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.

#### Contents

- Rational Equations - Basic
- Rational Equations - Intermediate
- Rational Equations - Advanced

## 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

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

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

See Also7.6: Solve Rational EquationsWhat Is An Extraneous Solution? (3 Key Concepts To Know) | jdmeducationalA Rational Equation, With and Without Extraneous Roots – The Math Doctors7.6: Solve Rational Equations\[\begin{align}(1-x)(2x+3) + 2x(2x+3) &= 11(1-x)(2x+3) + 3(1-x)\\-2x^2-x+3+4x^4+6x&=-22x^2-11x\\24x^2 + 19x - 33 &= 0.\end{align}\]

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

\[\begin{align}8 =& x - 2 \\10 =& x.\end{align}\]

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

\[\begin{align}x-5 =& 2x + 3 \\-8 =& x.\end{align}\]

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

\[\begin{align}x-6 =& (x+3)(x-2) \\x-6 =& x^2 +x -6 \\0 =& x^2.\end{align}\]

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

\[\begin{align}x^2 + 3x =& -2x -6 \\x^2 +5x +6 =& 0 \\(x+2)(x+3) =& 0 .\end{align}\]

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

\[\begin{align}-4x^2 -4x + a =& 4x + 1 \\-4x^2 - 8x + a - 1 =& 0.\end{align}\]

The discriminant is

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

If \(a = -3,\)

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

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

\[\begin{align}-4x^2 - 8x + a - 1 =& -1 + 4 + a - 1 \\=& a +2 \\=& 0.\end{align}\]

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

\[\begin{align}x^2 + 6x =& -2x + a \\x^2 + 8x - a =& 0.\end{align}\]

The discriminant is

\[\begin{align}\frac{D}4 = 4^2 + a = a + 16 .\end{align}\]

If \(a = -16,\)

\[\begin{align}x^2 + 8x - a = x^2 + 8x + 16

= (x+4)^2 .\end{align}\]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

\[\begin{align}x^2 + 8x - a =& 4 -16 - a \\=& -12 - a \\=& 0.\end{align}\]

If \(a = -12,\) then we have only one solution \(x = -6.\)

Assume one solution is \(x = -5.\) Then

\[\begin{align}x^2 + 8x - a =& 25 - 40 - a \\=& -15 - a \\=& 0.\end{align}\]

If \(a = -15,\) then we have only one solution \(x = -3.\)

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