Given $\frac{m}{n} = \frac{x^{2} + 2 y}{x^{2} - 2 y}$ ; make 'x' as the subject of the formula. Then find the value of x, if y = 2, m =5 and n =4.

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Solution

Given, $\frac{m}{n} - \frac{x^{2} + 2 y}{x^{2} - 2 y}$
$=> m x^{2} - 2 m y = n x^{2} + 2 n y$
$=> x^{2} (m - n) = 2 y (m + n)$
$x^{2} = \frac{2 y (m + n)}{(m - n)}$
$=> x = \sqrt{\frac{2 y (m + n)}{(m - n)}}$
When, $y = 2, m = 5, n = 4,$
$x = \sqrt{\frac{2 (2) (5 + 4)}{(5 - 4)}}$
$x = \sqrt{36}$
$x = 6$

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Similar questions

\frac{m}{n} = \frac{x^{2} + 2 y}{x^{2} - 2 y}

y = 2, m = 5 a n d n = 4

x : (m^{2} - 2 n) = Y : (m^{2} + 2 n)

3 m = 5 x = y = 15

\frac{10 - 3 x}{5 + 2 x} = \frac{m}{n}

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3 m - 4 n = 2

n = 2.5

\frac{10 - 3 x}{5 + 2 x} = \frac{m}{n}

3 m - 411 = 2

n = 2.5

m = 2 n + \sqrt{n^{2} + x},

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