Question

answer it please CORRECTLY​

Answer 1

Question:-

Solve:

$\sf \dfrac{5}{x + 4} + \dfrac{3}{x - 4} = \dfrac{2x}{ {x}^{2} - 16 }$

Solution:-

$\sf \dfrac{5}{x + 4} + \dfrac{3}{x - 4} = \dfrac{2x}{ {x}^{2} - 16 }$

$\implies\dfrac{5(x - 4) + 3(x + 4)} {(x + 4)(x - 4)} = \dfrac{2x}{ {x}^{2} - 16 }$

$\implies\dfrac{5x - 20 + 3x + 12} {( {x}^{2} - 16)} = \dfrac{2x}{ {x}^{2} - 16 }$

$\implies\dfrac{8x - 8} {( {x}^{2} - 16)} = \dfrac{2x}{ ({x}^{2} - 16 )}$

$\implies\dfrac{8x - 8} { \cancel{( {x}^{2} - 16)}} = \dfrac{2x}{ \cancel{ ({x}^{2} - 16 )} }$

$\implies \: 8(x - 1) = 2x$

$\implies(x - 1) = \dfrac{ \not2x}{ \not8} \\ \implies \: x - 1 = \dfrac{x}{4} \\ \implies4(x - 1) = x \\ \implies4x - 4 = x \\ \implies4x - x = 4 \\ \implies3x = 4 \\ \\ \implies\pink{\bold{\boxed{\boxed{ \orange{\therefore \: x = \dfrac{4}{3}}}}}}$

Correct answer:-

• $\large{\boxed{\boxed{ \bold{ \orange{x = \dfrac{4}{3}}}}}}$

Answer 2

$\Large{\underbrace{\underline{\sf{Understanding\: the\: Question}}}}$

Here in this question, concept of lowest common multiple (LCM) is used. The question is about finding value of x from the faction terms. We can find the value of x by using LCM and transportation of LHS and RHS methods.

So let's start!

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$\sf\longrightarrow\dfrac{5}{x+4}+\dfrac{3}{x-4}=\dfrac{2x}{x^2-16}$

Take LCM from denominators

$\sf\longrightarrow\dfrac{5(x-4)+3(x+4)}{(x+4)(x-4)}=\dfrac{2x}{x^2-16}$

Now simplify numerator

$\sf\longrightarrow\dfrac{5x-20+3x+12}{(x+4)(x-4)}=\dfrac{2x}{x^2-16}$

$\sf\longrightarrow\dfrac{8x-8}{(x+4)(x-4)}=\dfrac{2x}{x^2-16}$

Now we can see that denominator is in the form (a+b)(a-b)=a²-b² [an algeberic identity] , where a=x and b=4.

So:-

$\sf\longrightarrow\dfrac{8x-8}{[(x^2)-(4^2)]}=\dfrac{2x}{x^2-16}$

Now simplify denominator

$\sf\longrightarrow\dfrac{8x-8}{x^2-16}=\dfrac{2x}{x^2-16}$

Now transport LHS from denominator to numerator of RHS.

$\sf\longrightarrow 8x-8=\dfrac{2x(x^2-16)}{x^2-16}$

Now we can cancel out same numerator and denominator from RHS.

$\sf\longrightarrow 8x-8=2x$

Now transport -8 to RHS.

$\sf\longrightarrow 8x-2x=8$

$\sf\longrightarrow 6x=8$

Now transport 6 from LHS to RHS

$\sf\longrightarrow x=\dfrac{8}{6}$

Now make it into Simplest from

$\sf\longrightarrow x=\dfrac{\not{8}}{\not6}$

$\large\underline{\boxed{\sf {\red{x=\dfrac{4}{3}}}}}\bigstar$

$\large\sf{The\: Required \:value \:of\: x=\dfrac{4}{3}}$

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Important thing to remember while transportation of LHS and RHS.

• Whenever you transport a positive digit to another side, it will change into negative.

• Whenever you transport a negative digit to another side, it will change into positive.

• Whenever you transport denominator to other side, it will change into numerator.

• Whenever you transport numerator to another side, it will change into denominator.

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