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Fractions already have a common denominator. Addition and subtraction 50:50 split. Students shouldn't have to worry about reducing fractions by design.

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Express the following as a single fraction. Use / as the fraction bar and use brackets to group the numerator and denominator separately, e.g. $\\dfrac{m+1}{2n}$ is written (m+1)/(2n)

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Since both denominators are the same, i.e. $\\simplify{{ynden[0]*numbers[2]}x+{ynden[1]*numbers[3]}}$. We can simply addsubtract the numerators, and put that over the common denominator.

\n

\n

$\\begin{align*}\\dfrac{\\simplify{{ynnum1[0]*numbers[0]}-{ynnum1[1]*numbers[1]}y}}{\\simplify{{ynden[0]*numbers[2]}x+{ynden[1]*numbers[3]}}}+\\dfrac{\\simplify{{numbers[3]}x-{numbers[0]}y}}{\\simplify{{ynden[0]*numbers[2]}x+{ynden[1]*numbers[3]}}}&=\\dfrac{\\simplify{{ynnum1[0]*numbers[0]}+{addSub*numbers[3]}x+{-addSub*numbers[0]-ynnum1[1]*numbers[1]}y}}{\\simplify{{ynden[0]*numbers[2]}x+{ynden[1]*numbers[3]}}}\\end{align*}$

\n

$\\begin{align*}\\dfrac{\\simplify{{ynnum1[0]*numbers[0]}-{ynnum1[1]*numbers[1]}y}}{\\simplify{{ynden[0]*numbers[2]}x+{ynden[1]*numbers[3]}}}-\\dfrac{\\simplify{{numbers[3]}x-{numbers[0]}y}}{\\simplify{{ynden[0]*numbers[2]}x+{ynden[1]*numbers[3]}}}&=\\dfrac{\\simplify{ {ynnum1[0]*numbers[0]}-{ynnum1[1]*numbers[1]}y} -(\\simplify{{numbers[3]}x-{numbers[0]}y})}{\\simplify{{ynden[0]*numbers[2]}x+{ynden[1]*numbers[3]}}}\\\\[3pt]&=\\dfrac{\\simplify{{ynnum1[0]*numbers[0]}+{addSub*numbers[3]}x+{-addSub*numbers[0]-ynnum1[1]*numbers[1]}y}}{\\simplify{{ynden[0]*numbers[2]}x+{ynden[1]*numbers[3]}}}\\end{align*}$

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adding or subtracting

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yes/no for denominator. Used to turn terms on or off

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shuffle([0,1,1])

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$\\dfrac{\\simplify{{ynnum1[0]*numbers[0]}-{ynnum1[1]*numbers[1]}y}}{\\simplify{{ynden[0]*numbers[2]}x+{ynden[1]*numbers[3]}}}+\\dfrac{\\simplify{{numbers[3]}x-{numbers[0]}y}}{\\simplify{{ynden[0]*numbers[2]}x+{ynden[1]*numbers[3]}}}$ $\\dfrac{\\simplify{{ynnum1[0]*numbers[0]}-{ynnum1[1]*numbers[1]}y}}{\\simplify{{ynden[0]*numbers[2]}x+{ynden[1]*numbers[3]}}}-\\dfrac{\\simplify{{numbers[3]}x-{numbers[0]}y}}{\\simplify{{ynden[0]*numbers[2]}x+{ynden[1]*numbers[3]}}}$  $=$ [[0]]

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