Maths / Exponents and Power / Statement Sums Involving Exponent and Power

QUESTION

By what number should $\dpi{120} \fn_jvn \large (-24)^{-1}$ be divided so that the quotient may be $\dpi{120} \fn_jvn \large 3^{-1}$ ?

EXPLANATION
Explain TypeExplanation Content
Text

Let the required number be x. Then ,

$\dpi{120} \fn_jvn \large (-24)^{-1} \div x=3^{-1}$

$\dpi{80} \frac{(-24)}{x}=3^{-1}$

$\dpi{80} \frac{\frac{1}{-24}}{x}=\frac{1}{3}$

$\dpi{80} \Rightarrow\frac{\frac{1}{-24}}{x}=\frac{1}{3}\Rightarrow 3=-24x\Rightarrow x= \frac{3}{-24}\Rightarrow x= \frac{3}{-24}=-\frac{1}{8}$

Aliter, We know that

Dividend = Quotient $\dpi{80} \times$ Divisor

or, Divisor = Divident $\dpi{80} \div$ Quotient

Here, Dividend = $\dpi{80} (-24)^{-1}$ and Quotient = $\dpi{80} 3^{-1}$

Therefore, Divisor = $\dpi{80} (-24)^{-1} \div 3^{-1}=\begin{pmatrix}\frac{1}{-24} \end{pmatrix} n\div \begin{pmatrix}\frac{1}{3}\end{pmatrix}=\frac{1}{-24}\times \frac{3}{1}=\frac{1}{-8}=(-8)^{-1}$

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