number, = 1,2,3,4.. etc.


The number of groups will be with the
last group having people

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Let x & (x+2) be the 2 integers.
3x=2(x+2)+12
3x=2x+4+12
3x-2x=16
x=16 answer for the smaller integer.
16+2 for the larger integer.
Proof:
3*16=2*18+12
48=36=12
48=48

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1.5^2+b^2=13.5
b^2=12
b=sqrt(12)
=sqrt(4*3)
=2sqrt(3) yards
.
Ed

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Let x = first consecutive positive integer
then
x+1 = second consecutive positive integer
.
x^2 - 25 = 3(x+1)
x^2 - 25 = 3x+3
x^2 - 3 - 28 = 0
.
Factoring:
(x-7)(x+4) = 0
therefore,
x = {-4, 7}
.
Since we looking for POSITIVE integers then the solution MUST be:
7 and 8

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1) The lesser of two consecutive even integers is 10 more than one-half the greater. Find the integers.
1st: 2x
2nd: 2x+2
-----
EQUATION:
2x =(1/2)(2x+2)+10
2x = x + 11
x = 11
-------------
1st: 2x = 22
2nd: 2x + 2 = 24
===================================

2) The greater of two consecutive even integers is 6 less than the times the lesser. Find the integers.
I'll leave this to you. Something is missing is the statement of the problem.
=====================================
3) Find four consecutive integers such that twice the sum of the two greater integers exceeds three times the first by 91.
1st: x
2nd: x+1
3rd: x+2
4th: x+3
---------------
EQUATION:
2(x+2 + x+3) = 3x + 91
4x + 12 = 3x + 91
x = 79
x+1 = 80
etc.
======================
4) Find a set of four consecutive positive integers such that the greatest integer in the set is twice the least integer in the
I'll leave this to you.
=============================
Cheers,
Stan H.

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LET X & X+2 BE THE INTEGERS.
X+2=2X-6
X-2X=-6-2
-X=-8
X=8 FOR THE SMALLER INTEGER.
8+22=10 FOR THE LARGER INTEGER.
PROOF:
10=2*8-6
10=16-6
10=10

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1st integer = x
2nd integer = x+2
======================
x(x+2)=168


(x+14)(x-12)=0
x+14=0
x=-14
x-12=0
x=12
=================
So we end up with 2 possibilities;
x = 12 or x = -14
Now we need our second;
x + 2 = 12+2=14
x+2 = -14+2=-12
====================
When x = 12, the second integer = 14
When x = -14, second integer = -12
:)

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find the consecutive odd numbers such that twice the sum of the first and second is one more than three times the third.
.
Let x = 1st consecutive odd number
then
x+2 = 2nd consecutive odd number
x+4 = 3rd consecutive odd number
.
From: "twice the sum of the first and second is one more than three times the third" we get:
2(x + x+2) = 3(x+4)+1
2(2x+2) = 3(x+4)+1
4x+4 = 3x+12+1
4x+4 = 3x+13
x+4 = 13
x = 9 (1st number)
.
2nd number:
x+2 = 9+2 = 11
.
3rd number:
x+4 = 9+4 = 13
.
Ans: 9, 11, 13

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I'll use the "elmination method"...
.
Let x = smaller number
and
y = larger number
.
Since we have two unknowns, we'll need to find two equations.
Equation 1 can be extracted from "The sum of two integers is fifteen":
x + y = 15
.
Equation 2 can be extracted from "Eight times the smaller is 1 less than three times the larger.":
8x = 3y -1
rearranging:
8x - 3y = -1
.
The two equations now are:
x + y = 15
8x - 3y = -1
.
Multiplying the top equation by 3 to get:
3x + 3y = 45
8x - 3y = -1
.
Adding the two equations together:
3x + 3y = 45
8x - 3y = -1
-------------
11x = 44
dividing both sides by 11:
x = 4 (smaller number)
.
larger number is found by substituting the above into equation 1:
x + y = 15
4 + y = 15
y = 11 (larger number)
.
Ans: 4 and 11

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25^2=(2x-10)^2+x^2
625=4x^2-40x+100+x^2
5x^2-40x+100-625=0
5x^2-40x-525=0
5(x^2-8x-105=0
5(x-15)(x+7)=0
x-15=0
x=15 fot the short leg
2*15-10=30-10=20 for the long leg
Proof:
25^2=20^2+15^2
625=400+225
625=625

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15^2=(x+3)^2+x^2
225=x^2+6x+9+x^2
2x^2+6x+9-225=0
2x^2+6x-216=0
2(x^2=3x-108)=0
2(X=12)(x-9)=0
x-9=0
x=9 feet is the distance from the wall to the bottom of the ladder.
Proof:
9^2+(9+3)^2=15^2
81+12^2=225
81+144=225
225=225

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(2x-10)^2+x^2=25^2
4x^2-40x+100+x^2=625
5x^2-40x+100-625=0
5x^2-40x-525=0
5(x^2-8x-105)=0
5(x-15)(x+7)=0
x-15=0
x=15 for the short side.
2*15-10
30-10=20 for the long side.
Proof:
15^2+20^2=25^2
225+400=625
625=625

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A rectangle is 15 cm wide and 18 cm long. If both dimensions are decreased by the same amount, the area of the new rectangle formed is 116 cm^2 less than the area of the original. Find the dimensions of the new rectangle.
:
Original rectangle area: 15 * 18 = 270 sq/cm
:
Let x = amt each dimension is decreased:
:
Area of new rectangle
(15-x)(18-x) = 270 - 33x + x^2
:
Original rectangle - new rectangle = 116 sq/cm
270 - (270 - 33x + x^2) 116
270 - 270 + 33x - x^2 = 116
Arrange as a quadratic on the right
0 = x^2 - 33x + 116
Factor this to:
(x - 4)(x - 29) = 0
x = 4 cm; is the only solution that makes sense.
:
:
New rectangle 11 * 14 = 154
270 - 154 = 116; confirms our solution

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Call the number N.



Two solutions.


.
.
.


.
.
.
Check your answers.


.
.
.


.
.
.
Both answers lead to true statements.
Both answers are good answers.

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Let's call the first integer, N.
the next 5 consecutive integers would be





and their squares would be,






Now add them together,


You know the sum of the squares equals 1111.




Two solutions.
.
.
.


The integers are then 11,12,13,14,15,16.
.
.
.


The integers are then -16,-15,-14,-13,-12,-11.