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 $2(x+1) + 3(x-1) = 6$.

$5x - 1 = 6 \;\Rightarrow\; x = \frac{7}{5}$

The candidate $\frac{7}{5}$ is neither $1$ nor $-1$, so it is in the domain and is genuine.

With $x^2 - 4 = (x-2)(x+2)$, the excluded values are $2$ and $-2$, and the LCD is $(x-2)(x+2)$.

Multiplying every term by the LCD gives $x(x+2) - 4(x-2) = 16$. Expanding and collecting,

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

The candidate $-2$ is excluded, so it is extraneous. The candidate $4$ is in the domain: $\frac{4}{2} - \frac{4}{6} = \frac{4}{3}$, which matches $\frac{16}{12}$. The only solution is $x = 4$.

The excluded value is $x = 3$. Multiplying every term by $x - 3$ gives $x = 3 + k(x-3)$, and collecting the $x$ terms,

$x(1-k) = 3(1-k)$

If $k \neq 1$, divide by the nonzero number $1 - k$ to get $x = 3$, the excluded value, so the equation has no solution at all. If $k = 1$, both sides are $0$ and the statement is true for every $x$, so every value in the domain solves the original. Only $k = 1$ gives infinitely many solutions.

The excluded values are $-1$ and $1$, and the LCD is $(x+1)(x-1)$.

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

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

Both candidates are excluded, so both are extraneous and the equation has no solution.

The excluded value is $x = 1$. The two sides share a denominator, so multiplying by $x - 1$ leaves the numerators.

$x^2 + 2 = 3 \;\Rightarrow\; x^2 = 1 \;\Rightarrow\; x = 1 \text{ or } x = -1$

The candidate $1$ is excluded and drops out. The candidate $-1$ is in the domain, so it is genuine: both sides equal $-\frac{3}{2}$ there. Exactly one solution survives.

Let $v$ be the running speed, so the walking speed is $v - 2$. Time is distance over speed, and the two legs add to $5$ hours:

$\frac{12}{v} + \frac{12}{v-2} = 5$

The excluded values are $0$ and $2$. Multiplying every term by the LCD $v(v-2)$ gives $12(v-2) + 12v = 5v(v-2)$, so $5v^2 - 34v + 24 = 0$ and $(5v-4)(v-6) = 0$.

Both candidates, $v = 6$ and $v = \frac{4}{5}$, are in the domain and satisfy the equation. But $v = \frac{4}{5}$ makes the walking speed $\frac{4}{5} - 2$, which is negative, so the story rejects it. She ran at $6$ km/h, walked at $4$ km/h, and the legs took $2$ and $3$ hours.

Factor first: $x^2 - x = x(x-1)$, so the excluded values are $0$ and $1$, and the LCD is $x(x-1)$.

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

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

The candidate $\frac{1}{2}$ is in the domain, so it is genuine. Checking: the left side is $\frac{2}{-1/4} + 2 = -8 + 2 = -6$, and the right side is $\frac{3}{-1/2} = -6$.

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

$3(x+1)^2 + 3(x-1)^2 = 10(x^2-1)$

Expanding, the left side is $6x^2 + 6$, so $6x^2 + 6 = 10x^2 - 10$, giving $4x^2 = 16$ and $x^2 = 4$.

The candidates are $2$ and $-2$, and neither is excluded, so both are genuine. Checking: $\frac{3}{1} + \frac{1}{3} = \frac{10}{3}$, and $\frac{-1}{-3} + \frac{-3}{-1} = \frac{1}{3} + 3 = \frac{10}{3}$.

Multiplying both sides by the LCD preserves any equality that already held, so no solution can be lost.

$A(r) = B(r) \;\Rightarrow\; m(r)A(r) = m(r)B(r)$

That makes the fourth statement true: solutions of the original always survive into the cleared equation. The first is false, because a root of the cleared equation may be excluded. The second is false for the reason just given. The third is false because an excluded value cannot be substituted at all: at least one side of the original does not exist there.

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 $2(x-3) + 3(x+3) = 12$.

$5x + 3 = 12 \;\Rightarrow\; x = \frac{9}{5}$

The candidate $\frac{9}{5}$ is in the domain, so it is genuine.

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

$(x-3)(x-2) = (x+1)(x+2)$

Expanding both sides, $x^2 - 5x + 6 = x^2 + 3x + 2$. The quadratic terms cancel, leaving $-5x + 6 = 3x + 2$, so $8x = 4$ and $x = \frac{1}{2}$.

The candidate is in the domain, so it is genuine. Checking: both sides equal $-1$ at $x = \frac{1}{2}$.

The excluded value is $x = 2$. The two sides share a denominator, so multiplying by $x - 2$ leaves the numerators:

$x = c$

The only candidate is $x = c$. It is a genuine solution whenever it lies in the domain, which is whenever $c \neq 2$. It is excluded exactly when $c = 2$, and then the single candidate is extraneous and nothing survives. So the equation has no solution precisely when $c = 2$.