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

1) To solve simple rational equations, you can use cross multiplication. Use cross multiplication to solve the equations below.

a) \(\Large \frac{x-1}{5}=\frac{9}{7}\)

\(7(x-1) = 5•9\)

\(7x-7 = 45\)

\(7x = 52\)  or \(x = \frac{52}{7}\)

b) \(\Large \frac{x+2}{3}=\frac{4}{9}\)

\(9x + 18 = 12\)

\(9x = -6\)

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

2) 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.

a) \(\Large \frac{3}{x+4}+\frac{1}{4}=\normalsize 5\)

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 \(19x = -64\)

\(x = -64/9\) 0r \(x = -3   7/19\)

b) \(\Large \frac{x}{x-5}+\frac{3}{x+2}=\frac{7x}{x^2-3x-10}\)

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.

Making Connections

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