SOLUTION: Find two positive numbers that differ by eight and whose reciprocals sum is 1/6

Question 944234: Find two positive numbers that differ by eight and whose reciprocals sum is 1/6
Found 2 solutions by macston, MathLover1:
Answer by macston(5194) (Show Source): You can put this solution on YOUR website!
X=one positive number, Y is the other positive number
;
X=Y+8 Add Y to each side of Equation 1 and substitute in Equation 2
Multiply each side by 6
Subtract 6/Y from each side
convert right side to fraction

Multiply each side by Y+8
Multiply each side by Y
Multiply right side through
Subtract 6Y from each side

Solved by pluggable solver: SOLVE quadratic equation with variable

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=208 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 9.21110255092798, -5.21110255092798. Here's your graph:

The positive answer is 9.2111 so ANSWER 1 Y=9.2111
X-Y=8
X-9.2111=8 Add 9.2111 to each side
X=17.2111 ANSWER 2 X=17.2111
FINAL ANSWER THE TWO POSITIVE NUMBERS ARE 17.2111 and 9.2111
CHECK
1/X+1/Y=1/6
1/17.2111+1/9.2111=1/6
0.16667=1/6
0.16667=0.16667

Answer by MathLover1(20849) (Show Source): You can put this solution on YOUR website!

let two positive numbers be and
if they differ by eight, we have
=> ......eq.1
and if their reciprocals sum is , we have

.

.....cross multiply

.....eq.2 ....substitute from eq.1

.....use quadratic formula

y = 2-2 sqrt(13)=-5.2 we need positive number only
find :
=>

check the sum of reciprocals:

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