SOLUTION: hi im a 13 year old boy in 8th grade in algerbra 1 honors class. im really stumped on these questions and need some help please. Please help me and thank you very much. merry earl

Question 251564: hi im a 13 year old boy in 8th grade in algerbra 1 honors class. im really stumped on these questions and need some help please. Please help me and thank you very much. merry early christmas!!
(1)The sum of the reciprocals of two consecutive odd integers is 8/15. find the number.
(2)The sum of the reciprocals of two consecutive even integers is 11/60. find the integers.
(3) the sum of two numbers is 10 and the sum of there reciprocals is 5/12. find the numbers.
Found 3 solutions by richwmiller, stanbon, dabanfield:
Answer by richwmiller(17219)   (Show Source): You can put this solution on YOUR website!
(1)The sum of the reciprocals of two consecutive odd integers is 8/15. find the number.
1/x+1/(x+2)=8/15
1/3+1/5=8/15
5/15+3/15=8/15
(2)The sum of the reciprocals of two consecutive even integers is 11/60. find the integers.
1/x+1/(x+2)=11/60
1/10+1/12=11/60
6/60+5/60=11/60
(3) the sum of two numbers is 10 and the sum of there reciprocals is 5/12. find the numbers.
x+y=10
1/x+1/y=5/12
1/4+1/6=5/12
3/12+2/12=5/12
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
(1)The sum of the reciprocals of two consecutive odd integers is 8/15. find the number.
Odd numbers are always one more than or one less than an even number.
Symbols for an odd number are 2x-1 or 2x+1
-------------------
Your problem:
1st odd: 2x-1
Next consecutive odd: 2x+1
-------------------------------
Equation:
1/(2x-1) + 1/(2x+1) = 8/15
---
Multiply thru by 15(2x-1)(2x+1):
15(2x+1) + 15(2x-1) = 8(2x+1)(2x-1)
30x+15 + 30x-15 = 8(4x^2-1)
60x = 32x^2-8
32x^2-60x-8 = 0
8x^2-15x-2 = 0
8x^2-16x+x-2 = 0
8x(x-2)+(x-2) = 0
(x-2)(8x+1) = 0
Positive solution:
x = 2
----
1st odd: 2x-1 = 3
2nd odd: 2x+1 = 5
========================
Checking
1/3 + 1/5 = 8/15
Multiply thru by 15:
5 + 3 = 8
8 = 8
=================
(2)The sum of the reciprocals of two consecutive even integers is 11/60. find the integers.
Follow the same procedure as in #1 but use the followin symbols
for consecutive even numbers.
1st: 2x
2nd: 2x+2
===============================
(3) the sum of two numbers is 10 and the sum of their reciprocals is 5/12.
Find the numbers.
----
Equations:
x + y = 10
1/x + 1/y = 5/12
-----------------------
Rearrange the equations:
x = 10-y
12y + 12x = 5xy
-------------------
Substitute for "x" and solve for "y":
12y + 12(10-y) = 5(10-y)y
12y + 120 - 12y = 50y - 5y^2
5y^2 -50y+120 = 0
y^2 - 10y + 24 = 0
(y-6)(y-4) = 0
y = 6 or y = 4
---
Substitute into x 10-y
If y = 6, x= 4
If y = 4, y = 6
======================
Checking:
x+y = 10
4+6 = 10
---------
1/4 + 1/6 = 5/12
Multiply thru by 12 to get:
3+2 = 5
===================
Cheers,
Stan H.
Answer by dabanfield(803)   (Show Source): You can put this solution on YOUR website!
hi im a 13 year old boy in 8th grade in algerbra 1 honors class. im really stumped on these questions and need some help please. Please help me and thank you very much. merry early christmas!!
(1)The sum of the reciprocals of two consecutive odd integers is 8/15. find the number.
(2)The sum of the reciprocals of two consecutive even integers is 11/60. find the integers.
(3) the sum of two numbers is 10 and the sum of there reciprocals is 5/12. find the numbers.
(1.) Let x be the first of the two consecutive odd integers. Then the next one must be 2 more than the first or x+2.
So we have:
x + (x+2) = 8/15 or
2x + 2 = 8/15
Now you can solve for x and x+2.
(2.) Let x be the first of the two consecutive even integers. Then the next one must be two more or x+2.
The reciprical of a number, say k, is 1/k.
So we have:
1/x + 1/(x+2) = 11/60
We need to get a common denominator for the left side so we can combine the two terms. One common denoiminator (not necessarily the lowest though) is always the product of the denominators of all the terms. In this case x*(x+2).
To get the first term to have the common denominator we can multiply it by
(x+2)/(x+2) = 1. This won't change our solution since we are multiplying by 1.
To get the second term's demoninator we can multiply it by x/x = 1.
Doing this we then have:
(1/x)*((x+2)/(x+2)) + ((1/(x+2))*((x/x)) = 11/60
(x+2)/(x*(x+2)) + (x/((x*(x+2)) = 11/60
Now both denominators on the left are x*(x+2) so we can combine their numerators (remember a/c + b/c = (a+b)/c).
So we have:
((x+2) + x))/((x*(x+2)) = 11/60
(2x + 2)/((x*(x+2))= 11/60
Multiply both sides of the equation above by x*(x+2) to clear the fractions on the left side:
2x+2 = (11/60)*((x*(x+2))
Expand and solve for x.
(3.) Let x be one number and y the other.
Then we have two simultaneous equations as follows:
x + y = 10
1/x + 1/y = 5/12
From the first equation we have x = 10-y
We can then substitute 10-y in the second equation for x and then solve for y. You will need to use the common denominator trick here also. Once you have the value for y you can then calculate x from x = 10-y.
Good luck. Merry Christimas to you too :)

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