Math, asked by ushachandrawanshi077, 8 months ago

can anyone help me solving the equation....within 5 min ...

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Answered by Anonymous

$\fbox{Solution\:of\:17}$

$\frac{1}{x + 6} + \frac{1}{x - 10} = \frac{3}{x - 4} \\ = > \frac{1 + 1}{x + 6 + x - 10} = \frac{3}{x - 4} \\ = > \frac{2}{2x - 4} = \frac{3}{x - 4} \\ = > 2(x - 4) = 3(2x - 4) \\ = > 2x - 8 = 6x - 12 \\ = > 2x - 6x = - 12 + 8 \\ = > - 4x = - 4 \\ = > x = \frac{ - 4}{ - 4} \\ = > x = 1$

$\fbox{Solution\:of\:18}$

$\sqrt{3x + 4} = x \\ = > \frac{1}{2} (3x + 4) = x \\ = > 3x \times \frac{1}{2} + 4 \times \frac{1}{2} = x \\ = > \frac{3x}{2} + 2 = x \\ = > \frac{3x + 4}{2} = x \\ = > 3x + 4 = 2x \\ = > 4 = 2x - 3x \\ = > - x = 4 \\ = > x = - 4$

Answered by Anonymous

Answer:

$<body bgcolor="r"><font color="cyan">$

$\huge{\red{\bf{\mathfrak{Answer-:}}}}$

$\huge{\bold{❤}}$

$\huge{\blue{\bf{\mathfrak{solution\:1-:}}}}$

$\frac{1}{x + 6 } + \frac{1}{x - 10} = \frac{3}{x - 4} \\ \\ \frac{x - 10 + x + 6}{(x - 10)(x + 6)} = \frac{3}{x - 4} \\ \\ \frac{2x - 4}{ {x}^{2} + 6x - 10x - 60 } = \frac{3}{x - 4} \\ \\ \frac{2x - 4}{ {x}^{2} - 4x - 60 } = \frac{3}{x - 4} \\ \\ (2x - 4)(x - 4) = 3 {x}^{2} - 12x \\ - 180 \\ \\ 2 {x}^{2} - 8x - 4x + 16 = 3 {x}^{2} - \\ 12x - 180 \\ \\ 3 {x}^{2} - 12x - 180 -2 {x}^{2} + 8x \\ + 4x - 16 \\ \\ {x}^{2} - 196 = 0\\ \\$

$\huge{\blue{\bf{\mathfrak{solution\:2-:}}}}$

$\sqrt{3x + 4} = x \\ \\ squaring \: both \: side \\ \\ 3x + 4 = {x}^{2} \\ \\ {x}^{2} - 3x - 4 = 0$

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can anyone help me solving the equation....within 5 min ...