Solving Equations with Algebraic Fractions

Solving Algebraic Fractions

Solving algebraic fractions

How do I solve an equation that contains algebraic fractions?

• There are two methods for solving equations that contain algebraic fractions

• One method is to add or subtract the algebraic fractions first and then solve as usual

  • For example, to solve $\frac{8}{x+1}-\frac{5}{x+2}=1$

  • First subtract the fractions and simplify, $\frac{3x+11}{\left(x+1\right)\left(x+2\right)}=1$

  • Then cross-multiply, expand and solve

    $\begin{array}{rcl}3x+11& =& 1\left(x+1\right)\left(x+2\right)\\ 3x+11& =& x^2+3x+2\\ 0& =& x^2-9\\ 0& =& \left(x-3\right)\left(x+3\right)\\ x& =& 3 \mathrm{or} x=-3\end{array}$

• Alternatively, you can remove the fractions first by multiplying everything on both sides of the equation by each expression in the denominators and then solve

  • For example, to solve the equation $\frac{4}{x-3}+\frac{5}{x+1}=5$

  • First multiply every term in the equation by both $\left(x-3\right)$ and $\left(x+1\right)$ and cancel common factors where possible

    • Multiply every term by $\left(x-3\right)$ (this bracket goes in the numerator of any fractions)
      $\begin{array}{rcl}\frac{4}{\cancel{\left(x-3\right)}}\cancel{\left(x-3\right)}+\frac{5\left(x-3\right) }{x+1}& =& 5\left(x-3\right)\\ 4+\frac{5\left(x-3\right) }{x+1}& =& 5\left(x-3\right)\end{array}$

    • Then multiply every term by $\begin{array}{rcl}& & \left(x+1\right)\end{array}$

      $\begin{array}{rcl}4\left(x+1\right)+\frac{5\left(x-3\right)}{\left(\cancel{x+1}\right)}\cancel{\left(x+1\right)}& =& 5\left(x-3\right)\left(x+1\right)\\ 4\left(x+1\right)+5\left(x-3\right)& =& 5\left(x-3\right)\left(x+1\right)\end{array}$

  • Then solve

    $\begin{array}{rcl}4x+4+5x-15& =& 5\left(x^2-2x-3\right)\\ 9x-11& =& 5x^2-10x-15\\ 0& =& 5x^2-19x-4\\ 0& =& \left(5x+1\right)\left(x-4\right)\\ x& =& -\frac{1}{5} \mathrm{or} x=4\end{array}$