Math, asked by celinbiju15, 1 year ago

subtract 210 by 54 ... steps needed​

Answers

Answered by AbhijithPrakash
3

Answer:

54-210=-156

Step-by-step explanation:

54-210

\gray{\mathrm{To\:subtract\:a\:larger\:number\:from\:a\:smaller\:one\:switch\:the}} \gray{\mathrm{order\:of\:the\:numbers.\:\:Negate\:the\:result\:by\:attaching\:a\:minus\:sign}}

210-54

\gray{\mathrm{Line\:up\:the\:numbers}}

\begin{matrix}\space\space&2&1&0\\ -&0&5&4\end{matrix}

\gray{\mathrm{Subtract\:each\:column\:of\:digits,\:starting\:from\:the\:right\:and\:working\:left}}

\gray{\mathrm{If\:the\:digit\:being\:subtracted\:is\:larger\:than\:the\:digit\:above\:it,\:'borrow'\:a\:digit\:from\:the\:next}} \gray{\mathrm{\:column\:to\:the\:left.}}

\gray{\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}}

\begin{matrix}\space\space&2&1&\textbf{0}\\ -&0&5&\textbf{4}\end{matrix}

\gray{\mathrm{The\:bottom\:number\:is\:larger\:than\:the\:upper\:number.\:\:Try\:to\:'borrow'\:a\:digit\:from\:the\:left.}}

\gray{\mathrm{The\:top\:digit\:is\:not\:bigger\:than\:the\:bottom\:one.\:\:Try\:to\:'borrow'\:a\:digit\:from\:the\:left.}}

\begin{matrix}\space\space&2&\textbf{1}&0\\ -&0&\textbf{5}&4\end{matrix}

\gray{\mathrm{Borrow\:}1\mathrm{\:from\:}2\mathrm{.\:\:The\:remainder\:is\:}1}

\begin{matrix}\space\space&\textbf{1}&10&\space\space\\ \space\space&\textbf{x}&1&0\\ -&\textbf{0}&5&4\end{matrix}

\gray{\mathrm{Add\:}1\mathrm{\:ten\:to\:}1:\quad \:10+1=11}

\begin{matrix}\space\space&1&\textbf{11}&\space\space\\ \space\space&x&\textbf{x}&0\\ -&0&\textbf{5}&4\end{matrix}

\gray{\mathrm{Borrow\:}1\mathrm{\:from\:}11\mathrm{.\:\:The\:remainder\:is\:}10}

\begin{matrix}\space\space&1&\textbf{10}&10\\ \space\space&x&\textbf{x}&0\\ -&0&\textbf{5}&4\end{matrix}

\gray{\mathrm{Add\:}1\mathrm{\:ten\:to\:}0:\quad \:10+0=10}

\begin{matrix}\space\space&1&10&\textbf{10}\\ \space\space&x&x&\textbf{x}\\ -&0&5&\textbf{4}\end{matrix}

\gray{\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:10-4=6}

\frac{\begin{matrix}\space\space&1&10&\textbf{10}\\ \space\space&x&x&\textbf{x}\\ -&0&5&\textbf{4}\end{matrix}}{\begin{matrix}\space\space&\space\space&\space\space&\textbf{6}\end{matrix}}

\gray{\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:10-5=5}

\frac{\begin{matrix}\space\space&1&\textbf{10}&10\\ \space\space&x&\textbf{x}&x\\ -&0&\textbf{5}&4\end{matrix}}{\begin{matrix}\space\space&\space\space&\textbf{5}&6\end{matrix}}

\gray{\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:1-0=1}

\frac{\begin{matrix}\space\space&\textbf{1}&10&10\\ \space\space&\textbf{x}&x&x\\ -&\textbf{0}&5&4\end{matrix}}{\begin{matrix}\space\space&\textbf{1}&5&6\end{matrix}}

=156

\gray{\mathrm{Place\:a\:minus\:sign\:in\:front\:of\:the\:answer}}

=-156

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