- To solve simple rational equations, you can use cross multiplication. Use cross multiplication to solve the equations below.
- \(\dfrac{x-1}{5}=\dfrac{9}{7}\)
\(7(x-1) = 5•9\)
\(7x-7 = 45\)
\(7x = 52\) or \(x = \dfrac{52}{7}\)
- \(\dfrac{x+2}{3}=\dfrac{4}{9}\)
\(9x+18 = 12\)
\(9x = -6\)
\(x = \dfrac{-2}{3}\)
- 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.
- \(\dfrac{3}{x+4}+\dfrac{1}{4}=5\)
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 \(x = -3\dfrac{7}{19}\)
- \(\dfrac{x}{x-5}+\dfrac{3}{x+2}=\dfrac{7x}{x^2-3x-10}\)
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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