Solving Rational Equations: Free Response

One denominator is already a single linear factor, and it announces at sight that $x=6$ is excluded. The other announces nothing until it is factored:

$x^2-4x-12=(x-6)(x+2)$

Set each distinct factor to zero. From $x-6=0$ comes $x=6$ again, and from $x+2=0$ comes $x=-2$. That second value is excluded by an equation on which no printed denominator displays it, which is exactly why the factoring comes first.

Those two values are the excluded values of the original equation, and they are fixed by it. Nothing done later can change them, and after the next step nothing on the page will remember them.

The LCD is the product of the distinct factors, each to the highest power it reaches in any one denominator:

$(x-6)(x+2)$

The printed linear denominator divides it, and the printed quadratic one is it.

Multiply all three terms by $(x-6)(x+2)$, the lone $1$ included. The fraction on the left keeps $x+2$, the $1$ becomes the whole product, and the right side loses its denominator entirely:

$x(x+2)+(x-6)(x+2)=48$

Expand and collect:

$2x^2-2x-60=0$

Divide by $2$ and factor:

$x^2-x-30=(x-6)(x+5)=0$

So the candidates are $x=6$ and $x=-5$.

Now take each candidate, one at a time, to the list written down in part A. The value $6$ is on that list, so the original equation says nothing at all there and the candidate is extraneous; it is thrown out however correctly it was derived. The value $-5$ is not on the list, so by the lesson's theorem it is a genuine solution, and no substitution is needed to certify it.

Substituting is still a cheap way to catch a slip in the algebra. The left side at $x=-5$ is

$\frac{-5}{-11}+1=\frac{5}{11}+\frac{11}{11}=\frac{16}{11}$

and the right side is $\frac{48}{25+20-12}=\frac{48}{33}=\frac{16}{11}$.

So the solution set is $\{-5\}$.

Take the two halves separately.

Why a rejection can happen. At an excluded value the LCD is zero, and there the multiplication cannot be reversed: dividing by zero is not available, so nothing carries a verdict from the cleared equation back to the original. Cancelling then does something else at the same time. The expression $\frac{x}{x-6}\cdot(x-6)$ and the polynomial $x$ agree everywhere except at $x=6$, where the first is undefined and the second is $6$, so cancelling has extended each side to inputs the original equation never had.

The cleared equation is therefore defined at the excluded values, and being defined there, it is free to be true there. This one is. At $x=6$ it reads

$6(8)+0(8)=48$

that is $48=48$, a true statement about polynomials and one the original equation cannot even make, since its left side names no number at $x=6$.

Why no substitution is owed. Let $r$ be a candidate that is not excluded. Then the LCD is some number $m(r)$ that is not zero, and the cleared equation reads

$m(r)A(r)=m(r)B(r)$

where $A$ and $B$ are the two sides of the original. Dividing both sides by the nonzero number $m(r)$ gives $A(r)=B(r)$, so $r$ solves the original as well. Membership in the domain is therefore the complete test, and a substitution adds nothing to it except a guard against arithmetic slips.

Factor the quadratic denominator first:

$x^2-6x-7=(x-7)(x+1)$

So $x=7$ and $x=-1$ are excluded, and the LCD is $(x-7)(x+1)$.

Multiply every term by the LCD and cancel:

$5(x+1)+3(x-7)=40$

Expand and collect. Every numerator here is a constant, so nothing quadratic survives the cancelling and a linear equation is left:

$8x-16=40$

so $8x=56$ and $x=7$. That is the only candidate the clearing produced, and it is on the excluded list, so it is extraneous. Nothing else was produced to take its place.

The equation has no solution, and that is the answer rather than a sign of a mistake.

The only denominator is $2x-1$, which is zero at $x=\frac{1}{2}$, so that is the excluded value and the LCD is $2x-1$.

Multiply every term by it, the lone $3$ included:

$6x=3(2x-1)+5$

The right side is $6x-3+5$, that is $6x+2$, so the equation reads $6x=6x+2$. Subtract $6x$ from both sides:

$0=2$

The variable has cancelled entirely and what is left standing is false. No value of $x$ makes it true, so the cleared equation has no roots, and the clearing produced no candidate at all. There is nothing to test against the excluded value.

The equation has no solution.

A claim of the form "no solution forces an excluded candidate" is settled by producing one equation with no solution whose clearing hands over nothing.

The second equation is such a case. Clearing it gives

$0=2$

and a statement with no variable in it has no roots to offer. So the candidate list is empty. There is no candidate to compare with $x=\frac{1}{2}$, and no extraneous root anywhere in the work, yet the equation still has no solution. The claim is therefore false.

It is worth saying where the student's instinct is sound. The first equation does behave as the claim describes: its one candidate is the excluded value, and rejecting it empties the solution set. So both routes to an empty answer are real, and only one of them involves an extraneous root. What the two share is the verdict, not the reason.

