Solving Rational Equations - Algebra 3

1. For what value of $k$ is $x=-3$ an extraneous solution of $\frac{x}{x+3}-\frac{5}{x-1}=\frac{k}{x^2+2x-3}$?
   Answer: $12$ Factor $x^2+2x-3=(x+3)(x-1)$, so $x=-3$ and $x=1$ are excluded. Multiplying by the LCD gives $x(x-1)-5(x+3)=k$. Substituting $x=-3$ yields $(-3)(-4)-5(0)=12$, so $k=12$. With $k=12$, the cleared equation has roots $9$ and $-3$; $-3$ is extraneous, leaving $9$ valid.
2. What is the solution set of $\frac{1}{x-1}+\frac{2}{x^2+x+1}=\frac{3x}{x^3-1}$?
   Answer: $\{-1\}$ Factor $x^3-1=(x-1)(x^2+x+1)$, and $x^2+x+1$ is never zero, so $x=1$ is the only excluded value. Multiplying by $x^3-1$ gives $x^2+x+1+2(x-1)=3x$, which simplifies to $x^2-1=0$. Thus the candidates are $x=\pm1$; since $x=1$ is excluded, the solution set is $\{-1\}$.
3. A student solves $\frac{x}{x-1}+\frac{2}{x+1}=\frac{7x-5}{x^2-1}$ as follows: Step 1: $x(x+1)+2(x-1)=7x-5$. Step 2: $x^2+3x-2=7x-5$. Step 3: $x^2-4x+3=0$. Step 4: $(x-1)(x-3)=0$, so $x=1$ or $x=3$. Step 5: Since both values satisfy the multiplied equation, the solutions are $x=1$ and $x=3$. At which step is the first error?
   Answer: Step 5 Whenever you solve a rational equation, multiplying by the LCD clears denominators, but it can also introduce extraneous solutions that make an original denominator zero. Here the LCD is $(x-1)(x+1)$, so the domain requires $x\neq 1$ and $x\neq -1$. Step 1 correctly multiplies each term by the LCD to get $x(x+1)+2(x-1)=7x-5$. Step 2 expands correctly to $x^2+3x-2=7x-5$, and Step 3 correctly rearranges to $x^2-4x+3=0$. Step 4 factors correctly as $(x-1)(x-3)=0$, giving $x=1$ or $x=3$. All these algebraic steps are valid. The first error is in Step 5, where the student says both values are solutions because they satisfy the multiplied equation. But $x=1$ makes the original denominators $x-1$ and $x^2-1$ equal to zero, so it is extraneous and must be rejected. The only valid solution is $x=3$. The choice "Step 1" is a trap if you think the multiplication is wrong; it is correct, but you still must check domain restrictions afterward. "Step 3" and "Step 4" are not errors because the rearrangement and factoring are accurate. Study tip: after solving a rational equation, always plug each candidate into the original denominators; any value that gives zero is extraneous, no matter how nicely it satisfies the cleared equation.
4. How many real numbers satisfy $\frac{x^2-1}{x-1}=\frac{x^2+x}{x}$?
   Answer: Infinitely many The domain excludes $x=1$ because of $x-1$, and $x=0$ because of the denominator $x$. For every other real value, $\frac{x^2-1}{x-1}=x+1$ and $\frac{x^2+x}{x}=x+1$. Thus the equation is an identity for all real numbers except $0$ and $1$, so infinitely many real numbers satisfy it.
