Question

solve this pleaseeeee fast​

Answer 1

Answer:

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Answer 2

$\large\sf\underline{Given\::}$

• $\sf\:\frac{(x+1) (4x-3) -4x^{2}+3}{(4x+1) }=2$

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$\large\sf\underline{To\:find\::}$

• Value of x

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$\large\sf\underline{Solution\::}$

$\sf\:\frac{(x+1) (4x-3) -4x^{2}+3}{(4x+1) }=2$

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• Multiplying the terms at numerator

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$\sf\implies\:\frac{(4x^{2}+4x-3x-3)-4x^{2}+3}{(4x+1) }=2$

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• Opening the brackets at the numerator

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$\sf\implies\:\frac{4x^{2}+4x-3x-3-4x^{2}+3}{(4x+1) }=2$

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• Arranging the terms at numerator

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$\sf\implies\:\frac{4x^{2}-4x^{2}+4x-3x-3+3}{4x+1}=2$

$\sf\implies\:\frac{\cancel{4x^{2}}-\cancel{4x^{2}}+4x-3x-\cancel{3}+\cancel{3}}{4x+1}=2$

$\sf\implies\:\frac{4x-3x}{4x+1}=2$

$\sf\implies\:\frac{x}{4x+1}=2$

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• Cross multiplying

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$\sf\implies\:2(4x+1) =x$

$\sf\implies\:8x+2 =x$

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• Transposing x to other side

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$\sf\implies\:8x-x+2 =0$

$\sf\implies\:7x+2 =0$

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• Transposing 2 to the other side

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$\sf\implies\:7x =-2$

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• Transposing 7 also to the other side

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$\large{\mathfrak\red{\implies\:x\:=\:\frac{-2}{7}}}$

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!! Hope it helps !!