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.
Return to Solving Rational Equations.