Question

*If x/y = 2/3 then 2x² + 3y² / 2x² - 3y² = ?*

1️⃣ 35/19
2️⃣ 19/35
3️⃣ -19/35
4️⃣ -35/19​

Answer 1

{Answer:-} \\

\pink{\☆ Solution:- {\implies \frac{35}{19}}}

$</p><p></p><p> \huge \blue{Answer:-} \\</p><p></p><p> \huge\pink{☆ Solution: {\implies \frac{35}{19}}}$

Answer 2

Given : x/y = 2/3

To Find : 2x² + 3y² / 2x² - 3y²

1️⃣ 35/19

2️⃣ 19/35

3️⃣ -19/35

4️⃣ -35/19​

Solution:

x/y = 2/3

Squaring both sides

=> ( x/y)² = (2/3)²  = 4/9

2x² + 3y² / 2x² - 3y²

Dividing numerator and denominator by y²

= (2( x/y)² + 3) / (2( x/y)²  - 3)

= (2 (4/9) + 3) / (2 (4/9)  - 3)

multiplying numerator and denominator by 9

= ( 8 +  27) / ( 8 - 27)

= 35 / (-19)

= - 35/19

2x² + 3y² / 2x² - 3y²   = -35/19​

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