Solving Rational Equations: Additional Practice (Set 1)

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

Multiplying every term by the LCD gives $3(x+1) + 2(x-1) = 4$.

$5x + 1 = 4 \;\Rightarrow\; x = \frac{3}{5}$

The candidate $\frac{3}{5}$ is in the domain, so it is genuine.

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

Multiplying every term by the LCD gives $x^2 + 2(x+2) = 4$, and the constants cancel:

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

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

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 $5(x-1) - 4x = 1$.

$x - 5 = 1 \;\Rightarrow\; x = 6$

The candidate $6$ is in the domain, so it is genuine. Checking: $\frac{5}{6} - \frac{4}{5} = \frac{25 - 24}{30} = \frac{1}{30}$, which matches $\frac{1}{36-6}$.

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

Multiplying every term by the LCD gives $4 + 2(x-2) = 3(x+2)$.

$2x = 3x + 6 \;\Rightarrow\; x = -6$

The candidate $-6$ is in the domain, so it is genuine. Checking: $\frac{4}{32} + \frac{2}{-4} = \frac{1}{8} - \frac{1}{2} = -\frac{3}{8}$, which matches $\frac{3}{-8}$.

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

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

$x^2 - x = 0 \;\Rightarrow\; x(x-1) = 0 \;\Rightarrow\; x = 0 \text{ or } x = 1$

The candidate $1$ is excluded, so it is extraneous. The candidate $0$ is in the domain, so it is genuine: the left side is $0 - 2 = -2$ and the right side is $\frac{2}{-1} = -2$. The solution is $x = 0$, which is a reminder that $0$ is an ordinary value, excluded only when it makes a denominator vanish.

You need an equation whose cleared form produces the candidate $x = -1$ while $-1$ is excluded from the original.

$\frac{x}{x+1} = \frac{-1}{x+1} \;\Rightarrow\; x = -1, \text{ and } x = -1 \text{ is excluded}$

Now rule the others out. The equation $\frac{x}{x-1} = \frac{-1}{x-1}$ also clears to $x = -1$, but its excluded value is $1$, so there $-1$ is a genuine solution and both sides equal $\frac{1}{2}$. The equation $\frac{1}{x+1} = \frac{1}{2x}$ clears to $2x = x + 1$, giving the candidate $x = 1$. And $\frac{x^2}{x+2} = \frac{1}{x+2}$ clears to $x^2 = 1$, whose roots $1$ and $-1$ both lie in its domain, so neither of them is extraneous.

The excluded values are $2$ and $-2$, and the LCD is $(x-2)(x+2)$.

Multiplying every term by the LCD gives $(x+2) + (x-2) = 4$.

$2x = 4 \;\Rightarrow\; x = 2$

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

Call the number $x$. Its reciprocal exists only when $x \neq 0$, so $0$ is excluded, and the equation is $x + \frac{2}{x} = 3$. Multiply every term by $x$.

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

The candidates are $1$ and $2$. Neither is excluded and both are positive, so both survive every filter. Checking: $1 + 2 = 3$ and $2 + 1 = 3$.

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 $2(x-1) + 3x = 4$.

$5x - 2 = 4 \;\Rightarrow\; x = \frac{6}{5}$

The candidate $\frac{6}{5}$ is in the domain, so it is genuine.

Clearing the denominators is reversible at every value where the LCD is nonzero, so the original and the cleared equation agree exactly on the domain. That makes domain membership a complete test.

$\text{candidate in the domain} \iff \text{candidate solves the original}$

Every candidate satisfies the cleared equation by construction, so that check decides nothing. Being an integer is irrelevant. And making the LCD zero is precisely what marks a candidate as extraneous, so the last option has the logic backwards.

The excluded values are $1$ and $-1$. Multiplying both sides by the LCD $(x-1)(x+1)$ gives

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

Every term with an $x$ in it cancels, leaving $1 = -3$, which is false. The cleared equation has no roots, so there are no candidates and the equation has no solution.

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

Multiplying every term by the LCD gives $6 - 2(x+2) = x$.

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

The candidate $\frac{2}{3}$ is in the domain, so it is genuine. Checking: $x^2 + 2x = \frac{16}{9}$, so the left side is $\frac{27}{8} - 3 = \frac{3}{8}$, and the right side is $\frac{1}{8/3} = \frac{3}{8}$.