Rational Equations | Brilliant Math & Science Wiki

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

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

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. \( _\square \)