Question

Please help ASAP
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Answer 1

We are given $\sf{\dfrac{2}{x-3} +\dfrac{x}{x^2-9} =\dfrac{4}{x+3} }$.

How to solve if the denominator has an unknown?

This equation is called a rational equation.

To solve this equation no unknown should be in the denominator. So, we multiply to the denominator to remove it.

The denominator has HCF of $\sf{x^2-9}$. So we multiply $\sf{x^2-9}$ on both sides.

$\implies\sf{\dfrac{2(x^2-9)}{x-3} +\dfrac{x(x^2-9)}{x^2-9} =\dfrac{4(x^2-9)}{x+3} }$

$\implies\sf{2(x+3)+x=4(x-3)}$

We solve this equation and get $\sf{x=18}$ as the solution.

Also, we need to check whether the denominator becomes 0. The solution should not belong to $\sf{x^2-9=0}$.

$\sf{18\neq 3}$ and $\sf{18\neq -3}$, so the solution is valid.

Answer 2

To solve :

$\sf \dfrac{2}{x - 3} \: + \: \dfrac{x}{ {x}^{2} - 9 } \: = \: \dfrac{4}{x + 3}$

Solution :

$\sf \dfrac{2}{x - 3} \: + \: \dfrac{x}{ {x}^{2} - 9 } \: = \: \dfrac{4}{x + 3}$

• Now, x² - 9 = (x)² - (3)²

Note- using identity [ a² - b² = (a + b)(a - b) ]

➝ x² - 9 = (x+3)(x-3)

$\sf \implies \: \dfrac{2}{x - 3} \: + \: \dfrac{x}{ (x + 3)({x} - 3)} \: = \: \dfrac{4}{x + 3}$

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Taking LCM on LHS ,

LCM of (x-3) and (x-3)(x+3) = (x-3)(x+3)

$\sf \implies \: \: \dfrac{[ \: 2 \times (x+3) ] +\: (x \: \times \: 1)}{ (x + 3)({x} - 3)} \: = \: \dfrac{4}{x + 3}$

$\sf \implies \: \: \dfrac{2x + 6+\: x}{ (x + 3)({x} - 3)} \: = \: \dfrac{4}{x + 3}$

$\sf \implies \: \: \dfrac{ 3x + 6}{ (x + 3)({x} - 3)} \times (x + 3) \: \: = \: 4$

$\sf \implies \: \: \dfrac{ 3x + 6}{ ({x} - 3)} \: \: = \: 4$

$\sf \implies \: \: 3x + 6 \: = \: 4 \times (x - 3)$

$\sf \implies \: \: 3x + 6 \: = \: 4x - 12$

$\sf \implies \: \: 4x - 3x = 12 + 6 \:$

$\sf \implies \: \: \bold{x = 18}\:$

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ANSWER : x = 18