If 4 - x-2 - 1 - x-3 - 4 - 2x-1- then the values of x are

The correct option is C 2, 7 / 2 Given equation is: 4 / x+2 1 / x+3 = 4 / 2x+1 #160; ⇒ 4( x+3 ) 1( x+2 ) #160;/( x+2 ) ( x+3 ) #1 ...

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Byju's Answer

Standard XII

Mathematics

Parts of a linear equation

If 4 / x+2 ...

Question

If $\frac{4}{x + 2} - \frac{1}{x + 3} = \frac{4}{2 x + 1}$ , then the values of $x$ are

2, \frac{- 7}{2}

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3, \frac{- 7}{3}

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1, \frac{1}{2}

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2, \frac{1}{2}

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Solution

The correct option is C $2, \frac{- 7}{2}$
Given equation is: $\frac{4}{x + 2} - \frac{1}{x + 3} = \frac{4}{2 x + 1}$
$\Rightarrow \frac{4 (x + 3) - 1 (x + 2)}{(x + 2) (x + 3)} = \frac{4}{2 x + 1}$
$\Rightarrow \frac{4 x + 12 - x - 2}{(x + 2) (x + 3)} = \frac{4}{2 x + 1}$
$\Rightarrow \frac{3 x + 10}{(x + 2) (x + 3)} = \frac{4}{2 x + 1}$
$\Rightarrow (3 x + 10) (2 x + 1) = 4 (x + 2) (x + 3)$
$\Rightarrow {6 x}^{2} + 23 x + 10 = 4 (x^{2} + 5 x + 6)$
$\Rightarrow 2 x^{2} + 3 x - 14 = 0$
$\Rightarrow 2 x^{2} - 4 x + 7 x - 14 = 0$
$\Rightarrow 2 x (x - 2) + 7 (x - 2) = 0$
$\Rightarrow (x - 2) (2 x + 7) = 0$
$\Rightarrow x - 2 = 0$ and $2 x + 7 = 0$
$\Rightarrow x = 2$ and $x = \frac{- 7}{2}$
Hence, $x = \frac{- 7}{2}, 2$ are the required solutions.

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Q. If

∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{2} - 2 x + 3 & 7 x + 2 & x + 4 2 x + 7 & x^{2} - x + 2 & 3 x 3 & 2 x - 1 & x^{2} - 4 x + 7 \end{matrix} ∣ ∣ ∣ ∣ ∣ = a x^{6} + b x^{5} + c x^{4} + d x^{3} + e x^{2} + f x + g

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