Solve rational equations

You can't divide by $0$, so a denominator can never equal $0$. Take

$x-5$ is $0$ when $x=5$, and $x+2$ is $0$ when $x=-2$. So the restrictions are

Write them down before you cancel or multiply anything. They belong to the original equation, and they still count after the denominators are gone.

Any value your algebra gives you is only a possible solution. It becomes a real one when it passes the restrictions and makes the original equation true. A value that works in a rewritten equation but not in the original is called an extraneous solution.

So every rational equation follows the same four steps. Restrict first, check last:

How do you pick for the second step? Picture what the equation turns into once the fractions are gone.

• Clear with a short LCD when the denominators are short, like $x+4$, and one multiplication leaves something quick to finish, like a linear equation or a quadratic that factors. In the SAT example, multiplying by $x+4$ left a linear equation, and two more lines finished it.
• Graph the original sides when the intersections are quicker than the algebra, like when clearing would leave a long quadratic that doesn't factor, or when each answer choice has its own decimal you can match, like $\frac{1+\sqrt{145}}{8}\approx1.63$. Type the left side and the right side into Desmos exactly as written, and click where they meet.

Sometimes the question asks only for the sum or the product of the solutions. Then you don't need the solutions at all. Clear the fractions by hand and read the answer from the coefficients: the solutions of $ax^2+bx+c=0$ add up to $-\frac{b}{a}$ and multiply to $\frac{c}{a}$. These shortcuts are called Vieta's relationships.

Find the least common denominator, or LCD: the simplest expression that has every denominator's factors in it. Then multiply both whole sides by it.

For

the restrictions are $x\ne0$ and $x\ne1$, and the LCD is $x(x-1)$. Multiply every term by it:

Each denominator cancels with its matching factor, so no fractions are left:

Watch the $5$ on the right. It's easy to multiply only the fractions and leave the $5$ alone. That gives $2(x-1)+3x=5$, so $x=\frac{7}{5}$, and $\frac{7}{5}$ doesn't make the original equation true. The $5$ is a term too, so it gets multiplied by $x(x-1)$ like everything else.

Some hard questions put a constant, like $k$, into the equation and ask when there's exactly one valid solution. Here's one, where $k$ is a positive constant:

Start the usual way. The restrictions are $x\ne2$ and $x\ne-3$. Factor $x^2-4$ as $(x-2)(x+2)$ and cancel $x-2$, which is fine since $x\ne2$. Then multiply both sides by $x+3$:

Expand and move $k$ over, and the cleared quadratic is

So how can this end with exactly one solution? The first idea is usually a repeated root, where the discriminant is $0$. Remember what the discriminant's sign tells you about a quadratic: positive means two different real roots, zero means one repeated root, and negative means no real roots. Here the discriminant is $5^2-4(1)(6-k)=1+4k$. Since $k>0$, it's positive, so the cleared equation always has two different real roots. A repeated root can't happen here.

That's the tricky part. With two roots, the only way to end up with one valid solution is to throw one of them out, so one root has to be a forbidden value. Could it be $x=-3$? Plugging $-3$ into $(x+2)(x+3)$ gives $0$, so that would need $k=0$, and $k$ is positive. So it has to be the other forbidden value, $x=2$. Plug $x=2$ into $(x+2)(x+3)=k$:

When $k=20$, the cleared equation factors as

Throw out $x=2$ and keep $x=-7$. The original equation has exactly one valid solution.

So there are two ways to get exactly one valid solution, and a question with $k$ can use either one:

1. The cleared equation has one repeated real root, and that root is allowed.
2. The cleared equation has more than one real root, and all but one of them are forbidden values. That's what happened here.