Question

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Answer 1

★ Solution

$\dfrac{x - 2}{x - 3} + \dfrac{3x - 11}{x - 4} = \dfrac{4x + 13}{x + 1}$

Taking LCM in LHS , we get

$\dfrac{(x - 2)(x - 4) + (x - 3)(3x - 11)}{(x - 3)(x - 4)} = \dfrac{4x + 13}{x + 1}$

$\dfrac{ ({x}^{2} - 4x - 2x + 8) + ( {3x}^{2} - 11x - 9x + 33) }{( {x}^{2} - 4x - 3x + 12)} = \dfrac{4x + 13}{x + 1}$

$\dfrac{ {4x}^{2} - 26x + 41 }{ {x}^{2} - 7x + 12 } = \dfrac{4x + 13}{x + 1}$

$4 {x}^{3} + 4 {x}^{2} - 26 {x}^{2} - 26x + 41x + 41 = 4 {x}^{3 } + 13 {x}^{2} - 28 {x}^{2} - 91x + 48x + 156$

$4 {x}^{2} - 26 {x}^{2} - 13 {x}^{2} + 28 {x}^{2} - 26x + 41x + 91x - 48x + 41 - 156 = 0$

$- 7 {x}^{2} + 58x - 115 = 0$

$7 {x}^{2} - 58x + 115 = 0$

Using quadratic formula ,

$x = \dfrac{ - b± \sqrt{ {b}^{2} - 4ac} }{2a}$

Here a = 7 b = -58 c = 115

$x = \dfrac{58± \sqrt{ {( - 58)}^{2} - 4 \times 7 \times 115 } }{14}$

$x = \dfrac{58 + 12}{14} or \: \dfrac{58 - 12}{14}$

$x = 5 \: or \: \dfrac{23}{7}$

Answer 2

Solution

\dfrac{x - 2}{x - 3} + \dfrac{3x - 11}{x - 4} = \dfrac{4x + 13}{x + 1}

x−3

x−2

+

x−4

3x−11

=

x+1

4x+13

Taking LCM in LHS , we get

\dfrac{(x - 2)(x - 4) + (x - 3)(3x - 11)}{(x - 3)(x - 4)} = \dfrac{4x + 13}{x + 1}

(x−3)(x−4)

(x−2)(x−4)+(x−3)(3x−11)

=

x+1

4x+13

\dfrac{ ({x}^{2} - 4x - 2x + 8) + ( {3x}^{2} - 11x - 9x + 33) }{( {x}^{2} - 4x - 3x + 12)} = \dfrac{4x + 13}{x + 1}

(x

2

−4x−3x+12)

(x

2

−4x−2x+8)+(3x

2

−11x−9x+33)

=

x+1

4x+13

\dfrac{ {4x}^{2} - 26x + 41 }{ {x}^{2} - 7x + 12 } = \dfrac{4x + 13}{x + 1}

x

2

−7x+12

4x

2

−26x+41

=

x+1

4x+13

4 {x}^{3} + 4 {x}^{2} - 26 {x}^{2} - 26x + 41x + 41 = 4 {x}^{3 } + 13 {x}^{2} - 28 {x}^{2} - 91x + 48x + 1564x

3

+4x

2

−26x

2

−26x+41x+41=4x

3

+13x

2

−28x

2

−91x+48x+156

4 {x}^{2} - 26 {x}^{2} - 13 {x}^{2} + 28 {x}^{2} - 26x + 41x + 91x - 48x + 41 - 156 = 04x

2

−26x

2

−13x

2

+28x

2

−26x+41x+91x−48x+41−156=0

- 7 {x}^{2} + 58x - 115 = 0−7x

2

+58x−115=0

7 {x}^{2} - 58x + 115 = 07x

2

−58x+115=0

Using quadratic formula ,

x = \dfrac{ - b± \sqrt{ {b}^{2} - 4ac} }{2a}x=

2a

−b±

b

2

−4ac

Here a = 7 b = -58 c = 115

x = \dfrac{58± \sqrt{ {( - 58)}^{2} - 4 \times 7 \times 115 } }{14}x=

14

58±

(−58)

2

−4×7×115

x = \dfrac{58 + 12}{14} or \: \dfrac{58 - 12}{14}x=

14

58+12

or

14

58−12

x = 5 \: or \: \dfrac{23}{7}x=5or

7

23