Math Doubts

Division of Unlike Algebraic Terms

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

Introduction

The unlike algebraic terms have different literal factors. So, the quotient of division of two unlike algebraic terms is also an algebraic term.

The quotient rules of exponents are used when the unlike algebraic terms contain one or more literals commonly.

Example

$6xy$ and $3x^2$ are two unlike algebraic terms.

01

Divide an algebraic term by another

Take, the algebraic term $6xy$ is divided by the algebraic term $3x^2$. The division of them is written in mathematical form as follows.

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

The literal factor of the term $6xy$ is $xy$ and the literal factor of $3x^2$ is $x^2$. The literal factors of both terms are different but $x$ is a common factor in both terms.

02

Get Quotient of them

It can be written in product form to understand the division easily.

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

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

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

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

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

The example has proved that the division of two unlike algebraic terms is an algebraic term.

More Examples

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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