Breuken vermenigvuldigen
$
\eqalign{
& {4 \over {a - 1}} \cdot {{a + 2} \over 3} = {{4\left( {a + 2} \right)} \over {(a - 1) \cdot 3}} = {{4a + 8} \over {3a - 3}} \cr
& 3x \cdot {x \over {x - 1}} = {{3x^2 } \over {x - 1}} \cr
& {{a + 5} \over {a - 1}} \cdot 4 = {{\left( {a + 5} \right) \cdot 4} \over {a - 1}} = {{4a + 20} \over {a - 1}} \cr
& x\left( {x - 1} \right) \cdot {1 \over {x + 2}} = {{x\left( {x - 1} \right)} \over {x + 2}} = {{x^2 - x} \over {x + 2}} \cr
& \left( {2a - 3} \right) \cdot {{4a} \over {2a - 3}} = 4a \cr}
$
Breuken optellen
$
\eqalign{
& {5 \over {3x}} + {3 \over {2y}} = {{5 \cdot 2y} \over {6xy}} + {{3 \cdot 3x} \over {6xy}} = {{9x + 10y} \over {6xy}} \cr
& {5 \over {3x}} + {3 \over {2x}} = {{5 \cdot 2} \over {6x}} + {{3 \cdot 3} \over {6x}} = {{10 + 9} \over {6xy}} = {{19} \over {6xy}} \cr
& 5 + {3 \over {2x}} = {{5 \cdot 2x} \over {2x}} + {3 \over {2x}} = {{10x + 3} \over {2x}} \cr
& {5 \over {3x^2 y}} + {3 \over {2xy^2 }} = {{5 \cdot 2y} \over {6x^2 y^2 }} + {{3 \cdot 3x} \over {6x^2 y^2 }} = {{9x + 10y} \over {6x^2 y^2 }} \cr
& {{7xy} \over {4x}} + {{xy} \over {2x}} = {{7y} \over 4} + {y \over 2} = 1{3 \over 4}y + {1 \over 2}y = 2{1 \over 4}y \cr}
$