Solutions to Problem Set 19

Math 101-06

10-23-2017

[Fraction equations]

1. Solve $\dfrac{2}{x^2} = 1 +
   \dfrac{1}{x}$ .

$$\eqalign{ \dfrac{2}{x^2} & = 1 + \dfrac{1}{x} \cr \noalign{\vskip2pt} x^2 \cdot \dfrac{2}{x^2} & = x^2 \cdot 1 + x^2 \cdot \dfrac{1}{x} \cr \noalign{\vskip2pt} 2 & = x^2 + x \cr 0 & = x^2 + x - 2 \cr 0 & = (x + 2)(x - 1) \cr}$$

The possible solutions are $x =
   1$ and $x = -2$ , and both check.


2. Solve $\displaystyle
   \dfrac{1}{x + 3} + \dfrac{1}{x} = \dfrac{25}{x^2 + 3 x}$ .

$$\eqalign{ \dfrac{1}{x + 3} + \dfrac{1}{x} & = \dfrac{25}{x^2 + 3 x} \cr \noalign{\vskip2pt} \dfrac{1}{x + 3} + \dfrac{1}{x} & = \dfrac{25}{x(x + 3)} \cr \noalign{\vskip2pt} x(x + 3) \cdot \dfrac{1}{x + 3} + x(x + 3) \cdot \dfrac{1}{x} & = x(x + 3) \cdot \dfrac{25}{x(x + 3)} \cr \noalign{\vskip2pt} x + (x + 3) & = 25 \cr 2 x + 3 & = 25 \cr 2 x + 3 - 3 & = 25 - 3 \cr 2 x & = 22 \cr \dfrac{1}{2} \cdot 2 x & = \dfrac{1}{2} \cdot 22 \cr x & = 11 \cr}$$

The solution is $x = 11$ .


3. Solve $\displaystyle 1 =
   \dfrac{3}{x - 2} - \dfrac{x + 7}{x^2 - x - 2}$ .

$$\eqalign{ 1 & = \dfrac{3}{x - 2} - \dfrac{x + 7}{x^2 - x - 2} \cr \noalign{\vskip2pt} 1 & = \dfrac{3}{x - 2} - \dfrac{x + 7}{(x - 2)(x + 1)} \cr \noalign{\vskip2pt} (x - 2)(x + 1) \cdot 1 & = (x - 2)(x + 1) \cdot \dfrac{3}{x - 2} - (x - 2)(x + 1) \cdot \dfrac{x + 7}{(x - 2)(x + 1)} \cr \noalign{\vskip2pt} (x - 2)(x + 1) & = 3(x + 1) - (x + 7) \cr x^2 - x - 2 & = 3 x + 3 - x - 7 \cr x^2 - x - 2 & = 2 x - 4 \cr x^2 - x - 2 - 2 x + 4 & = 2 x - 4 - 2 x + 4 \cr x^2 - 3 x + 2 & = 0 \cr (x - 1)(x - 2) & = 0 \cr}$$

The possible solutions are $x =
   1$ and $x = 2$ , but $x = 2$ causes division by 0 in the original equation. The only solution is $x = 1$ .


4. Solve $\displaystyle
   \dfrac{8}{x + 3} - \dfrac{6}{x^2 + 5 x + 6} = 1$ .

$$\eqalign{ \dfrac{8}{x + 3} - \dfrac{6}{x^2 + 5 x + 6} & = 1 \cr \noalign{\vskip2pt} \dfrac{8}{x + 3} - \dfrac{6}{(x + 2)(x + 3)} & = 1 \cr \noalign{\vskip2pt} (x + 2)(x + 3) \cdot \dfrac{8}{x + 3} - (x + 2)(x + 3) \cdot \dfrac{6}{(x + 2)(x + 3)} & = (x + 2)(x + 3) \cdot 1 \cr \noalign{\vskip2pt} 8(x + 2) - 6 & = (x + 2)(x + 3) \cr 8x + 16 - 6 & = x^2 + 5 x + 6 \cr 8x + 10 & = x^2 + 5 x + 6 \cr 8x + 10 - 8x - 10 & = x^2 + 5 x + 6 - 8x - 10 \cr 0 & = x^2 - 3 x - 4 \cr 0 & = (x - 4)(x + 1) \cr}$$

The solutions are $x = 4$ and $x = -1$ .


5. Solve $\displaystyle
   \dfrac{1}{x + 1} + \dfrac{19}{x^2 - x - 2} = \dfrac{x + 2}{x - 2}$ .

$$\eqalign{ \dfrac{1}{x + 1} + \dfrac{19}{x^2 - x - 2} & = \dfrac{x + 2}{x - 2} \cr \noalign{\vskip2pt} \dfrac{1}{x + 1} + \dfrac{19}{(x - 2)(x + 1)} & = \dfrac{x + 2}{x - 2} \cr \noalign{\vskip2pt} (x - 2)(x + 1) \cdot \dfrac{1}{x + 1} + (x - 2)(x + 1) \cdot \dfrac{19}{(x - 2)(x + 1)} & = (x - 2)(x + 1) \cdot \dfrac{x + 2}{x - 2} \cr \noalign{\vskip2pt} (x - 2) + 19 & = (x + 1)(x + 2) \cr x + 17 & = x^2 + 3 x + 2 \cr x + 17 - x - 17 & = x^2 + 3 x + 2 - x - 17 \cr 0 & = x^2 + 2 x - 15 \cr 0 & = (x + 5)(x - 3) \cr}$$

The solutions are $x = -5$ and $x = 3$ .


6. Solve $\displaystyle \dfrac{3
   x}{x - 1} = \dfrac{3}{x - 1} - 2$ .

$$\eqalign{ \dfrac{3 x}{x - 1} & = \dfrac{3}{x - 1} - 2 \cr \noalign{\vskip2pt} (x - 1) \cdot \dfrac{3 x}{x - 1} & = (x - 1) \cdot \dfrac{3}{x - 1} - (x - 1) \cdot 2 \cr \noalign{\vskip2pt} 3 x & = 3 - 2(x - 1) \cr 3 x & = 3 - 2 x + 2 \cr 3 x & = 5 - 2 x \cr 3 x + 2 x & = 5 - 2 x + 2 x \cr 5 x & = 5 \cr x & = 1 \cr}$$

However, $x = 1$ causes division by zero in the original equation. Hence, there are no solutions.


7. Solve $\displaystyle x + 1 -
   \dfrac{4}{x + 1} = \dfrac{4 x}{x + 1}$ .

$$\eqalign{ x + 1 - \dfrac{4}{x + 1} & = \dfrac{4 x}{x + 1} \cr \noalign{\vskip2pt} (x + 1) \cdot x + 1 - (x + 1) \cdot \dfrac{4}{x + 1} & = (x + 1) \cdot \dfrac{4 x}{x + 1} \cr \noalign{\vskip2pt} (x + 1)^2 - 4 & = 4 x \cr x^2 + 2 x + 1 - 4 & = 4 x \cr x^2 + 2 x - 3 & = 4 x \cr x^2 + 2 x - 3 - 4 x & = 4 x - 4 x \cr x^2 - 2 x - 3 & = 0 \cr (x - 3)(x + 1) & = 0 \cr}$$

The possible solutions are $x =
   3$ and $x = -1$ , but $x = -1$ causes division by 0 in the original equation. The only solution is $x = 3$ .


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