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rewona [7]
3 years ago
13

Solve for x. 3/x-1 - 1/x^-1=5/x-1 Please show all work

Mathematics
2 answers:
navik [9.2K]3 years ago
5 0

3/(x-1) - 1/(x^2-1) = 5/(x-1)

Subtract 3/(x-1) from both sides

-1/(x^2-1) = 2/(x-1)

Factor x^2 - 1

-1/[(x-1)(x+1] = 2/(x-1)

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

-1 = 2 (x+1)

-1 = 2x + 2

Subtract 2 from both sides

-3 = 2x

Divide by 2 on both sides

-3/2 = x

Kitty [74]3 years ago
3 0

Answer:

x = -3/2

Step-by-step explanation:

\frac{3(x+1)}{(x-1)(x+1)} - \frac{1}{(x-1)(x+1)} = \frac{5(x+1)}{(x-1)(x+1)} \\  Find the common denominator

3(x+1) - 1 = 5(x+1)   Distribute

3x + 3 -1 = 5x +5    Combine Like terms

3x + 2 = 5x + 5      Solve for x

  2 = 2x +5

  -3 = 2x

  x = -3/2

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7 0
3 years ago
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Answer:

We have the equation

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Then, the augmented matrix of the system is

\left[\begin{array}{cccc}0&0&0&8\\0&0&4&4\\0&3&3&3\\1&1&1&1\end{array}\right]

We exchange rows 1 and 4 and rows 2 and 3 and obtain the matrix:

\left[\begin{array}{cccc}1&1&1&1\\0&3&3&3\\0&0&4&4\\0&0&0&8\end{array}\right]

This matrix is in echelon form. Then, now we apply backward substitution:

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3c_2+3c_3+3c_4=0\\3c_2+3*0+3*0=0\\c_2=0

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dlinn [17]

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8 0
2 years ago
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