Solving Rational Equations: Practice

A value is excluded when it makes a denominator zero. The only denominator here is $x - 4$.

$x - 4 = 0 \;\Rightarrow\; x = 4$

So $4$ is excluded and every other real number is allowed. The excluded value is the root of the denominator, not its opposite.

The denominators are $x$, $2x$, and $4$. The LCD is the smallest expression each of them divides into.

$4x \div x = 4, \qquad 4x \div 2x = 2, \qquad 4x \div 4 = x$

Each division comes out clean, so the LCD is $4x$. Multiplying the denominators together would give $8x^2$, which works but is not least.

The excluded value is $x = 5$. Both sides share the denominator $x - 5$, so multiplying by it leaves the numerators.

$x = 5$

The only candidate is exactly the value the original equation excludes, so it is extraneous and must be rejected. Nothing survives, and the equation has no solution. The clearing step did not blunder: at $x = 5$ it multiplied both sides by zero, and $0 = 0$ is true of anything.

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

$4(x-1) = 2(x+1) \;\Rightarrow\; 4x - 4 = 2x + 2 \;\Rightarrow\; x = 3$

The candidate $3$ is in the domain, so it is genuine. Checking: $\frac{4}{4} = 1$ and $\frac{2}{2} = 1$.

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

$3(x+4) = 5x \;\Rightarrow\; 3x + 12 = 5x \;\Rightarrow\; x = 6$

The candidate $6$ is in the domain, so it is genuine. Checking: $\frac{3}{6} = \frac{1}{2}$ and $\frac{5}{10} = \frac{1}{2}$.

The excluded value is $x = 0$. Multiplying every term by the LCD $x$ gives $2 + x = 5$.

$x = 3$

The candidate $3$ is in the domain, so it is genuine. Checking: $\frac{2}{3} + 1 = \frac{5}{3}$.

The excluded value is $x = 3$. The two fractions have the same denominator, so multiplying both sides by $x - 3$ leaves the numerators.

$x + 2 = 5 \;\Rightarrow\; x = 3$

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

Factor every denominator first.

$x^2 - 4 = (x-2)(x+2)$

The three denominators are then $x-2$, $x+2$, and $(x-2)(x+2)$. The last one is already a multiple of the other two, so the LCD is $(x-2)(x+2)$. Multiplying the unfactored denominators together would repeat factors needlessly.

Look at what sits underneath each fraction bar. In $\frac{x}{4}$ the denominator is the constant $4$, which is never zero, and the two polynomial equations have no denominators at all.

$\frac{4}{x} \text{ is undefined when } x = 0$

Only $\frac{4}{x} + 1 = 3$ has the variable underneath, so only that equation excludes a value, namely $x = 0$.

First find the excluded values from the original denominators.

$x - 4 = 0 \;\Rightarrow\; x = 4, \qquad x + 2 = 0 \;\Rightarrow\; x = -2$

The candidate $x = 4$ is excluded, so it is extraneous no matter how correctly it was derived. The candidate $x = -1$ is in the domain, and a candidate in the domain is always a genuine solution, so $x = -1$ is the only solution.

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

$3x = x - 2 \;\Rightarrow\; 2x = -2 \;\Rightarrow\; x = -1$

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

The excluded value is $x = 4$. Both sides share the denominator $x - 4$, so multiplying by it leaves the numerators.

$x^2 = 16 \;\Rightarrow\; x = 4 \text{ or } x = -4$

The candidate $4$ is excluded, so it is extraneous. The candidate $-4$ is in the domain, so it is genuine: both sides equal $\frac{16}{-8} = -2$ there. The solution is $x = -4$ alone.