please solve

Question

8/1-x^2+5/x+1= 4/x-1

Answer 1

8 / (1 - x^2) + 5 / (x + 1) = 4 / (x - 1)

-8 / (x^2 - 1) + 5 / (x + 1) = 4 / (x - 1)

-8 /((x + 1)(x - 1)) + 5 / (x + 1) = 4 / (x - 1)

Can’t divide by zero, so x ≠ ± 1.

Multiply both sides by (x + 1)(x - 1):

-8 + 5(x - 1) = 4(x + 1)

-8 + 5x - 5 = 4x + 4

Add -4x + 13 to both sides:

x = 17

check:

8 / (1 - (17)^2) + 5 / ((17) + 1) =? 4 / ((17) - 1)

8/((1 + 17)(1 - 17)) + 5/18 =? 4/16

8/((18)(-16)) + 5/18 =? 1/4

-1/((18)(2)) + 2*5/(2*18) =? 1/4

9/(2*18) = 1/4 √

Answer 2

Hi Mondrea;

I believe Kay's interpretation is correct.  Please let me know if we are mistaken.

8/1-x^2+5/x+1= 4/x-1

[8/(1-x²)]+[5/(x+1)]=4/(x-1)

[8/(1-x²)]+[5/(x+1)]=4/(x-1)

(1-x²)=(1+x)(1-x)

Let's take [5/(x+1)], and multiply it by (1-x)/(1-x) such that the denominator is also (1-x²)

[8/(1+x)(1-x)]+[5(1-x)/(1+x)(1-x)]=4/(x-1)

On the left side, the denominators are identical.  Let's add these together...

[8+5(1-x)]/[(1-x)(1+x)]=4/(x-1)

8+5(1-x)=8+5-5x=13-5x

On the right side, let's multiply the numerator and denominator by -1/-1...

(13-5x)/[(1+x)(1-x)]=[4/(x-1)](-1/-1)

(13-5x)/[(1+x)(1-x)]=-4/(1-x)

(1-x) is now in the denominator of both sides.  It cancels...

(13-5x)/(1+x)=-4

Cross-multiply...

13-5x=-4(1+x)

13-5x=-4-4x

13-5x=-4-4x

Add 4 to both sides...

4+13-5x=-4-4x+4

17-5x=-4x

Add 5x to both sides...

5x+17-5x=-4x+5x

17=x

 

Let's check our work with the original equation...

[8/(1-x^2)]+[5/(x+1)]= 4/(x-1)

[8/(1-17²)]+[5/(17+1)]=4/(17-1)

[8/(1-289)]+[5/(17+1)]=4/(17-1)

(8/-288)+(5/18)=4/16

(-8/288)+(5/18)=4/16

(-1/36)+(5/18)=1/4

Let's multiply (5/18) by (2/2)=10/36

Let's multiply (1/4)(9/9)=9/36

(-1/36)+(10/36)=9/36

9/36=9/36

 

 

 

 

 

Answer 3

8/(1-x²)    + 5/(x+1)  = 4/(x-1)

 

8/(1-x)(1+ x)     + 5/(x+1)  = 4/(x-1)

 

multiply both sides with (1+x)(x-1)

 

-8 + 5(x-1)  = 4(x+1)(-1)

 

-8 + 5x - 5 = -4x - 4

 

9x = 9

 

x = 1