Question

Solve question number m.​

Answer 1

Answer:

$x \: = a \: or \: b$

Step-by-step explanation:

$Given \: = > \\ \dfrac{ {a}^{2} \: (x - b) }{(a - b)} - {x}^{2} \: = \dfrac{ {b}^{2} \: (a - x) }{(b - a)}$

$To \: find \: = > \\ value \: of \: x$

$Solution = > \\ \dfrac{ {a}^{2}(x - b) }{a - b} - {x}^{2} = \dfrac{ {b}^{2}(a - x) }{(b - a)}$

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

$(a - b) \: cancel \: out \: from \: each \: side \: we \: get$

$= > {a}^{2} (x - b) - {x}^{2} (a - b) = - {b}^{2} (a - x)$

$= > {a}^{2} x - {a}^{2} b - {x}^{2}(a - b) = - {b}^{2}a + {b}^{2} x$

$= > - {x}^{2} (a - b) + {a}^{2} x - {b}^{2} x - {a}^{2} b + {b}^{2} a \: = 0$

$= > - {x}^{2} (a - b) + ( {a}^{2} - {b}^{2} )x - ab(a - b) = 0$

$dividing \: whole \: equation \: by \: (a - b)$

$= > - \dfrac{ {x}^{2}(a - b) }{a - b} + \dfrac{(a + b) \: (a - b)x}{a - b} - \dfrac{ab(a - b)}{a - b} = 0$

$= > - {x}^{2} + (a + b)x - ab = 0$

$changing \: sign \: of \: whole \: equation$

$= > {x}^{2} - (a + b)x + ab = 0$

$= > {x}^{2} - ax - bx + ab = 0$

$= > x \: ( \: x - a \: ) \: - b \: ( \: x - a \: ) = 0$

$= > ( \: x - a \: ) \: ( \: x - b \: ) \: = 0$

$if \\ x - a = 0 \\ = > x = a$

$if \\ x - b = 0 \\ = > x = b$

$so \\ x \: = \: a \: or \: b$

Answer 2

$\mathfrak{\huge{\orange{ANSWER}}}$