Question

Calculation the roots of the equation,
$\frac{x - 1}{x - 2} + \frac{x - 3}{x - 4} = 2 \times \frac{1}{3}$
,(where x is not equal 2,4)...​

Answer 1

Answer:

$\frac{x - 1}{x - 2} + \frac{x - 3}{x - 4} = \frac{2}{3} \\ \rightarrow \: \frac{(x - 1)(x - 4) + (x - 3)(x - 2)}{(x - 2)(x - 4)} = \frac{2}{3} \\ = > \frac{ {x}^{2} - 5x + 4 + {x}^{2} - 5x + 6 }{ {x}^{2} - 6x + 8 } = \frac{2}{3} \\ = > \frac{ {2x}^{2} - 10x + 10 }{ {x}^{2} - 6x + 8 } = \frac{2}{3} \\ 6 {x}^{2} - 30x + 30 = 2 {x}^{2} - 12x + 16 \\ = > 4 {x}^{2} - 18x + 14 = 0 \\ hope \: it \: helps \: you...$

Answer 2

Answer:

x−2

x−1

+

x−4

x−3

=

3

2

→

(x−2)(x−4)

(x−1)(x−4)+(x−3)(x−2)

=

3

2

=>

x

2

−6x+8

x

2

−5x+4+x

2

−5x+6

=

3

2

=>

x

2

−6x+8

2x

2

−10x+10

=

3

2

6x

2

−30x+30=2x

2

−12x+16

=>4x

2

−18x+14=0

hopeithelpsyou...

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