Question

(4-x)/(x^2-5x+4)+(2)/(x-1)=1 show all steps please!!

Answer 1

Answer:

$x = 2, 4$

Step-by-step explanation:

Given :

$\frac{4-x}{x^{2} -5x+4} + \frac{2}{x-1} =1$

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Factorize the denominator of the first term :

⇒ x² - 5x + 4

⇒ x² - x - 4x + 4

⇒ x(x - 1) - 4(x - 1)

⇒ (x - 4)(x - 1)

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Hence, the equation now is :

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

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Multiply the second term by (x - 4) in the numerator and denominator :

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

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Combine the numerator of both terms and bring the denominator to the other side :

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

$4 - x+2x-8 = x^{2} - 5x + 4$

$x - 4 = x^{2} - 5x + 4$

$x^{2} -5x-x+4+4=0$

$x^{2} -6x+8=0$

$(x-4)(x-2)=0$

$x = 2, 4$