5. Solve the rational equation: $\frac{x}{x+2}+\frac{2}{x-2}=\frac{x^2+4}{x^2-4}.$
   Answer: All real numbers except $-2$ and $2$. Whenever you solve a rational equation, your first move should always be to identify the domain: any value that makes a denominator zero cannot be a solution. Here the denominators are $x+2$, $x-2$, and $x^2-4=(x+2)(x-2)$, so $x=-2$ and $x=2$ are immediately excluded. Now combine the left side: $\frac{x}{x+2}+\frac{2}{x-2}=\frac{x(x-2)+2(x+2)}{x^2-4}=\frac{x^2-2x+2x+4}{x^2-4}=\frac{x^2+4}{x^2-4}.$ That is exactly the right side. So the equation is an identity for every allowable value of $x$. Since the only excluded values are $-2$ and $2$, the solution set is all real numbers except those two. Why are the other choices traps? "The only solution is $x=0$" is tempting because $0$ does satisfy the equation, but it is not the only solution — any allowable number works. "There is no solution" might come from noticing the excluded values, but that confuses "not defined" with "no solution." "The only solution is $x=2$" is doubly wrong: $x=2$ makes the denominator $x-2$ and $x^2-4$ zero, so it is not even in the domain. Study tip: before doing any algebra in a rational equation, write down the domain restrictions. Then simplify both sides. If they become identical, the solution is every domain value — and if your simplification creates an answer outside the domain, discard it as extraneous.
6. For what value of $b$ does $\frac{x-1}{x-3}-\frac{4}{x+3}=\frac{b}{x^2-9}$ have exactly one valid solution, $x=5$?
   Answer: $24$ The domain excludes $x=3$ and $x=-3$. Multiplying by $x^2-9$ gives $(x-1)(x+3)-4(x-3)=b$, which simplifies to $x^2-2x+9=b$. For $b=24$, the equation becomes $x^2-2x-15=0$, with roots $5$ and $-3$. Since $-3$ is excluded, the only valid solution is $5$.
7. For what value of $a$ will $x=1$ be an extraneous solution of $\frac{ax+1}{x-1}+\frac{2}{x+1}=\frac{6}{x^2-1}$?
   Answer: $2$ The domain excludes $x=1$ and $x=-1$. Multiplying by $x^2-1$ gives $(ax+1)(x+1)+2(x-1)=6$. Substituting $x=1$ yields $2(a+1)=6$, so $a=2$. With $a=2$, the cleared equation has roots $1$ and $-\frac72$, and $1$ is excluded, making it extraneous. The other values follow from sign or substitution errors.
8. Which equation has $x=2$ as an extraneous solution?
   Answer: $\frac{x}{x-2}+2=\frac{2}{x-2}$ Whenever you solve a rational equation, extraneous solutions are numbers that appear from the algebra but make a denominator zero in the original equation. Your first move should be to note which $x$-values are excluded. For $\frac{x}{x-2}+2=\frac{2}{x-2}$, the denominator $x-2$ excludes $x=2$. If you multiply both sides by $x-2$, you get $x+2(x-2)=2$, which simplifies to $3x=6$, so $x=2$. But $x=2$ was excluded, so it is extraneous. Now check the others. In $\frac{x^2-4}{x+2}=0$, factoring gives $\frac{(x-2)(x+2)}{x+2}=0$, and after canceling, $x=2$. Since $x+2\neq 0$ at $x=2$, it is a valid solution. Similarly, $\frac{x^2-3x+2}{x-1}=0$ factors to $\frac{(x-1)(x-2)}{x-1}=0$, giving $x=2$, and $x-1\neq0$ at $x=2$. And $\frac{x^2-5x+6}{x-3}=0$ factors to $\frac{(x-2)(x-3)}{x-3}=0$, again giving $x=2$, with $x-3\neq0$ at $x=2$. Each of these has a different excluded value, but none excludes $2$. Study tip: before solving a rational equation, write down excluded values. After solving, reject any candidate that is excluded. Extraneous solutions come from multiplying by a variable denominator, not from the factoring itself.
