A mathematical operation of dividing an algebraic term by its unlike term is called the division of unlike algebraic terms.

The division of any two unlike algebraic terms is written by displaying a division sign between them for calculating their quotient. The quotient of the unlike terms is actually calculated by eliminating the equivalent literal factors in literal coefficients and also calculating the numerical coefficients of them. The quotient of any two unlike algebraic terms is always an algebraic term.

$6xy$ and $3x^2$ are two unlike algebraic terms, and divide the term $6xy$ by $3x^2$.

Write the division of unlike algebraic terms in mathematical form.

$6xy \div 3x^2$

$\implies$ $\dfrac{6xy}{3x^2}$

Factorize the each algebraic term for eliminating the equivalent factors.

$\implies$ $\dfrac{6xy}{3x^2} \,=\, \dfrac{6 \times x \times y}{3 \times x^2}$

$\implies \dfrac{6xy}{3x^2} \,=\, \dfrac{6}{3} \times \dfrac{x}{x^2} \times y$

Find the quotient of the unlike terms by cancelling equivalent literal factors by the quotient rules of exponents, and then finding the quotient of the numerical coefficients.

$\implies$ $\dfrac{6xy}{3x^2} \,=\, \require{cancel} \dfrac{\cancel{6}}{\cancel{3}} \times \dfrac{\cancel{x}}{\cancel{x^2}} \times y$

$\implies$ $\dfrac{6xy}{3x^2} \,=\, 2 \times \dfrac{1}{x} \times y$

$\therefore \,\,\,\,\,\,$ $\dfrac{6xy}{3x^2} \,=\, \dfrac{2y}{x} \,\,$ (or) $\,\, 2x^{-1}y$

In this way, the division of any two unlike algebraic terms can be calculated in algebraic mathematics in three simple steps. The quotient of any two unlike terms is an algebraic temr.

Look at the following examples to learn how to divide an algebraic term by its unlike term.

$(1) \,\,\,\,\,\,$ $\dfrac{-b}{2a}$ $\,=\,$ $-\dfrac{1}{2}a^{-1}b$

$(2) \,\,\,\,\,\,$ $\dfrac{5cd^2}{cd}$ $\,=\,$ $\require{cancel} \dfrac{5 \times \cancel{c} \times \cancel{d^2}}{\cancel{c} \times \cancel{d}}$ $\,=\, 5d$

$(3) \,\,\,\,\,\,$ $\dfrac{14e}{7f^2}$ $\,=\,$ $\require{cancel} \dfrac{\cancel{14}}{\cancel{7}} \times \dfrac{e}{f^2}$ $\,=\,$ $2ef^{-2}$

$(4) \,\,\,\,\,\,$ $\dfrac{0.5gh}{5g}$ $\,=\,$ $\require{cancel} \dfrac{\cancel{0.5}}{\cancel{5}} \times \dfrac{\cancel{g}}{\cancel{g}} \times h$ $\,=\,$ $0.1h$

$(5) \,\,\,\,\,\,$ $\dfrac{ij^4}{k}$ $\,=\,$ $ij^4k^{-1}$

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