Question

Plz help question 3

Answer 1

Answer:

$\large{\underline{\boxed{\sf x = \dfrac{1}{2} \: \: and \: \: y =- \dfrac{3}{2}}}}$

Step-by-step explanation:

Given that ;

$\tt \frac{2xy}{x + y} = \frac{3}{2} \: \: ....(i) \\ \\ \tt \frac{xy}{2x - y} = - \frac{3}{10} \: \: ....(ii)$

On cross-multiplying (i) and (ii),

$\tt \frac{2xy}{x + y} = \frac{3}{2} \\ \\ \implies \tt 3x + 3y = 4xy \: \sf ...(iii)$

$\tt \frac{xy}{2x - y} = - \frac{3}{10} \\ \\ \implies \tt - 6x + 3y = 10xy \: \sf \: ....(iv)$

On Subtracting eqⁿ (iii) from (iv),

$\tt 3x + 3y - ( - 6x + 3y) = 4xy - 10xy \\ \\ \tt 3x + 3y + 6x - 3y = - 6xy \\ \\ \tt 9x = - 6xy \\ \\ \tt y = \frac{ - 6x}{9x} \\ \\ \boxed{\therefore \: \bf y = - \frac{3}{2}}$

Substituting the value of y in (iii),

$\tt 3x + 3y = 4xy \\ \\ \implies \tt 3x + 3 \left( \frac{ - 3}{2} \right) = 4x \left( \frac{ - 3}{2} \right ) \\ \\ \implies \tt 3x - \frac{9}{2} = \frac{ - 12x}{2} \\ \\ \implies \tt \frac{6x - 9}{2} = - 12x$

On cancelling the denominator,

$\implies \tt 6x - 9 = - 12x \\ \\ \implies \tt 6x + 12x = 9 \\ \\ \implies \tt 18x = 9 \\ \\ \implies \tt x = \frac{9}{18} \\ \\ \boxed{ \therefore \bf \: x = \frac{1}{2} }$