Question

$3x + 2 + \frac{3}{3x + 2} = - 4$
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Answer 1

$\huge\underline{\overline{\mid{\bold{\mathcal{\blue{Answer}\mid}}}}}$

$→3x + 2 + \frac{3}{3x + 2} = - 4 \\ → \frac{[3x + 2(3x) ]+[ 3x + 2(2)] + 3}{3x + 2} = - 4 \\ → \frac{ {9x}^{2} + 6x + 6x + 4+ 3 }{3x + 2} = - 4 \\→ {9x}^{2} + 12x + 7 = 3x + 2( - 4) \\ →{9x}^{2} + 12x + 7 = - 12x - 8 \\ →{9x}^{2} + 12x + 12x + 7 + 8 = 0 \\ → {9x}^{2} + 24x + 15 = 0 \\ \\Dividing \: each \: term \: by \: 3 \\ → {3x}^{2} + 8x + 5 = 0 \\ → {3x}^{2} + 3x + 5x + 5 = 0 \\→3x(x + 1) + 5(x + 1) = 0 \\ →(3x + 5)(x + 1) = 0$

$Equating \: each \: t o \: 0 \\ →x + 1 = 0 \\{\boxed{ x = - 1}} \\ →3x + 5 = 0 \\ 3x = - 5 \\ {\boxed{x = \frac{ - 5}{3}}}$

$Hence, \: value \: of \: x \: is \: -1 \: or \: \frac{ - 5}{3}$