Question

Please solve the 10th question... Plz

Answer 1

$Given : \frac{3}{x - 1} + \frac{1}{x + 1}=\frac{4}{x}$

LCM = x(x - 1)(x + 1)

= > 3(x)(x + 1) + (x)(x - 1) = 4(x - 1)(x + 1)

= > 3x^2 + 3x + x^2 - x = 4(x^2 - 1)

= > 4x^2 + 2x = 4x^2 - 4

= > 4x^2 + 2x - 4x^2 = -4

= > 2x = -4

= > x = -2.

Hope this helps!

Answer 2

• 3/x+1 + 4/x-1 =29/4x-1
• =>3(x-1)+4(x+1)/x2-1 =29/4x-1
• =>3x-3+4x+4/x2-1 =29/4x-1
• =>7x+1/x2-1 = 29/4x-1
• =>28x2+4x-7x-1 =29x2-29
• =>-x2-3x+28 = 0
• =>x2+3x-28=0
• =>x2+7x-4x-28 = 0
• =>x(x+7)-4(x+7)=0
• =>(x+7)(x-4)=0
• =>x+7=0
• =>x= -7
• or,
• =>x-4=0
• =>x=4
• this is the answer of your question
• please mark me as a brilliant
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