RATL 2 | Lesson 4 | Practice (Solving Rational Equations) Solutions

  1. To solve simple rational equations, you can use cross multiplication. Use cross multiplication to solve the equations below.
  1. \(\dfrac{x-1}{5}=\dfrac{9}{7}\)

\(7(x-1) = 5•9\)

\(7x-7 = 45\)

\(7x = 52\)  or \(x = \dfrac{52}{7}\)

  1. \(\dfrac{x+2}{3}=\dfrac{4}{9}\)

\(9x+18 = 12\)

\(9x = -6\)

\(x = \dfrac{-2}{3}\)

  1. What about more complicated equations?

Try finding common denominators as a first step in solving the next equations. Once all of the terms have the same denominator, solve the equation of the numerators.

  1. \(\dfrac{3}{x+4}+\dfrac{1}{4}=5\)

LCD is the product of the two denominators or \(4x+16\) because the denominators have no common factors.

Multiply the equation by \(4x+16\). The resulting equation is

\(12+x+4 = 20x+80\) or \(19x =-64\)

\(x = -\dfrac{64}{9}\) or \(x = -3\dfrac{7}{19}\)

  1. \(\dfrac{x}{x-5}+\dfrac{3}{x+2}=\dfrac{7x}{x^2-3x-10}\)

In order to find the LCD we need to factor \(x^2-3x-10\), the factors are \(x-5\) and \(x+2\) therefore in looking at the three denominators we determine that the LCD is

\((x-5)(x+2)\).   Multiplying through by the LCD, the equation becomes:

\(x^2+2x+3x-15 = 7x\).  Solving for \(x\):  \(x^2-2x-15 = 0\)

Factoring: \((x+3)(x-5) = 0\) or \(x = -3\) or \(x = 5\).   But the restricted values are 5 and -2, so the only solution is \(x = -3\).  See next page as to a brief explanation as to why the the equation had restricted values.

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