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

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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Answer 2

Given :- x/y = 2/3 .

To Find :- 2x² + 3y² / 2x² - 3y² = ?

1) 35/19

2) 19/35

3) -19/35

4) -35/19

Answer :- (4) (-35/19) .

Explanation :-

→ (x/y) = (2/3)

squaring both sides,

→ (x/y)² = (2/3)²

→ x²/y² = 4/9 -------- Eqn.(1)

now,

→ (2x² + 3y²) / (2x² - 3y²)

divide both numerator and denominator by y²,

→ {(2x² + 3y²) / y² } /{ (2x² - 3y²) / y²}

→ {(2x²/y²) + 3} / {(2x²/y²) - 3}

putting value from Eqn.(1),

→ {(2 * 4/9) + 3} / {(2 * 4/9) - 3}

→ (8/9 + 3) / (8/9 - 3)

→ {(8 + 27)/9} / {(8 - 27)/9}

→ (35/9) /( -19/9)

→ (35/9) * (-9/19)

→ (-35/19) (Option (4) (Ans.)

Shortcut :-

→ x/y = 2/3 .

• Assume x = 2 and y = 3,

putting value,

→ (2x² + 3y²) / (2x² - 3y²)

→ (2*2² + 3*3²) / (2*2² - 3*3²)

→ (2*4 + 3*9) / (2*4 - 3*9)

→ (8 + 27) / (8 - 27)

→ (35/(-19)

→ (-35/19) (Option (4) (Ans.)

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