The distinction matters because it says what to look at next. If the variable survives the clearing, its roots are candidates and each one must be tested against the excluded values. If the variable cancels, read the statement left standing, and here the two cases part company. A false statement leaves no candidates at all, so there is nothing to test. A true statement such as $3x=3x$ makes every value a candidate, and then the excluded values must still be struck out, which leaves the whole domain rather than the whole number line.

Let $n$ be the number of rows in the hall as it stands. The chairs are shared equally, so each row holds $\frac{240}{n}$ chairs.

With $4$ more rows there are $n+4$ of them, and each would hold $\frac{240}{n+4}$ chairs. That is $10$ fewer than before, so the two row sizes differ by $10$:

$\frac{240}{n}-\frac{240}{n+4}=10$

Any arrangement of that same comparison will do, since each rearranges into the others. What matters is that the larger row size belongs to the smaller number of rows.

The denominators are $n$ and $n+4$, so the equation excludes $n=0$ and $n=-4$. Note that these are excluded from the equation, which is a statement about arithmetic, not about the hall; the hall will have its own opinion later.

Multiply every term by the LCD $n(n+4)$:

$240(n+4)-240n=10n(n+4)$

The left side collapses to $960$, and the right side expands, so after dividing through by $10$:

$n^2+4n-96=0$

That factors as $(n+12)(n-8)=0$, giving the candidates $n=8$ and $n=-12$.

Neither candidate is an excluded value, so by the lesson's theorem both are genuine solutions of the equation. Confirm the second one to see that this is not a technicality: at $n=-12$ the left side is $\frac{240}{-12}-\frac{240}{-8}=-20+30=10$.

Only one of them can count rows in a hall. A row count cannot be negative, so $n=-12$ is set aside for that reason and $n=8$ describes the hall: $8$ rows of $\frac{240}{8}=30$ chairs. The check the caretaker cares about is that $12$ rows would hold $\frac{240}{12}=20$ chairs each, which is $10$ fewer.

An excluded value can only cause trouble by arriving among the candidates, and the candidates are the roots of the cleared equation. So test the excluded values there. At $n=0$,

$0^2+4(0)-96=-96$

and at $n=-4$,

$(-4)^2+4(-4)-96=-96$

Neither gives zero, so neither excluded value is a root of the cleared equation and neither could ever have appeared on the candidate list. In this model the domain check was owed, was performed, and found nothing, which is a result rather than a formality: it is what licenses accepting both candidates as genuine solutions of the equation.

The two rejections are different acts. The domain rejects a value the equation cannot evaluate: at such a value one side of the original names no number, so the value could not solve it whatever the arithmetic said. That is the only way a root of the LCD-cleared equation can be extraneous, and it applies to every rational equation cleared that way, modelled or not.

The context rejects a value the equation evaluates perfectly well. Here $n=-12$ satisfies the original equation exactly, and nothing mathematical is wrong with it; what rules it out is that a hall cannot have a negative number of rows. Call that root impossible for the hall, not extraneous, and name which filter you are using when you drop it. Saying "reject the negative root" is a statement about halls, and it would be equally true if the equation had no denominators at all.

The check. It claims that squaring each candidate returns $49$, and it does. But the candidates were produced by solving $x^2=49$, so they satisfy that equation by construction: the test was passed before it was set. A test that no candidate can fail distinguishes nothing, and in particular it cannot tell an extraneous root from a genuine one, since both satisfy the cleared equation equally.

The rejection. It claims that because the two candidates arise from one factorisation, a verdict on one carries to the other. Nothing supports that. Being excluded is a property of a single number held against a single list, the list of values that make an original denominator zero. The list here is $x=-6$ and $x=7$, written correctly at the top of the page.

Hold each candidate against it separately. The candidate $7$ is on the list, so it is extraneous. The candidate $-7$ is not on the list, and the original denominators say why:

$-7+6=-1, \qquad -7-7=-14$

Neither is zero, so the original equation is perfectly well defined at $-7$. That candidate is thrown out for no reason at all, and with it the equation's only solution.

The two errors pull in opposite directions and are worth naming that way. The check is too weak to reject anything, and the rule that follows it rejects too much.

The clearing was correct, so the candidates stand: $x=7$ and $x=-7$. The candidate $7$ is an excluded value and is rejected. The candidate $-7$ is not, so it is a genuine solution.

A check capable of failing is a substitution into the original equation, since a wrong candidate would produce two different numbers there. At $x=-7$ the left side is

$\frac{-7}{-1}+\frac{7}{-14}=7-\frac{1}{2}=\frac{13}{2}$

and the right side is

$\frac{91}{49+7-42}=\frac{91}{14}=\frac{13}{2}$

The two sides agree, so the algebra as well as the verdict is sound, and the solution set is $\{-7\}$.

The test is short because the LCD makes it short: a root of the LCD-cleared equation is a genuine solution of the original exactly when it lies in the original domain, that is, exactly when it is not one of the excluded values. Say which multiplier the sentence is about, because the guarantee is the LCD's and not every multiplier's.

