Question

if a/b=7/3 then find 5a+2b/5a-2b​

Answer 1

Solution :

It is given that , a/b = 7/3 .

Let a = 7k and b = 3k

So;

5a = 5 × 7k = 35k

2a = 2 × 7k = 14k

.

5b = 5 × 3k = 15k

2b = 2 × 3k = 6k

( 5a + 2b)

=> 35k + 6k

=> 41k

( 5a - 2b )

=> 35k - 6k

=> 29k

Required ratio :

=> ( 5a + 2b)/( 5a - 2b)

=> 41k/29k

=> 41 : 29 .

This is the required answer.

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Additional Information :

(a + b)² = a² + 2ab + b²

(a + b)² = (a - b)² + 4ab

(a - b)² = a² - 2ab + b²

(a - b)² = (a + b)² - 4ab

a² + b² = (a + b)² - 2ab

a² + b² = (a - b)² + 2ab

2 (a² + b²) = (a + b)² + (a - b)²

4ab = (a + b)² - (a - b)²

ab = {(a + b)/2}² - {(a-b)/2}²

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

(a + b)³ = a³ + 3a²b + 3ab² b³

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - 3a²b + 3ab² - b³

a³ + b³ = (a + b)( a² - ab + b² )

a³ + b³ = (a + b)³ - 3ab( a + b)

a³ - b³ = (a - b)( a² + ab + b²)

a³ - b³ = (a - b)³ + 3ab ( a - b )

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Answer 2

$\mathsf \red{given} \\ \\ \mathsf{ \frac{a}{b} = \frac{7}{3} } \\ \\ \mathsf{let \: \: a = 7k \:, \: b = 3k } \\ \\ \mathsf \red{substitute \: \: a \: and \: b \: \: in \: \: \frac{5a + 2b}{5a - 2b} } \\ \\ \mathsf{ \frac{5(7k) + 2(3k)}{5(7k) - 2(3k)} = \frac{35k + 6k}{35k - 6k} } \\ \\ = = > \mathsf{ \frac{41k}{29k} = \frac{41}{29} } \\ \\ \boxed{\mathsf \red{ \therefore \: the \: ratio \: of \: \:\frac{5a + 2b}{5a - 2b} \: \: is \: \: \frac{41}{29} }}$