Solve for 3x-4 x-1 - 3x-2 x-1- x- -1 and x- 1

The correct option is D 32 3x+4x+1=3x+2x 1⇒ (3x+4)(x 1)=(3x+2)(x+1) #160; ⇒ 3x^2 3x+4x 4=3x^2+3x+2x+2 ⇒ 4x 4=2 ⇒x= 32 ...

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Byju's Answer

Standard X

Mathematics

Modulus Function

Solve for 3...

Question

Solve for $\frac{3 x + 4}{x + 1} = \frac{3 x + 2}{x - 1}$ , $x \neq - 1$ and $x \neq 1$

\frac{- 3}{2}

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\frac{2}{3}

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\frac{- 1}{3}

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None of these

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Solution

The correct option is D $\frac{- 3}{2}$
$\frac{3 x + 4}{x + 1} = \frac{3 x + 2}{x - 1} \Rightarrow (3 x + 4) (x - 1) = (3 x + 2) (x + 1)$

$\Rightarrow 3 x^{2} - 3 x + 4 x - 4 = 3 x^{2} + 3 x + 2 x + 2$
$\Rightarrow - 4 x - 4 = 2$
$\Rightarrow x = \frac{- 3}{2}$

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x + 2) ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ 2 x^{3} - 4 x^{2} - 3 x - 1

Solve :

$(i) \frac{1}{3} x - 6 = \frac{5}{2} (i i) \frac{2 x}{3} - \frac{3 x}{8} = \frac{7}{12} (i i i) (x + 2) (x + 3) + (x - 3) (x - 2) - 2 x (x + 1) = 0 (i v) \frac{1}{10} - \frac{7}{x} = 35 (v) 13 (x - 4) - 3 (x - 9) - 4 (x + 4) = 0 (v i) x + 7 - \frac{8 x}{3} = \frac{17 x}{6} - \frac{5 x}{8} (v i i) \frac{3 x - 2}{4} - \frac{2 x + 3}{3} = \frac{2}{3} - x (v i i i) \frac{x + 2}{6} - (\frac{11 - x}{3} - \frac{1}{4}) = \frac{3 x - 4}{12} (i x) \frac{2}{5 x} - \frac{5}{3 x} = \frac{1}{15} (x) \frac{x + 2}{3} - \frac{x + 1}{5} = \frac{x - 3}{4} - 1 (x i) \frac{3 x - 2}{3} + \frac{2 x + 3}{2} = x + \frac{7}{6} (x i i) x - \frac{x - 1}{2} = 1 - \frac{x - 2}{3} (x i i i) \frac{9 x + 7}{2} - (x - \frac{x - 2}{7}) = 36 (x i v) \frac{6 x + 1}{2} + 1 = \frac{7 x - 3}{3}$

Q. Solve:

{lim}_{x \to 1} \frac{x^{4} - {3 x}^{3} + 2}{x^{3} - {5 x}^{2} + 3 x + 1}

Q. Solve for

x

\frac{3 x}{4} - \frac{(x - 1)}{3} = \frac{(x - 2)}{2}

Q. Solve for

x

3 x - \frac{x - 2}{3} = 4 - \frac{x - 1}{4}

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Solve for 3x+4x+1=3x+2x−1, x≠−1 and x≠1

The correct option is D −323x+4x+1=3x+2x−1⇒(3x+4)(x−1)=(3x+2)(x+1) ⇒3x2−3x+4x−4=3x2+3x+2x+2⇒−4x−4=2⇒x=−32

Solve for $\frac{3 x + 4}{x + 1} = \frac{3 x + 2}{x - 1}$ , $x \neq - 1$ and $x \neq 1$

The correct option is D $\frac{- 3}{2}$
$\frac{3 x + 4}{x + 1} = \frac{3 x + 2}{x - 1} \Rightarrow (3 x + 4) (x - 1) = (3 x + 2) (x + 1)$

$\Rightarrow 3 x^{2} - 3 x + 4 x - 4 = 3 x^{2} + 3 x + 2 x + 2$
$\Rightarrow - 4 x - 4 = 2$
$\Rightarrow x = \frac{- 3}{2}$