Question

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Answer 1

Question in English:-

• Given that, $\sf \dfrac{a}{4-a} + \dfrac{b}{4-b} + \dfrac{c}{4-c} = 1 ,$ show that $\sf \dfrac{1}{4-a} + \dfrac{1}{4-b} + \dfrac{1}{4-c} = 1.$

Proof:-

First considering what is given,

$\sf \dfrac{a}{4 - a} + \dfrac{b}{4 - b} + \dfrac{c}{4 - c} = 1$

The main concept is to somehow cancel the "a", "b" and "c" in the numerator term and make it 1 in order to reach the proof.

For this,

Adding 3 to both sides,

$\sf \longrightarrow \dfrac{a}{4 - a} + \dfrac{b}{4 - b} + \dfrac{c}{4 - c} + 3 = 1 + 3$

Now, in LHS we will split 3 in this way,

$\sf \longrightarrow \dfrac{a}{4 - a} + 1 + \dfrac{b}{4 - b} + 1+ \dfrac{c}{4 - c} + 1 = 4$

Taking LCM for the terms in LHS,

$\sf \longrightarrow \dfrac{a + 4 - a}{4 - a} + \dfrac{b + 4 - b}{4 - b} + \dfrac{c + 4 - c}{4 - c} = 4$

$\sf \longrightarrow \dfrac{ 4}{4 - a} + \dfrac{ 4 }{4 - b} + \dfrac{ 4 }{4 - c} = 4$

Now dividing LHS and RHS by 4,

$\sf \longrightarrow \dfrac{1}{4} \bigg( \dfrac{ 4}{4 - a} + \dfrac{ 4 }{4 - b} + \dfrac{ 4 }{4 - c} \bigg) = \dfrac{1}{4} \times 4$

$\sf \longrightarrow \dfrac{ 4}{4(4 - a)} + \dfrac{ 4 }{4(4 - b)} + \dfrac{ 4 }{ (4 - c )} = 1$

On cancelling 4 from numerator and denominator of each term in LHS,

$\sf \longrightarrow \dfrac{ \cancel{4}}{ \cancel{4} (4 - a)} + \dfrac{ \cancel{4} }{ \cancel{4}(4 - b)} + \dfrac{ \cancel{4} }{ \cancel{4}(4 - c )} = 1$

$\longrightarrow \underline{ \underline{ \sf{ \green{ \dfrac{ 1}{ 4 - a} + \dfrac{ 1 }{ 4 - b} + \dfrac{ 1 }{ 4 - c } = 1}}}}$

Hence proved.