Solving Linear Equations: Free Response

Multiplying both sides by $-9$ is undone by dividing both sides by $-9$.

$-9 \cdot A = -9 \cdot B \quad\Longrightarrow\quad \dfrac{-9A}{-9} = \dfrac{-9B}{-9}$

Because $-9$ never changes and is never zero, that division is legal no matter which equation the multiplication started from, so every solution that survives the multiplication survives the division straight back. Nothing about the equation itself, its degree, or its variable ever affects this: the number multiplied by is fixed in advance and it is not zero, and that is the entire guarantee.

The denominator $x+8$ is zero at $x=-8$, so that value is excluded before any clearing.

Multiply both sides by $x+8$; the denominator cancels on each side.

$3x + 2 = -4$

Solve the resulting linear equation.

$3x = -6, \qquad x = -2.$

The one candidate is $x=-2$.

Multiplying both sides by $x+8$ is undone by dividing both sides by $x+8$, and that division is legal at every $x$ for which $x+8 \neq 0$, that is every $x \neq -8$.

$(x+8) \neq 0 \iff x \neq -8$

At every such $x$ the step is reversible, so it cannot add or lose a solution there: whatever satisfied the original equation still satisfies the cleared one, and whatever satisfies the cleared one still satisfies the original.

The candidate found in part B is $x=-2$, and $-2 \neq -8$, so it falls exactly among the values where the step is trustworthy. Nothing about the reasoning above depended on how the numbers $3$, $2$, and $-4$ were chosen, only on the fact that the multiplier $x+8$ is not zero there. So $x=-2$ is a genuine solution of the original equation, not a number that only appears to solve it because the denominators were cleared.

Distribute and combine like terms on the left.

$8(x-1) - 2x = 8x - 8 - 2x = 6x - 8$

The equation now reads $6x - 8 = 6x - 8$. Subtract $6x$ from both sides.

$-8 = -8$

That is true for every $x$, so the reduced form is $0x = 0$ and the equation is an identity: the solution set is $\mathbb{R}$.

Distribute and combine like terms on the left exactly as before.

$8(x-1) - 2x = 6x - 8$

The equation now reads $6x - 8 = 6x + 5$. Subtract $6x$ from both sides.

$-8 = 5$

That is false for every $x$, so the equation is a contradiction and the solution set is $\varnothing$.

Every linear equation reachable by distributing and combining like terms reduces to the form $ax=b$ for fixed numbers $a$ and $b$, exactly as parts A and B did.

If $a \neq 0$, dividing both sides by $a$ is legal and gives

$x = \dfrac{b}{a},$

a single number, so the solution set has exactly one member.

If $a = 0$, the equation reads $0 \cdot x = b$, which is really just the numeric statement $0 = b$ with no $x$ left in it at all. Either $b = 0$, in which case the statement is true for every real number and the solution set is $\mathbb{R}$, or $b \neq 0$, in which case the statement is false for every real number and the solution set is $\varnothing$.

Those are the only two possibilities once $a=0$, and together with the single-solution case they exhaust every value $a$ could take: $a$ is either zero or it is not, with nothing in between. A solution set of exactly one member, or of every real number, or of none, are the only outcomes this covers, and none of them has exactly two members.

$a = 0,\ b = 0 \ \Rightarrow\ \mathbb{R}, \qquad a = 0,\ b = 13 \ \Rightarrow\ \varnothing$

That is why part A, where the constants reduced to $a=0$ and $b=0$, gave every real number, while part B, where they reduced to $a=0$ and $b=13$, gave none. The left side never changed between the two parts, so the coefficient $a$ stayed fixed at $0$ in both; only the constant on the right did, and once $a=0$ is already fixed, that single changed constant is exactly what decides between the two branches still available, the identity and the contradiction outcomes, since the variable term has already canceled.

A fraction is undefined exactly where its denominator is zero, so set each denominator to zero on its own.

$x + 1 = 0 \ \Rightarrow\ x = -1, \qquad x - 3 = 0 \ \Rightarrow\ x = 3$

Both $x=-1$ and $x=3$ are excluded from the domain before any clearing takes place: at $x=-1$ the left side $\frac{4}{x+1}$ is undefined, and at $x=3$ the right side $\frac{6}{x-3}$ is undefined, so at each of these two values the equation as a whole has no comparison to make.

Multiply both sides by $(x+1)(x-3)$; each denominator cancels against its own factor.

$4(x-3) = 6(x+1)$

Distribute and collect the variable on one side.

$4x - 12 = 6x + 6, \qquad -18 = 2x, \qquad x = -9.$

The value $x=-9$ does not appear on the excluded list $\{-1, 3\}$ from part A, so nothing rules it out. Checking in the original equation, the left side is $\dfrac{4}{-9+1} = -\dfrac{1}{2}$ and the right side is $\dfrac{6}{-9-3} = -\dfrac{1}{2}$, which agree, so $x=-9$ is a genuine solution and the solution set is $\{-9\}$.

