Question

Solve for x :-

$\sf{1) \: \dfrac{a}{x - a} + \dfrac{b}{x - b} = \dfrac{a + b}{x - a - b}}$
$\sf{2) \: \dfrac{x - a}{ {a}^{2} - {b}^{2} } = \dfrac{x - b}{ {b}^{2} - {a}^{2} } }$
$\sf{3) \: (3 + \sqrt{ 3})x + 2 = 5 + 3\sqrt{3}}$
​

Answer 1

Question:-

Solve for x:-

$\sf{1) \: \dfrac{a}{x - a} + \dfrac{b}{x - b} = \dfrac{a + b}{x - a - b}}$

$\\ \sf{2) \: \dfrac{x - a}{ {a}^{2} - {b}^{2} } = \dfrac{x - b}{ {b}^{2} - {a}^{2} } }$

$\\ \sf{3) \: (3 + \sqrt{ 3})x + 2 = 5 + 3\sqrt{3}}$

Required Answers:-

$\sf{ \bold{1) \: \dfrac{a}{x - a} + \dfrac{b}{x - b} = \dfrac{a + b}{x - a - b}}}$

$\\ \sf{ \implies{ \dfrac{a}{x - a} + \dfrac{b}{x - b} = \dfrac{a}{x - a - b} + \dfrac{b}{x - a - b}}}$

$\\ \sf{ \implies{ \dfrac{a}{x - a} - \dfrac{a}{x - a - b} = \dfrac{b}{x - a - b} - \dfrac{b}{x - b} }}$

$\\ \sf{ \implies{ \dfrac{ \cancel{ax - {a}^{2}} - ab \: \cancel{ - ax + {a}^{2} }}{(x - a - b)(x - a)} = \dfrac{ \cancel{bx - {b}^{2} - bx} + ab \: \cancel{ + {b}^{2} }}{(x - a - b)(x - b)}}}$

$\\ \sf{ \implies{ \dfrac{ - ab}{(x - a)(x - a - b)} = \dfrac{ab}{(x - a - b)(x - b)}}}$

$\\ \sf{ \implies{ \dfrac{ - 1}{(x - a)(x - a - b)} = \dfrac{1}{(x - a - b)(x - b)}}}$

$\\ \sf{ \implies{ \dfrac{ - 1}{x - a} = \dfrac{1}{x - b}}}$

$\\ \sf{ \implies{x - a = - x + b}}$

$\\ \sf{ \implies{x + x = a + b}}$

$\\ \sf{ \implies{2x = a + b}}$

$\\ \therefore{ \sf{x = \red{ \dfrac{a + b}{2}}}}$

$\\ \sf{ \bold{2) \: \dfrac{x - a}{ {a}^{2} - {b}^{2} } = \dfrac{x - b}{ {b}^{2} - {a}^{2} }}}$

$\\ \sf{ \implies{\dfrac{x - a}{ {a}^{2} - {b}^{2} } = \dfrac{x - b}{ - ( {a }^{2} - {b}^{2})}}}$

$\\ \sf{ \implies{\dfrac{x - a}{ 1 } = \dfrac{x - b}{ - 1}}}$

$\\ \sf{ \implies{x - b = - x - a}}$

$\\ \sf{ \implies{x + x = a + b}}$

$\\ \sf{ \implies{2x = \dfrac{a + b}{2}}}$

$\\ \sf{ \therefore{x = \red{ \dfrac{a + b}{2}}}}$

$\\ \sf{ \bold{{3) \: (3 + \sqrt{ 3})x + 2 = 5 + 3\sqrt{3}}}}$

$\\ \sf{ \implies{(3 + \sqrt{3} )x = 5 + 3 \sqrt{3} - 2}}$

$\\ \sf{ \implies{(3 + \sqrt{3}) x = 3 + 3\sqrt{3}}}$

$\\ \sf{ \implies{(3 + \sqrt{3}) x = \sqrt{3} (\sqrt{3} +3)= \sqrt{3} (3+\sqrt{3} )}}$

Dividing both sides by 3+√3 we get,

$\\ \sf{ \therefore{x = \red{3}}}$

Answer 2

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