Math, asked by Rumi123456789, 1 year ago

If (6x^2-xy):(2xy-y^2)=6:1, find x:y

Answers

Answered by MarkAsBrainliest
52
Answer :

Now,

(6x² - xy) : (2xy - y²) = 6 : 1

⇒ (6x² - xy) / (2xy - y²) = 6 / 1

⇒ 1 (6x² - xy) = 6 (2xy - y²)

⇒ 6x² - xy = 12xy - 6y²

⇒ 6x² - 13xy + 6y² = 0

⇒ 6x² - (9 + 4)xy + 6y² = 0

⇒ 6x² - 9xy - 4xy + 6y² = 0

⇒ 3x (2x - 3y) - 2y (2x - 3y) = 0

⇒ (2x - 3y) (3x - 2y) = 0

So, either 2x - 3y = 0 or, 3x - 2y = 0

⇒ 2x = 3y & 3x = 2y

⇒ x/y = 3/2 & x/y = 2/3

⇒ x : y = 3 : 2 & x : y = 2 : 3

Therefore, the required ratio of x and y be

x : y = 3 : 2 or, x : y = 2 : 3

#MarkAsBrainliest
Answered by NoNameBuffer
17

(6x² - xy) : (2xy - y²) = 6 : 1

⇒ (6x² - xy) / (2xy - y²) = 6 / 1

⇒ 1 (6x² - xy) = 6 (2xy - y²)

⇒ 6x² - xy = 12xy - 6y²

⇒ 6x² - 13xy + 6y² = 0

⇒ 6x² - (9 + 4)xy + 6y² = 0

⇒ 6x² - 9xy - 4xy + 6y² = 0

⇒ 3x (2x - 3y) - 2y (2x - 3y) = 0

⇒ (2x - 3y) (3x - 2y) = 0

So, either 2x - 3y = 0 or, 3x - 2y = 0

⇒ 2x = 3y & 3x = 2y

⇒ x/y = 3/2 & x/y = 2/3

⇒ x : y = 3 : 2 & x : y = 2 : 3

Therefore, the required ratio of x and y be

x : y = 3 : 2 or, x : y = 2 : 3

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