Math, asked by Mrtoppwr, 9 months ago

box + box + =8 down both sides + and + down box _ box =6 down =13=8​

Answers

Answered by Anonymous
11

ᴀɴsᴡᴇʀ :

\setlength{\unitlength}{0.2 cm}\begin{picture}(12,6)\thicklines\put(6,6){\circle{5}}\put(14,6){\circle{5}}\put(6,-2){\circle{5}}\put(14,-2){\circle{5}}\put(9.3,5.7){$+$}\put(9.3,-2.3){$-$}\put(5.2,1.6){$+$}\put(13.2,1.6){$+$}\put(17,5.5){$=\:8$}\put(17,-2.5){$=\:6$}\put(5.3,-6.5){$||$}\put(13.4,-6.5){$||$}\put(4.9,-9){$13$}\put(13.4,-9){$8$}\end{picture}

\normalsize\quad\bullet\:\sf\ Let \: the \: number \: in \: box_{1} \: be \: \bf\ P \\ \normalsize\quad\bullet\:\sf\ Let \: the \: number \: in \: box_{2} \: be \: \bf\ Q \\ \normalsize\quad\bullet\:\sf\ Let \: the \: number \: in \: box_{3} \: be \: \bf\ R \\ \normalsize\quad\bullet\:\sf\ Let \: the \: number \: in \: box_{4} \: be \: \bf\ S

  \rule{100}1

\normalsize\sf\ Thus, \: the \: equations \: formed:

\normalsize\twoheadrightarrow\sf\ P + Q = 8 \: ---(eq.1)

\normalsize\twoheadrightarrow\sf\ R - S  = 6 \: ---(eq.2)

\normalsize\twoheadrightarrow\sf\ P + R = 13  \: ---(eq.3)

\normalsize\twoheadrightarrow\sf\ Q + S = 8 \: ---(eq.4)

 \rule{170}2

\boxed{\begin{minipage}{5.3 cm}\underline{\qquad\quad\sf Basic\:Concept\qquad\quad}\\\quad\swarrow\qquad\qquad\ \\\sf Equation(3 -1)\\\swarrow\qquad\qquad\searrow\\\sf Equation(2 + 4)\quad Equation(5 + 6)\end{minipage}}

\underline{\maltese\:\textbf{According \: to \: the \: question \: now:}}

\underline{\dag\:\textsf{Subtracting \: equation \: (3) \: from \: (1) }}

\normalsize\ : \implies\sf\ Equation_{3} \: - \: Equation_{1}

\normalsize\ : \implies\sf\cancel{P} + R - \cancel{P} - Q = 13 - 8

\normalsize\ : \implies\sf\ R - Q = 5 \: ---(eq.5)

\underline{\dag\:\textsf{Adding \: equation \: (2) \: and \: (4) }}

\normalsize\ : \implies\sf\ Equation_{2} \:  + \: Equation_{4}

\normalsize\ : \implies\sf\ R - \cancel{S} + Q + \cancel{S} = 6 + 8

\normalsize\ : \implies\sf\ R + Q = 14 \: ---(eq.6)

\underline{\dag\:\textsf{Adding \: equation \: (5) \: and \: (6) }}

\normalsize\ : \implies\sf\ Equation_{5} \: + \: Equation_{6}

\normalsize\ : \implies\sf\ R - \cancel{Q} + R + \cancel{Q} = 14 + 9

\normalsize\ : \implies\sf\ 2R = 19

\normalsize\ : \implies\sf\ R = \frac{\cancel{19}}{\cancel{2}} = 9.5

\normalsize\ : \implies{\underline{\boxed{\sf \pink{R = 9.5}}}}

\underline{\dag\:\textsf{Block \: the \: values \: in \: available \: data }}

\scriptsize\sf{\quad\star\ Putting \: the \: value \: of \: R \: in \: eq.3}

\normalsize\ : \implies\sf\ P + R = 13

\normalsize\ : \implies\sf\ P + 9.5 = 13

\normalsize\ : \implies\sf\ P =  13 - 9.5

\normalsize\ : \implies\sf\ P = 3.5

\normalsize\ : \implies{\underline{\boxed{\sf \green{P  = 3.5 }}}}

\scriptsize\sf{\quad\star\ Putting \: the \: value \: of \: P \: in \: eq.1}

\normalsize\ : \implies\sf\ P + Q = 8

\normalsize\ : \implies\sf\ 3.5 + Q = 8

\normalsize\ : \implies\sf\ Q = 8 - 3.5

\normalsize\ : \implies\sf\ Q = 4.5

\normalsize\ :  \implies{\underline{\boxed{\sf \red{Q  = 4.5 }}}}

\scriptsize\sf{\quad\star\ Putting \: the \: value \: of \: R \: in \: eq.2}

\normalsize\ : \implies\sf\  R - S = 6

\normalsize\ : \implies\sf\ 9.5 - S = 6

\normalsize\ : \implies\sf\ -S = 6 - 9.5

\normalsize\ : \implies\sf\ -S = -3.5

\scriptsize\sf{\quad\star\ Cancel \: negative \: signs \: both \: side}

\normalsize\ : \implies\sf\ S = 3.5

\normalsize\ : \implies{\underline{\boxed{\sf \blue{S  = 3.5 }}}}

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