The Sum Of 2 Numbers Is 10. The Sum Of Their Reciprocals Is 5/12.Can You Find The Numbers?

Let A , and B represent the numbers.


A  +   B   = 10

1/A  + 1/B  = 5/12

Find the common denominator for
A, B   which is AB   by multiplying B by 1 and A, and multiplying A by 1 and B  

The second equation will be

(A+B)/AB = 5/12

we know that A+B = 10

You now have  10/AB = 5/12

cross multiply and solve for AB

10x12=5AB
120 = 5AB   divide both sides by 5

24 = AB

We now have A + B = 10
  AB  = 24
Divide both sides by B

A = 24/B
Substitute A in Equation 1

24/B + B =10
Multiply each term by B
24 + B^2 = 10B
Subtract 10B from both sides
B^2 -10B +24 = 0
(B-6)(B-4) = 0

B= 6, and B=4

If B = 6, then A =4

Your two numbers are 6 and 4.

(B + A)/AB = 5/12

Let the two numbers are a and b.
Then, a+b = 10......(1)
1/a+1/b = 5/12
(a+b)/ab = 5/12
12(a+b) = 5ab....(2)
substitute the value from eq. (1) into eq. (2)
12(10) = 5ab

5ab = 120
ab = 24
a = 24/b....(3)
substitute the value of a from eq. (3) into eq. (1)
(24/b)+b=10
24+b^2= 10b
b^2-10b+24=0
b^2-6b-4b+24=0
b(b-6)-4(b-6)=0
(b-4)(b-6)=0
b = 4,6
Put the value of b into eq. (1)
a = 6,4

Let two numbers are a and b
then, a+b=10
1/a+1/b=5/12
Solving,
(a+b)/ab=5/12
12(a+b)=5ab
After putting the value of a+b
12(10)=5ab
5ab=120
ab=24
a=24/b
After putting the velue of a
(24/b)+b=10
b^2-10b+24=0
b^2-6b-4b+24=0
b(b-6)-4(b-6)
(b-4)(b-6)
b=4,6
a=6,4
so, numbers are (6,4) and (4,6)

Let two numbers are x and y
then, x+y= 10
1/x+1/y=5/12
Then Second equation will be,
(x+y)/xy=5/12
10/xy=5/12
5xy=120
xy=24
x=24/y
then, 24/y+y=10
y^2-10y+24=0
y^2-6y-4y+24=0
y(y-6)-4(y-6)=0
(y-4)(y-6)=0
y=4,6
then, x=12,4
So numbers are (4,6)