# Rational Equations

A **rational equation** is an equation containing at least one fraction whose numerator and denominator are polynomials, $\frac{P(x)}{Q(x)}.$ These fractions may be on one or both sides of the equation. A common way to solve these equations is to reduce the fractions to a common denominator and then solve the equality of the numerators. While doing this, we have to make sure to note cases where indeterminate forms like $\frac{0}{0}$ or $\frac{1}{0}$ may arise.

## Rational Equations - Basic

Solve $\frac{1}{x} = 2 .$

Looking at the equation, we can see that it's asking which reciprocal gives $2$. This is $\frac{1}{2}$ and we can conclude that it is the solution. $_\square$

While it is possible to use this inspection method, it is easier to use a more general method. In general, if an equation is in the form of an irreducible proportion $\frac{a}{b}=\frac{c}{d}$, one can cross multiply to obtain a polynomial $ad - bc = 0$. This polynomial can then be solved using whatever appropriate method necessary while noting that $b \neq 0$ and $d \neq 0$.

Solve

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

Using the cross-multiplying method described above gives

$\begin{aligned} 3 + x &= 2(2-x) \\ 3 + x &= 4 - 2x \\ 3x& = 1 \\ x &= \frac{1}{3}.\ _\square \end{aligned}$

This method can be extended to any rational equation. However, for expressions with more terms, instead of cross-multiplying we multiply both sides of the equation by the LCM of the denominators.

Find all the solutions of

$\frac{1}{x} + \frac{2}{1-x} = \frac{11}{x} + \frac{3}{x(2x+3)}.$

First note that $x \neq 0, x \neq 1,$ and $x \neq \frac{-3}{2},$ as they all lead to a zero denominator. When multiplying the whole expression by the LCM of the denominators $x(1-x)(2x+3),$ we get

$\begin{aligned} (1-x)(2x+3) + 2x(2x+3) &= 11(1-x)(2x+3) + 3(1-x)\\ -2x^2-x+3+4x^4+6x&=-22x^2-11x\\ 24x^2 + 19x - 33 &= 0. \end{aligned}$

Using the quadratic formula to solve this equation, we get

$x = \frac{-19 \pm \sqrt{3529}}{48}.\ _\square$

Solve the equation $\frac{1}{x-2}=\frac{1}{8}.$

Multiplying both sides by $8(x-2)$ gives

$\begin{aligned} 8 =& x - 2 \\ 10 =& x. \end{aligned}$

Substituting $x=10$ satisfies the given equation, so the answer is 10. $_\square$

Solve the equation $\frac{1}{2x+3}=\frac{1}{x-5}.$

Multiplying both sides by $(2x+3)(x-5)$ gives

$\begin{aligned} x-5 =& 2x + 3 \\ -8 =& x. \end{aligned}$

Substituting $x=-8$ satisfies the given equation, so the answer is -8. $_\square$

## Solve the equation $\frac{1}{(x+3)(x-2)}=\frac{1}{x-6}.$

Multiplying both sides by $(x+3)(x-2)(x-6)$ gives

$\begin{aligned} x-6 =& (x+3)(x-2) \\ x-6 =& x^2 +x -6 \\ 0 =& x^2. \end{aligned}$

Substituting $x=0$ satisfies the given equation, so the answer is 0. $_\square$

## Solve the equation $\frac{x^2 + 3x}{x + 2}=\frac{-2x -6}{x + 2}.$

Multiplying both sides by $x+2$ gives

$\begin{aligned} x^2 + 3x =& -2x -6 \\ x^2 +5x +6 =& 0 \\ (x+2)(x+3) =& 0 . \end{aligned}$

Observe that $x=-2$ is not a solution because the given equation has zeros in the denominator.

Substituting $x=-3$ satisfies the given equation, so the answer is -3. $_\square$

## Rational Equations - Intermediate

If the equation

$\frac{-4x^2 -4x + a}{2x + 1}=\frac{4x + 1}{2x + 1}$

has only one solution, what is $a?$

Multiplying both sides by $2x+1$ gives

$\begin{aligned} -4x^2 -4x + a =& 4x + 1 \\ -4x^2 - 8x + a - 1 =& 0. \end{aligned}$

The discriminant is

$\begin{aligned} \frac{D}4 =& (-4)^2 -(-4)(a-1) \\ =& 16 + 4a - 4 \\ =& 4a + 12 \\ =& 4(a+3). \end{aligned}$

If $a = -3,$

$\begin{aligned} -4x^2 - 8x + a - 1 =& -4x^2 - 8x - 4 \\ =& -4 (x^2 +2x +1) \\ =& -4(x+1)^2 \\&= 0.\\ \end{aligned}$

Substituting $x=-1$ satisfies the given equation.

If $a > -3,$ then $-4x^2 - 8x + a - 1 = 0$ has two solutions. However, if one solution is $-\frac{1}{2},$ the other one solution will be left because substituting $x = -\frac{1}{2}$ makes the denominators in the given equation zero.

Assume one solution is $x = -\frac{1}{2}.$ Then

$\begin{aligned} -4x^2 - 8x + a - 1 =& -1 + 4 + a - 1 \\ =& a +2 \\ =& 0. \end{aligned}$

If $a = -2,$ then we have only one solution.

Therefore, $a = -2 \text{ or } -3.\ _\square$

## Rational Equations - Advanced

If the equation

$\frac{x^2 + 6x }{x^2 + 7x + 10}=\frac{-2x + a}{x^2 + 7x + 10}$

has only one solution, what is $a?$

Multiplying both sides by $x^2 + 7x + 10$ gives

$\begin{aligned} x^2 + 6x =& -2x + a \\ x^2 + 8x - a =& 0. \end{aligned}$

The discriminant is

$\begin{aligned} \frac{D}4 = 4^2 + a = a + 16 . \end{aligned}$

If $a = -16,$

$\begin{aligned} x^2 + 8x - a = x^2 + 8x + 16 = (x+4)^2 . \end{aligned}$

Substituting $x=-4$ satisfies the given equation.

If $a > -16,$ $x^2 + 8x - a = 0$ has two solutions. However, if one solution is -2 or -5, the other one solution will be left because substituting $x = -2 \text{ or } -5$ makes the denominators in the given equation zero.

Assume one solution is $x = -2.$ Then

$\begin{aligned} x^2 + 8x - a =& 4 -16 - a \\ =& -12 - a \\ =& 0. \end{aligned}$

If $a = -12,$ then we have only one solution $x = -6.$

Assume one solution is $x = -5.$ Then

$\begin{aligned} x^2 + 8x - a =& 25 - 40 - a \\ =& -15 - a \\ =& 0. \end{aligned}$

If $a = -15,$ then we have only one solution $x = -3.$

Therefore, $a = -12, -15, -16.\ _\square$

**Cite as:**Rational Equations.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/rational-equations/