Question

those who are able to solve properly please solve this i will mark the right answer as brainliest

Answer 1

$\boxed{Hy\:Mate}$

$<marquee direction="left"></p><p><strong>✌ </strong>Follow Me ✌ </marquee>$

Answer 2

AnswEr :

$\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$

$\rule{170}1$

$\large\qquad{\boxed{\sf \green{P}}} \: \large\sf\ + \: \large{\boxed{\sf \red{Q}}} \: \large\sf\ = 13 \\ \: \:\qquad\ + \: \: \quad\quad\ + \\\large\qquad{\boxed{\sf \pink{R}}} \: \large\sf\ - \: \: \large{\boxed{\sf \blue{S}}} = 8$

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

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

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

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

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

$\rule{170}1$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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