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Question

one # is three times another #. the sum of their reciprocal is 4/9. find the two #'s.

Answer 1

Let the numbers be x and other be y,
We have : x = 3y
Also : 1/x + 1/y = 4/9
or
1/3y + 1/y = 4/9
(1 + 3)/(3y) = 4/9
4/(3y) = 4/9
y = 3
x = 9

So 3 , 9 are your two numbers.
Have a nice day...
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Answer 2

The two numbers are x and 3x. 1/x + 1/3x = 4/9 so 3/3x + 1/3x = 4/9 so 4/3x = 4/9 which means 3x=9 so x=3. So the numbers (x and 3x) are 3 and 9. Hope this helps, but its difficult to express algebra without the correct symbols, etc.

Answer 3

let's say "one" is x, "another" is y.

"one is 3 times another"=> x =3y (E1)
"the sum of their reciprocal is 4/9" => 1/x + 1/y = 4/9 (E2)
you got your system of 2 equations, 2 unknowns.

substitute x from E1 in E2 => 1/(3y) + 1/y = 4/9
<=> 1/(3y) + 3/(3y) = 4/9
<=> 4/(3y) = 4/9
<=> 3y/4 = 9/4
<=> 3y = 9
<=> y = 3
substitue y in E1 => x = 3 * 3 = 9

so x = 9 and y = 3

Answer 4

Let one is x another is 3x
according to condition
1/x+1/3x=4/9
(3+1)/3x =4/9
4/3x=4/9
x= 9/3=3
so another # is 9 Ans. is 3 & 9

Answer 5

3 and 9

3 is 3 times 9, and

1/3 + 1/9 = 3/9 + 1/9 = 4/9

Answer 6

x = 3y
1/x + 1/y = 4/9

Substitute:
1/3y + 1/y = 4/9
1/3y + 3/3y = 4/9
4/3y = 4/9
3y = 9
y = 3
x = 9

Check:
1/3 + 1/9 = 4/9
3/9 + 1/9 = 4/9

Sho 'nuff!

Answer 7

1/3x + 1/x = 4/9
1/3x + 3/3x = 4/9
4/3x = 4/9
cross multiply
36 = 12x
x = 3

Therefore, your numbers are 3 and 9 (3 x3 =9 ).

Answer 8

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