9. Solve the rational equation: $\frac{x^2-4}{x-2}=0.$
   Answer: $x=-2$ When solving a rational equation, always keep the denominator in mind: a value that makes the denominator zero is not allowed, even if it makes the numerator zero. Here you have $\frac{x^2-4}{x-2}=0.$ Factor the numerator as a difference of squares: $x^2-4=(x-2)(x+2).$ So the equation becomes $\frac{(x-2)(x+2)}{x-2}=0.$ You might be tempted to cancel $x-2$, but cancellation is only valid when $x\neq 2$, since plugging $x=2$ into the original equation gives a denominator of zero. Assuming $x\neq 2$, the expression simplifies to $x+2=0,$ so $x=-2.$ Since $-2$ does not make the original denominator zero, it is the valid solution. The choice "$x=2$ and $x=-2$" captures the trap of solving the numerator alone: $x=2$ makes the numerator zero, but it also makes the denominator zero, so the original expression is undefined there. The choice "$x=2$" alone has the same flaw. The choice "there is no solution" is incorrect because $-2$ satisfies the original equation: $\frac{(-2)^2-4}{-2-2}=\frac{0}{-4}=0.$ On rational equation problems, write down excluded values from the denominator before simplifying. Then solve the simplified equation and check that your answer is not one of those excluded values. That single habit prevents the most common errors on this exam.
10. What is the solution set of $\frac{x}{x-2}+\frac{3}{x+4}=\frac{6x}{x^2+2x-8}$?
    Answer: $\{-3\}$ When you solve a rational equation, the key is to remember that multiplying by a common denominator can introduce extraneous solutions — values that satisfy the transformed equation but make an original denominator zero. So first note the domain: $x\neq 2$ and $x\neq -4$, because either would make a denominator undefined. The denominator $x^2+2x-8$ factors as $(x-2)(x+4)$, so the least common denominator is $(x-2)(x+4)$. Multiply every term by this: $$x(x+4)+3(x-2)=6x$$ $$x^2+4x+3x-6=6x$$ $$x^2+7x-6=6x$$ $$x^2+x-6=0$$ $$(x+3)(x-2)=0$$ So the potential solutions are $x=-3$ and $x=2$. But $x=2$ is excluded from the domain because it makes $x-2$ zero, so it is extraneous. The only valid solution is $-3$, giving $\{-3\}$. The choice $\{-3,2\}$ includes the extraneous value $2$, so it fails the domain check. The choice $\{2\}$ is entirely extraneous — it comes from solving but violates the original equation. The choice $\varnothing$ misses that $-3$ is perfectly valid after substitution. A good habit: before solving any rational equation, write down the excluded values. Then after solving, always check your answers against that list. This simple step prevents the most common trap on this exam.
11. For what value of $b$ does $\frac{x-2}{x+2}+\frac{3}{x-3}=\frac{b}{(x+2)(x-3)}$ have exactly one valid solution, $x=4$?
    Answer: $20$ The domain excludes $x=-2$ and $x=3$. Multiplying by $(x+2)(x-3)$ gives $(x-2)(x-3)+3(x+2)=b$, which simplifies to $x^2-2x+12=b$. For $b=20$, the equation becomes $x^2-2x-8=0$, with roots $4$ and $-2$. Since $-2$ is excluded, the only valid solution is $4$.
12. A bus rental costs ($120). If $n$ students attend, the cost per student is $\frac{120}{n}$ dollars. If 4 additional students attend, the cost per student decreases by ($1). This situation is modeled by $\frac{120}{n}=\frac{120}{n+4}+1$. How many students are in the original group?
    Answer: $20$ Multiplying by $n(n+4)$ gives $120(n+4)=120n+n(n+4)$, so $120n+480=120n+n^2+4n$, or $n^2+4n-480=0$. Factoring gives $(n+24)(n-20)=0$, so $n=-24$ or $n=20$. Since a student count cannot be negative, the original group has $20$ students.
13. A rational equation has domain restrictions $x\neq 1$ and $x\neq -2$. After solving, a student obtains $x=1$, $x=-2$, and $x=3$. Which statement is true?
