Question

Sᴏʟᴠᴇ:
$\frac{x - b - c}{a} \: \: + \frac{x - c - a}{c} \: \: + \frac{x - a - b}{c} = 3$
ᴡᴀɴᴛ ϙᴜᴀʟɪᴛʏ ᴀɴsᴡᴇʀs!!​

Answer 1

$\LARGE\underline{\boxed{\red{ǫᴜᴇsᴛɪᴏɴ :-}}}$

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$\frac{x - b - c}{a} \: \: + \frac{x - c - a}{c} \: \: + \frac{x - a - b}{c} = 3,\:then\:find\:x$

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$\LARGE\underline{\boxed{\pink{Sᴏʟᴜᴛɪᴏɴ :-}}}$

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• $\large{\tt{Rewrite\:\:3\:\:as\:\:1+1+1}}$

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$\longmapsto\frac{(x - b - c)}{a} \: \: + \frac{(x - c - a)}{c} \: \: + \frac{(x - a - b)}{c} = 1+1+1$

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$\longmapsto[ \frac{(x - b - c ) - 1}{a}]$+$[\frac{(x - c - a) - 1}{b}]$+$[ \frac{(x - a - b)- 1}{c} ]=0$

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$\longmapsto( \frac{x - b - c - a}{a}) + ( \frac{x - c - a - b}{b}) + ( \frac{x - a - b - c}{c}) = 0$

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• Take out $\large{\tt{(x-b-c-a)}}$ as common

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$\longmapsto(x - b - c - a) \: \: [ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0 \: ]$

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$\longmapsto(x - b - c - a) = 0 \: or \: \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \neq \: 0 \\ ∴(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \neq \: 0,given)$

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$\longmapsto(x - b - c - a) = 0$

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$\longmapsto∴ {\boxed{\red{x = a + b + c}}}$

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ㅤHope it Helps uh Dear!! ❤

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ㅤᕼᗩᑭᑭY ᒪᗴᗩᖇᑎíᑎᘜ ☃️