Question

Given a and b are two natural numbers.

$\frac{290}{a \: b} - \frac{3}{b} - \frac{5}{a } = 1$
Calculate
$\frac{a}{b} and \: a + b.$

Answer 1

Heya User,

[tex] \frac{290}{ab} - \frac{3}{b} - \frac{5}{a} = 1 \\\\ =\ \textgreater \ \frac{290}{ab} = \frac{3}{b} + \frac{5}{a} +1\\\\ =\ \textgreater \ \frac{290}{ab} = \frac{3a + 5b + ab}{ab} \\\\ =\ \textgreater \ ab \:[\: 3a + 5b + ab - 290 \:][/tex]

Now, ab ≠ 0;
=> [tex]3a + 5b +ab - 290 = 0\\giving \: us:-\ \textgreater \ \\ 3a + 5b +ab = 290;[/tex]

We know that (a,b) > 0;
--> a[ b + 3 ] + 5[ b + 3 ] = 290 + 15 ----> (i)

=> [ a + 5 ][ b + 3 ] = 305

--> Factors of 305 --> 1*305, 5*61

--> Only possible comparison is :->
--> [ a + 5 ] = 61 => a = 56
--> [ b + 3 ] = 5 ==> b = 2;

Hence, [ (a+b) , (a/b) ] = (58,28) <--- Ans.... :)

Answer 2

290/ab - 3/b - 5/a = 1

→ 290/ab = 3/b + 5/a + 1

→ 290/ab = (3a+5b+ab)/ab

→ 290 = 3a+5b+ab

adding 15 on both sides -

290+15 = 3a+5b+ab+15

→ 305 = a(3+b)+5(3+b)

→ 305 = (a+5)(3+b)

Now we can express 305 as 61×5

→ 61×5 = (a+5)(3+b)

On comparing both sides , we get-

a+5 = 61
→ a = 56

3+b = 5
→ b =2

Now a+b = 56+2 = 58

a/b = 56/2 = 28

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