Question

x & y are two positive integers such that(4x2 – 3y2): (2x2 + 3y2) = 37:59. What is the value of x:y?
04:3​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

The value of x:y is 4:3.

Step-by-step explanation:

Given:

x & y are two positive integers.

The equation $\frac{( 4x^{2} -3y^{2} )}{(2x^{2}+3y^{2} )} =\frac{37}{59}$.

To Find:

The value of x:y.

Solution:

As given - $\frac{( 4x^{2} -3y^{2} )}{(2x^{2}+3y^{2} )} =\frac{37}{59}$.

$\frac{( 4x^{2} -3y^{2} )}{(2x^{2}+3y^{2} )} =\frac{37}{59}$

$59( 4x^{2} -3y^{2} )=37(2x^{2} +3y^{2} )$

$236x^{2} -177y^{2} =74x^{2} +111y^{2}$

$236x^{2} -74x^{2} =177y^{2} +111y^{2}$

$162x^{2} =288y^{2}$

$\frac{x^{2} }{y^{2} } =\frac{288}{162}$

$\frac{x^{2} }{y^{2} } =\frac{16}{9}$

$(\frac{x}{y} )^{2} =(\frac{4}{3} )^{2}$

$\frac{x}{y} =\frac{4}{3}$

$x:y=4:3$

Thus, the value of x:y is 4:3.

PROJECT CODE #SPJ3