For What value of ‘x’ Which Satisfies this equation,
[tex](x – a) \div(x-b) = (x + a) \div (x+b)[/tex]
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1 thought on “For What value of ‘x’ Which Satisfies this equation,<br />[tex](x – a) \div(x-b) = (x + a) \div (x+b)[/tex]<br />​”

1. Step-by-step explanation:

   (x-a)/(x-b) = (x+a)/(x+b)

   (x-a)/(x-b) = (x+a)/(x+b)(x-a)(x-b)=(x+a)(x-b)

   (x-a)/(x-b) = (x+a)/(x+b)(x-a)(x-b)=(x+a)(x-b)x^2-ax+bx-ab = x^2 +ax-bx-ab

   (x-a)/(x-b) = (x+a)/(x+b)(x-a)(x-b)=(x+a)(x-b)x^2-ax+bx-ab = x^2 +ax-bx-ab-ax+bx-ax+bx=0

   (x-a)/(x-b) = (x+a)/(x+b)(x-a)(x-b)=(x+a)(x-b)x^2-ax+bx-ab = x^2 +ax-bx-ab-ax+bx-ax+bx=0-2ax+2bx=0

   (x-a)/(x-b) = (x+a)/(x+b)(x-a)(x-b)=(x+a)(x-b)x^2-ax+bx-ab = x^2 +ax-bx-ab-ax+bx-ax+bx=0-2ax+2bx=0-2x(a-b)=0

   (x-a)/(x-b) = (x+a)/(x+b)(x-a)(x-b)=(x+a)(x-b)x^2-ax+bx-ab = x^2 +ax-bx-ab-ax+bx-ax+bx=0-2ax+2bx=0-2x(a-b)=0x=0

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