Solving Rational Equations: Additional Practice (Set 2)

An extraneous root is a candidate that the clearing step produces and the domain then rejects, so you need an equation whose cleared form gives $x = 3$ while $3$ is itself excluded.

$\frac{x}{x-3} = \frac{3}{x-3} \;\Rightarrow\; x = 3, \text{ and } x = 3 \text{ is excluded}$

Now rule the others out. The equation $\frac{x}{x-3} = \frac{2}{x-3}$ clears to $x = 2$, which is in its domain and therefore genuine. The equation $\frac{x}{x+3} = \frac{3}{x+3}$ does clear to $x = 3$, but its excluded value is $-3$, so there $x = 3$ is a genuine solution. And $\frac{3}{x} = \frac{1}{x-3}$ clears to $3(x-3) = x$, giving $x = \frac{9}{2}$, so it never produces $3$ as a candidate at all.

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

$(x+2)(x-2) = x^2 \;\Rightarrow\; x^2 - 4 = x^2 \;\Rightarrow\; -4 = 0$

The variable cancels completely and leaves a false statement, so the cleared equation has no roots. There are no candidates to check, so the equation has no solution.

Factor the quadratic denominator: $x^2 + 4x + 3 = (x+1)(x+3)$. The excluded values are $-1$ and $-3$, and the LCD is $(x+1)(x+3)$.

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

$2x + 11 = 3x + 3 \;\Rightarrow\; x = 8$

The candidate $8$ is in the domain, so it is genuine. Checking: $\frac{5}{99} + \frac{2}{9} = \frac{5}{99} + \frac{22}{99} = \frac{27}{99} = \frac{3}{11}$, which matches the right side.

Let $t$ be the hours the slower crew needs alone, so the faster crew needs $t - 6$, and their rates add:

$\frac{1}{t} + \frac{1}{t-6} = \frac{1}{4}$

The excluded values are $0$ and $6$. Multiplying every term by the LCD $4t(t-6)$ gives $4(t-6) + 4t = t(t-6)$, so $t^2 - 14t + 24 = 0$ and $(t-12)(t-2) = 0$.

Both candidates are in the domain and satisfy the equation, but $t = 2$ would make the faster crew take $2 - 6 = -4$ hours, which is impossible, so the story rejects it. The slower crew takes $12$ hours, the faster takes $6$, and $\frac{1}{12} + \frac{1}{6} = \frac{1}{4}$ confirms it.

Factor first: $x^2 - 4x = x(x-4)$, so the excluded values are $0$ and $4$, and the LCD is $x(x-4)$.

Multiplying every term by the LCD gives $x^2 - 4(x-4) = 16$, and the constants cancel:

$x^2 - 4x + 16 = 16 \;\Rightarrow\; x^2 - 4x = 0 \;\Rightarrow\; x(x-4) = 0$

The candidates are $0$ and $4$, and both are excluded. Every candidate is extraneous, so the equation has no solution.

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

Multiplying every term by the LCD gives $3(x-2) + (x+2) = 8$.

$4x - 4 = 8 \;\Rightarrow\; x = 3$

The candidate $3$ is in the domain, so it is genuine. Checking: $\frac{3}{5} + \frac{1}{1} = \frac{8}{5}$, which matches $\frac{8}{9-4}$.

Let $t$ be the hours tap B needs alone, so tap A needs $t + 5$, and the rates add:

$\frac{1}{t} + \frac{1}{t+5} = \frac{1}{6}$

The excluded values are $0$ and $-5$. Multiplying every term by the LCD $6t(t+5)$ gives $6(t+5) + 6t = t(t+5)$, so $t^2 - 7t - 30 = 0$ and $(t-10)(t+3) = 0$.

Both candidates are in the domain and satisfy the equation, but a tap cannot take $-3$ hours, so the story rejects that one. Tap B takes $10$ hours and tap A takes $15$, and $\frac{1}{10} + \frac{1}{15} = \frac{1}{6}$ confirms it.

Call the smaller integer $n$, so the larger is $n + 1$, and the excluded values are $0$ and $-1$:

$\frac{1}{n} + \frac{1}{n+1} = \frac{7}{12}$

Multiplying every term by the LCD $12n(n+1)$ gives $12(n+1) + 12n = 7n(n+1)$, so $7n^2 - 17n - 12 = 0$ and $(7n+4)(n-3) = 0$.

The candidates are $n = 3$ and $n = -\frac{4}{7}$. Both lie in the domain, but only $3$ is a positive integer, so the story rejects the other. Checking: $\frac{1}{3} + \frac{1}{4} = \frac{7}{12}$.

Let $v$ be the speed in km/h. The actual time is $\frac{150}{v}$ and the faster time is $\frac{150}{v+10}$, and the first exceeds the second by half an hour:

$\frac{150}{v} - \frac{150}{v+10} = \frac{1}{2}$

The excluded values are $0$ and $-10$. Multiplying every term by the LCD $2v(v+10)$ gives $300(v+10) - 300v = v(v+10)$, so $v^2 + 10v - 3000 = 0$ and $(v-50)(v+60) = 0$.

Both candidates satisfy the equation, but a speed of $-60$ km/h cannot describe this car, so the story rejects it. The speed was $50$ km/h: the trip took $3$ hours, and at $60$ km/h it would have taken $2.5$ hours.

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

Multiplying every term by the LCD gives $2(x+2) + x(x-2) = x^2 + 4$, and expanding the left side,

$x^2 + 4 = x^2 + 4$

The variable cancels into a statement true for every value, so every value in the domain solves the original equation. The solution set is every real number except the two excluded values, which stay excluded because the original equation says nothing at all there.

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 $x^2 + 2(x-2) = 4$. Collecting terms,

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

The candidate $2$ is excluded, so it is extraneous. The candidate $-4$ is in the domain, so it is genuine: the left side is $\frac{2}{3} - \frac{1}{2} = \frac{1}{6}$, and the right side is $\frac{4}{24} = \frac{1}{6}$.

The excluded value is $x = 4$. Multiplying every term by $x - 4$, including the $1$, gives $a + (x-4) = x$, and the variable cancels:

$a - 4 = 0$

If $a \neq 4$ this is a false statement, so the cleared equation has no roots and the original has no solution. If $a = 4$ it is true for every $x$, so every value in the domain solves the original. Only $a = 4$ gives infinitely many solutions.