Question

plz answer the question in factorization method. question (v)​

Answer 1

Answer:

-7 & 6/7

Step-by-step explanation:

let (3x-1)/(2x+3) = y.......(1)

now the given equatíon....

3y -2/y = 5

3y^2 - 2 = 5y

3y^2 - 5y - 2 = 0

upon solving...

y = 2 & y = -1/3

now put in (1)

(3x-1)/(2x+3) = 2 ; (3x-1)/(2x+3) = -1/3

solving these we get....

x = -7 & x = 6/7

Answer 2

$\large{ \red{ \bold{ \underline{ \underline{Question....}}}}}$

$\large{ \sf{factorize \: \: 3( \frac{3x - 1}{2x + 3}) - 2( \frac{2x + 3}{3x - 1} ) = 5}}$

$\large{ \bold{ \green{ \underline{ \underline{answer...}}}}}$

x= 0 or -7

$\large{ \red{ \bold{ \underline{ \underline{Solution...}}}}}$

$\large{ \sf{ \to \: we \: can \: observe \: that \: the \: terms}} \\ \large{ \sf{in \: lhs \: are \: reciprocal \: of \: each \: other}} \\ \large{ \sf{so \: we \: will \: apply \: this...}} \\ \\ \large{ \sf{ \to let \: \frac{3x - 1}{2x + 3} \: be \: \: \red{a}}} \\ \large{ \sf{ \to \: then \: \frac{2x + 3}{3x - 1} = \frac{1}{a} }} \\ \\ \large{ \sf{ \to \: 3a - \frac{2}{a} = 5(polynomial \: now...)}} \\ \\ \large{ \sf{ \to \: \frac{3 {a}^{2} - 2 }{a} = 5}} \\ \\ \large{ \sf{ \to \: 3 {a}^{2} - 5a - 2 = 0}} \\ \\ \large{ \sf{ \to \: 3 {a}^{2} - 6a + a - 2 = 0}} \\ \\ \large{ \sf{ \to \: 3a(a - 2) + 1(a - 2) = 0}} \\ \\ \large{ \sf{ \to \: (3a + 1)(a - 2) = 0}} \\ \\ \large{ \sf{ \green{ \to \: then \: a = \frac{ - 1}{3} or \: 2}}}$

$\large{ \sf{ \: so \: now...there \: are \: two \: cases...}} \\ \\ \red{ \bold{ \underline{case = 1)}}} \: \large{ \sf{ \frac{3x - 1}{2x + 3} = \frac{ - 1}{3} }} \\ \\ \large{ \sf{ \to \: - 2x - 3 = 9x - 3}}(cross \: multiplication) \\ \\ \large{ \sf{ \to \: - 2x - 9x = - 3 + 3}} \\ \\ \large{ \sf{ \to \: - 11x = 0}} \\ \\ \large{ \sf{ \red{ \to \: x = 0} \blue{(ans - (1))}}} \\ \\ \red{ \bold{ \underline{case = 2)}}} \large{ \sf{ \frac{3x - 1}{2x + 3} = 2}} \\ \\ \large{ \sf{ \to \: 3x - 1 = 4x + 6}} \\ \\ \large{ \sf{ \to \: 3x - 4x = 6 + 1}} \\ \\ \large{ \sf{ \to \: - x = 7}} \\ \\ \large{ \sf{ \red{ \to \: x = - 7} (\blue{ans - (2))}}}$

$\large{ \bold{ \purple{required \: answer(x = 0 \: or \: - 7)}}}$