    Answer: The solution set is $\{3\}$. When you solve a rational equation, your first step is always to note the domain restrictions: any value that makes a denominator zero cannot be a solution. Here, the restrictions are $x\neq 1$ and $x\neq -2$. Once you solve and get candidate answers, you must filter them through those restrictions. The student found three candidates: $x=1$, $x=-2$, and $x=3$. But $x=1$ and $x=-2$ are exactly the values that are excluded from the domain, so they cannot be solutions. That leaves $x=3$, which is allowed, so the solution set is indeed $\{3\}$. Now look at the wrong choices. The set $\{1,-2\}$ contains only the two forbidden values, so it is not a valid solution set. The set $\{1,-2,3\}$ incorrectly includes the restricted values along with the valid one. And the claim that the equation has no solution is false because $x=3$ satisfies the equation and is permitted by the domain. The trap here is forgetting to check your solutions against the original domain restrictions — especially when the algebra itself produces those restricted values. Your study takeaway: after solving any rational equation, always substitute your answers back into the original denominators or compare them to the stated domain. A value that makes a denominator zero is never a solution, even if it appears in your factored work. That simple check is often the difference between a correct solution set and a careless mistake.
14. A boat travels 10 miles upstream and 10 miles downstream in a total of 5 hours. The current is 2 miles per hour. If $b$ is the boat's speed in still water, the situation is modeled by $\frac{10}{b-2}+\frac{10}{b+2}=5$. What is $b$?
    Answer: $2+2\sqrt{2}$ miles per hour Whenever you see a distance-rate-time word problem modeled by a rational equation, your two-step plan should be: solve the equation, then check every solution against the physical situation. Here the boat must make positive progress upstream, so $b>2$; any root less than 2 is invalid. Multiplying $\frac{10}{b-2}+\frac{10}{b+2}=5$ by $(b-2)(b+2)$ gives $10(b+2)+10(b-2)=5(b^2-4)$. Simplify to $20b=5b^2-20$, so $b^2-4b-4=0$. The quadratic formula yields $b=\frac{4\pm\sqrt{32}}{2}=2\pm2\sqrt{2}$. The positive value $2+2\sqrt2$ satisfies $b>2$ and is the correct speed. The other algebraic root, $2-2\sqrt2$, is a negative number, so it cannot be a boat speed; it also violates $b>2$. Choosing $4$ mph might come from assuming the current cancels completely, but substituting gives $10/2+10/6=6\frac23$ hours, not 5. Choosing $2+\sqrt2$ is an arithmetic slip: $\sqrt{32}/2$ is $2\sqrt2$, not $\sqrt2$, so this underestimates the correction to the still-water speed. On exam day, after solving a rational equation, always plug your answer back into the original word-problem conditions. Time must be positive, upstream speed must exceed the current, and any algebraic extraneous or negative root should be discarded.
15. What is the solution set of $\frac{2}{x+2}+\frac{x}{x^2-2x+4}=\frac{2x+4}{x^3+8}$?
    Answer: $\varnothing$ Factor $x^3+8=(x+2)(x^2-2x+4)$, and $x^2-2x+4$ is never zero, so $x=-2$ is excluded. Multiplying by $x^3+8$ gives $2(x^2-2x+4)+x(x+2)=2x+4$, which simplifies to $3x^2-4x+4=0$. Its discriminant is $(-4)^2-4(3)(4)=16-48=-32$, so there are no real solutions.
16. What is the solution set of $\frac{x+1}{x-2}-\frac{2x}{x+3}=\frac{5x-6}{x^2+x-6}$?
    Answer: $\left\{\frac{3-3\sqrt5}{2},\frac{3+3\sqrt5}{2}\right\}$ Factor $x^2+x-6=(x-2)(x+3)$, so $x\neq 2,-3$. Multiplying by the LCD gives $(x+1)(x+3)-2x(x-2)=5x-6$, which simplifies to $x^2-3x-9=0$. The quadratic formula gives $\frac{3\pm3\sqrt5}{2}$; neither value is $2$ or $-3$. The other forms come from sign errors, dropping the factor of 3 from $\sqrt{45}$, or incorrectly simplifying the discriminant.