The excluded values found in part A, $x=-1$ and $x=3$, do not depend on which numbers sit in the numerators; they come only from where the two denominators $x+1$ and $x-3$ are zero, which never changes across equations sharing those denominators.

If clearing produced the single candidate $x=-1$, that candidate would be exactly one of the excluded values. A candidate on the excluded list is never a solution of the original equation, however cleanly it emerged from the cleared one, because the original is undefined there. This hypothetical stipulates that clearing produces exactly that one candidate (a cleared linear equation need not always land on a single candidate; the previous question's pair showed a cleared equation can instead be an identity or a contradiction), so here, once the single candidate $x=-1$ is discarded, there is nothing left to replace it with.

$x = -1 \in \{-1, 3\} \ \Rightarrow\ \text{discarded}$

The solution set of that equation would therefore be $\varnothing$, for the same reason the opening example of this lesson had none: the one number the algebra points to is precisely the number the domain forbids.

Add $5$ to both sides, then divide by $2$.

$2x = 6, \qquad x = 3.$

The solution set is $\{3\}$.

Adding $\dfrac{1}{x-3}$ to both sides of $2x-5=1$ forms the new equation

$2x - 5 + \dfrac{1}{x-3} = 1 + \dfrac{1}{x-3}.$

Away from $x=3$, adding the same expression to both sides changes neither side's truth: whatever the two original sides were worth there, adding the identical quantity to each still leaves them equal exactly when they were equal before. So for any $x \neq 3$, the new equation holds precisely when $2x-5=1$ held, which by part A is only at $x=3$, a value already excluded from this case.

At $x=3$ itself, the added term $\dfrac{1}{x-3}$ is undefined, so the new equation gives no value to compare there at all.

$\{x \neq 3 : 2x-5=1\} = \varnothing, \qquad x=3 \text{ is undefined in the new equation}$

No number satisfies the new equation, so its solution set is $\varnothing$.

Nothing about the argument in part B depended on the particular numbers $2$, $5$, or $1$; it depended only on the fact that the term added, $\dfrac{1}{x-3}$, is undefined exactly where the original equation's one solution sat.

Replace $3$ by an arbitrary $k$, and suppose $x=k$ is the only solution of some original equation. Add an expression undefined exactly at $x=k$ to both sides. Away from $k$, adding the same quantity to both sides changes neither side's truth, so the new equation holds at any $x \neq k$ exactly when the original did, which is nowhere, since $k$ was the only solution. At $x=k$ itself, the added expression is undefined, so the new equation gives no comparison to make there either.

$\{x : \text{new equation holds}\} = \varnothing$

So this is not a special feature of $2x-5=1$ or of the number $3$: it happens for any equation with a single solution once an expression undefined exactly there is added to both sides.

Substitute $x=-6$ into the original equation and evaluate the denominator on each side.

$x + 6 = -6 + 6 = 0$

Both sides carry the same denominator, which is $0$ at $x=-6$, so both $\dfrac{2(-6)+7}{-6+6}$ and $\dfrac{-5}{-6+6}$ are undefined. There is no value on either side to compare, so $x=-6$ cannot be checked as true or false in the original equation at all.

Nothing is wrong with solving $2x+7=-5$; that gives $x=-6$ correctly, and no arithmetic anywhere in the student's work is mistaken.

The unreliable step is the one before that: multiplying both sides of $\dfrac{2x+7}{x+6} = \dfrac{-5}{x+6}$ by $x+6$. That step is undone by dividing back by $x+6$, and division by $x+6$ is only legal where $x+6 \neq 0$, that is $x \neq -6$.

$x + 6 = 0 \iff x = -6$

At $x=-6$ the multiplier is zero, so the step is not reversible there, and it is exactly at that point that a false candidate is free to appear. What the student skipped was listing the excluded value $x=-6$ before clearing the denominator, and then checking the candidate against that list once it appeared.

The cleared equation $2x+7=-5$ gives the single candidate $x=-6$.

$2x + 7 = -5 \quad\Longrightarrow\quad x = -6$

That is exactly the value excluded in parts A and B, where the shared denominator $x+6$ vanishes, so the candidate must be discarded. No other candidate exists to replace it, so the solution set of the original equation is $\varnothing$: the equation genuinely has no solution, not merely an unchecked one.

The check that settles this in general needs nothing beyond what parts A and B already used. Whenever a step clears a denominator by multiplying by an expression that can be zero, list the values making that expression zero before solving, then compare every candidate the cleared equation produces against that list. A candidate that appears on the list is extraneous and must be discarded however cleanly the arithmetic that produced it. A candidate that does not appear on the list still has to be confirmed in the equation as originally written, since the check that matters is always the original, never a line written after the denominators were cleared.