Why it holds was settled once and does not need proving again here: on the domain the LCD is a nonzero number, so the clearing reverses and the two equations agree, and off the domain the original has no value to agree with. Take that as given.

What is worth noticing is the input the test takes: one number, and the list of excluded values. It does not take the equation the candidate came from, nor its companions, nor how that equation was factored. So a verdict cannot transfer. The two candidates here arrive from the single equation

$x^2=49$

yet one of them makes an original denominator zero and the other does not. Two numbers can share every step of their derivation and still stand in completely different relations to the original domain, because the domain was fixed before any of those steps were taken.

The practical form of this is a habit: run the test once per candidate, and write the comparison down each time.

The denominators are $x-1$ and $x+3$, so the excluded values are $x=1$ and $x=-3$, and the LCD is $(x-1)(x+3)$.

Multiply every term by it, the lone $1$ on the right included:

$5(x+3)-2(x-1)=(x-1)(x+3)$

The left side is $3x+17$ and the right side is $x^2+2x-3$, so collecting gives

$x^2-x-20=0$

which factors as $(x-5)(x+4)=0$. The candidates are $x=5$ and $x=-4$, and neither is on the excluded list, so both are genuine solutions.

Confirming: at $x=5$ the left side of $E$ is $\frac{5}{4}-\frac{2}{8}=1$, and at $x=-4$ it is $\frac{5}{-5}-\frac{2}{-1}=-1+2=1$.

Every term is multiplied by the whole expression, so the extra factor $x+2$ appears in all three products and can be taken outside:

$(x+2)\bigl[5(x+3)-2(x-1)\bigr]=(x+2)(x-1)(x+3)$

Move everything to one side and keep $x+2$ outside:

$(x+2)\bigl[3x+17-(x^2+2x-3)\bigr]=0$

The bracket is the clearing from part A with its sign turned round, since $3x+17-(x^2+2x-3)=-(x^2-x-20)$. So the polynomial equation is

$-(x+2)(x-5)(x+4)=0$

and an overall factor of $-1$ changes no root.

Its roots are $x=-2$, $x=5$ and $x=-4$. The two familiar ones are still there, and one more has appeared: the zero of the factor that was added.

Take the roots one at a time.

The roots $x=5$ and $x=-4$ were shown in part A to satisfy $E$, and neither is excluded.

The root $x=-2$ is the new one. Put it into $E$:

$\frac{5}{-3}-\frac{2}{1}=-\frac{5}{3}-2=-\frac{11}{3}$

which is not $1$, so $-2$ does not satisfy $E$. Now the second test. The denominators of $E$ at $x=-2$ are $-3$ and $1$, neither of them zero, so $E$ is perfectly well defined at $-2$ and its value there is an ordinary number. The excluded values of $E$ are $1$ and $-3$, and $-2$ is neither.

So here is a candidate that passes the domain check and is still not a solution, which is precisely what the rule on trial forbids. The rule is false as stated for this multiplier, and it fails in the direction that matters: it would have certified $-2$ without ever looking at $E$.

What went wrong is visible in the multiplier. At $x=-2$ the factor $x+2$ is zero, so both sides became $0$ and agreed for a reason having nothing to do with $E$. The domain has no opinion about $-2$, because no denominator of $E$ vanishes there, so nothing in the domain check was ever going to catch it.

The property. The LCD is built out of the factors of the denominators, so the values where it is zero and the values excluded from the equation are the same set. That coincidence is the engine of the whole method: it is what turns "check your candidates" into a single question with a yes or no answer.

The half that survives. Suppose $r$ genuinely solves the original equation, and suppose the multiplier $m$ is defined at $r$. Both sides are equal numbers at $r$, and multiplying two equal numbers by $m(r)$ keeps them equal:

$m(r)A(r)=m(r)B(r)$

The argument never asks which multiplier $m$ is, so no multiplier that is defined at the solutions can lose one. The proviso is not idle: multiplying $x=0$ through by $\frac{1}{x}$ loses its only solution, because the multiplier does not exist there.

The half that fails, and in which direction. Going the other way means dividing by $m(r)$, which is legitimate only where $m(r)$ is not zero. So a root of the cleared equation that does not solve the original has to be a zero of the multiplier. Read that in the direction it is proved: it bounds where a gained root can appear, and it promises nothing about a zero of the multiplier becoming a root. Part A is the proof of the difference. The LCD there is zero at $1$ and at $-3$, and neither is a root of $x^2-x-20=0$.

What is owed. For the LCD the zeros of the multiplier are exactly the excluded values, so the excluded list already covers every place a gained root could sit, and reading a candidate against it settles the matter. A multiplier with a real zero of its own puts a possible root where that list says nothing, and there the domain check is no longer complete: each candidate has to go back into the original equation and both sides be compared. Note the word real. Multiplying through by $x^2+1$ adds no real zero at all, so it adds no candidate and costs nothing.