17. Which equation has solution set $\{-1\}$ after extraneous solutions are rejected?
    Answer: $\frac{x}{x-2}-\frac{4}{x+3}=\frac{10}{(x-2)(x+3)}$ For the correct equation, multiplying by $(x-2)(x+3)$ gives $x(x+3)-4(x-2)=10$, so $x^2-x-2=0$, with roots $2$ and $-1$. The value $2$ makes $x-2$ zero, so it is rejected; the valid solution is $-1$. The other equations produce different roots, or produce $-3$ and $2$, both of which are excluded.
18. For what value of $a$ does $\frac{2x+3}{x-1}-\frac{x}{x+2}=\frac{a}{x^2+x-2}$ have no real solution?
    Answer: $-12$ Since $x^2+x-2=(x-1)(x+2)$, the domain excludes $x=1$ and $x=-2$. Multiplying by the LCD gives $(2x+3)(x+2)-x(x-1)=a$, or $x^2+8x+6=a$, so $x^2+8x+(6-a)=0$. Its discriminant is $64-4(6-a)=40+4a$. For no real solution, $40+4a<0$, so $a<-10$. Only $-12$ satisfies this among the choices.
19. What is the solution set of $\frac{x+1}{x-1}-\frac{2}{x+2}=\frac{6}{x^2+x-2}$?
    Answer: $\varnothing$ Whenever you solve a rational equation, start by finding domain restrictions: denominators cannot be zero. Here $x-1$ and $x+2$ appear, so $x\neq 1$ and $x\neq -2$. Since $x^2+x-2=(x-1)(x+2)$, the right side gives the same restrictions. Multiply through by the common denominator $(x-1)(x+2)$. The left side becomes $(x+1)(x+2)-2(x-1)=x^2+x+4$. Setting this equal to $6$ gives $x^2+x+4=6$, so $x^2+x-2=0$, which factors as $(x+2)(x-1)=0$. The algebra gives $x=-2$ or $x=1$. But both of these values are excluded from the domain, so neither can be a solution. Therefore the solution set is the empty set $\varnothing$. The choices $\{-2\}$ and $\{1\}$ each contain one excluded value, and $\{-2,1\}$ contains both, but all three ignore the fact that these values make original denominators zero. The empty set is the only valid answer. Study tip: on rational equations, write the excluded values before solving, then discard any solution that matches one of them. Checking the domain is not optional—it is exactly what separates the correct answer from these traps.
20. For which value of $k$ does $\frac{x}{x-3}+\frac{k}{x+3}=\frac{18}{x^2-9}$ have exactly one distinct algebraic solution, and that solution is extraneous?
    Answer: $-9$ When you solve a rational equation like this, the first move is always: identify the excluded values, then multiply by the LCD $(x-3)(x+3)$ to clear denominators. Here $x=3$ and $x=-3$ are forbidden. After multiplying, you get $$x^2+(k+3)x-3k-18=0.$$ For the equation to have exactly one distinct algebraic solution, this quadratic must have a double root. Its discriminant is $$(k+3)^2-4(-3k-18)=(k+9)^2,$$ so the only way to get one distinct solution is $k=-9$. Substituting $k=-9$ gives $$x^2-6x+9=(x-3)^2=0,$$ so $x=3$. But $x=3$ makes the original denominator $x-3$ zero, so the single solution is extraneous. Thus $k=-9$ satisfies both conditions. For $k=-3$, the cleared equation is $x^2-9=0$, giving $x=\pm3$, both extraneous — two algebraic solutions, not one. For $k=3$, you get $(x+9)(x-3)=0$, so $x=3$ is extraneous but $x=-9$ is valid. For $k=9$, you get $(x-3)(x+15)=0$, so again $x=3$ is extraneous but $x=-15$ is valid. Each of those choices leaves a valid solution, so they miss the "extraneous-only" requirement. When you see "exactly one distinct algebraic solution and it is extraneous," think: the cleared equation must have a repeated root, and that repeated root must be a forbidden value. Check the discriminant and the domain.

Start this quiz