Solving Rational Equations: Practice

Since $x^2 - 1 = (x-1)(x+1)$, the excluded values are $1$ and $-1$, and the LCD is $(x-1)(x+1)$.

Multiplying every term by the LCD gives $x(x+1) - (x-1) = 2$, and the linear terms cancel:

$x^2 + 1 = 2 \;\Rightarrow\; x^2 = 1 \;\Rightarrow\; x = 1 \text{ or } x = -1$

Both candidates are excluded, so both are extraneous. Nothing survives the check, and the equation has no solution.

The excluded values are $x = -1$ and $x = 0$, and the LCD is $x(x+1)$. Multiply every term by it, remembering the $1$ on the right.

$x^2 + 2(x+1) = x(x+1) \;\Rightarrow\; x^2 + 2x + 2 = x^2 + x \;\Rightarrow\; x = -2$

The $x^2$ terms cancel, leaving a linear equation. The candidate $-2$ is in the domain, so it is genuine: $\frac{-2}{-1} + \frac{2}{-2} = 2 - 1 = 1$.

With $x^2 - x = x(x-1)$, the excluded values are $0$ and $1$, and the LCD is $x(x-1)$.

Multiplying every term by the LCD gives $3 + 2x = 5(x-1)$.

$3 + 2x = 5x - 5 \;\Rightarrow\; 8 = 3x \;\Rightarrow\; x = \frac{8}{3}$

The candidate $\frac{8}{3}$ is neither $0$ nor $1$, so it is in the domain and is genuine.

The excluded value is $x = 4$, and the LCD is $x - 4$. Multiply every term by it, including the $1$.

$2x = 8 + (x - 4) \;\Rightarrow\; 2x = x + 4 \;\Rightarrow\; x = 4$

The only candidate is the excluded value, so it is extraneous. Rejecting it leaves nothing, so the equation has no solution.

In one hour Maya paints $\frac{1}{6}$ of the room and Sam paints $\frac{1}{3}$ of it, so together they paint $\frac{1}{t}$ of it, where

$\frac{1}{6} + \frac{1}{3} = \frac{1}{t} \;\Rightarrow\; \frac{1}{2} = \frac{1}{t} \;\Rightarrow\; t = 2$

They take $2$ hours. Sanity check the answer against the story: together they must beat the faster of them working alone, so any answer of $3$ hours or more would be impossible.

The excluded values are $x = -2$ and $x = 0$. Multiply both sides by the LCD $x(x+2)$.

$5x = 3(x+2) \;\Rightarrow\; 5x = 3x + 6 \;\Rightarrow\; x = 3$

The candidate $3$ is in the domain, so it is genuine. Checking: $\frac{5}{5} = 1$ and $\frac{3}{3} = 1$.

The excluded values are $x = 1$ and $x = 0$, and the LCD is $x(x-1)$. Multiplying every term by it gives $4x - 3(x-1) = x(x-1)$.

$x + 3 = x^2 - x \;\Rightarrow\; x^2 - 2x - 3 = 0 \;\Rightarrow\; (x-3)(x+1) = 0$

The candidates are $3$ and $-1$, and neither is excluded, so both are genuine. Checking: $\frac{4}{2} - \frac{3}{3} = 1$ and $\frac{4}{-2} - \frac{3}{-1} = -2 + 3 = 1$. Not every quadratic hands you an extraneous root.

The excluded values are $x = 0$ and $x = -4$. Multiplying both sides by the LCD $x(x+4)$ gives $2(x+4) = x^2$.

$x^2 - 2x - 8 = 0 \;\Rightarrow\; (x-4)(x+2) = 0 \;\Rightarrow\; x = 4 \text{ or } x = -2$

Neither candidate is excluded, so both are genuine. Checking: $\frac{2}{4} = \frac{4}{8}$ and $\frac{2}{-2} = \frac{-2}{2}$.

Multiply each of the three terms by $x - 1$ separately.

$(x-1) \cdot \frac{3}{x-1} = 3, \qquad (x-1) \cdot 2 = 2(x-1), \qquad (x-1) \cdot \frac{x}{x-1} = x$

So the cleared equation is $3 + 2(x-1) = x$. The fractions lose their denominators, while the term that had none picks up a factor of $x - 1$.

Since $x^2 - 9 = (x+3)(x-3)$, the excluded values are $-3$ and $3$, and the LCD is $(x+3)(x-3)$.

Multiplying every term by the LCD gives $x(x-3) - 3(x+3) = 18$. Expanding and collecting,

$x^2 - 6x - 27 = 0 \;\Rightarrow\; (x-9)(x+3) = 0 \;\Rightarrow\; x = 9 \text{ or } x = -3$

The candidate $-3$ is excluded, so it is extraneous. The candidate $9$ is in the domain: $\frac{9}{12} - \frac{3}{6} = \frac{1}{4}$, which matches $\frac{18}{72}$. The only solution is $x = 9$.

The step is reversible wherever the multiplier is nonzero, since you can divide by it to get back. It fails only at the root of the multiplier.

$x = 5 \;\Rightarrow\; x - 5 = 0 \;\Rightarrow\; 0 \cdot A = 0 \cdot B \text{ is true for any } A \text{ and } B$

So $x = 5$ can satisfy the multiplied equation for a reason that has nothing to do with the original. Multiplying by a negative number is perfectly reversible, multiplying by a nonzero constant changes no solutions, and the degree of the result is not what causes the trouble.

Downstream the boat makes $v + 3$ and upstream it makes $v - 3$, and each time is distance divided by speed. Equal times give

$\frac{24}{v+3} = \frac{12}{v-3} \;\Rightarrow\; 24(v-3) = 12(v+3) \;\Rightarrow\; 12v = 108 \;\Rightarrow\; v = 9$

The excluded values are $v = 3$ and $v = -3$, and $9$ is neither, so the candidate is genuine. Checking the story: downstream at $12$ km/h takes $2$ hours, and upstream at $6$ km/h also takes $2$ hours.