Step-by-step explanation:
Answer:
\begin{gathered}\qquad \:\boxed{\begin{aligned}& \qquad \:\sf \: Price \: of \: notebook=Rs \: 45 \qquad \: \\ \\& \qquad \:\sf \:Price \: of \:instrument \: box =Rs \: 75 \end{aligned}} \qquad \\ \\ \end{gathered}
Priceofnotebook=Rs45
Priceofinstrumentbox=Rs75
Step-by-step explanation:
Let assume that
Price of one notebook be Rs x
Price of one instrument box be Rs y
According to statement, price of 1 notebook and one instrument box is Rs 120.
It means,
\begin{gathered}\sf\implies x + y = 120 \: \: \: - - - (1) \\ \\ \end{gathered}
⟹x+y=120−−−(1)
Further given that, the price of 3 notebooks and one instrument box is Rs 210.
\begin{gathered}\sf\implies 3x + y = 210 \: \: \: - - - (2) \\ \\ \end{gathered}
⟹3x+y=210−−−(2)
On Subtracting equation (1) from (2), we get
\begin{gathered}\sf \:2x = 90 \\ \\ \end{gathered}
2x=90
\begin{gathered}\sf\implies \bf \:x = 45 \\ \\ \end{gathered}
⟹x=45
On substituting the value of x in equation (1), we get
\begin{gathered}\sf \:45 + y = 120 \\ \\ \end{gathered}
45+y=120
\begin{gathered}\sf \:y = 120 - 45\\ \\ \end{gathered}
y=120−45
\begin{gathered}\sf\implies \bf \:y = 75\\ \\ \end{gathered}
⟹y=75
Hence,
\begin{gathered}\sf\implies \:\boxed{\begin{aligned}& \qquad \:\bf \: Price \: of \: notebook=Rs \: 45 \qquad \: \\ \\& \qquad \:\bf \:Price \: of \:instrument \: box =Rs \: 75 \end{aligned}} \qquad \\ \\ \end{gathered}
⟹
Priceofnotebook=Rs45
Priceofinstrumentbox=Rs75
\rule{190pt}{2pt}
Additional Information
\begin{gathered}\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}\end{gathered}
Moreidentities
Moreidentities
★(x+y)
2
=x
2
+2xy+y
2
★(x−y)
2
=x
2
−2xy+y
2
★x
2
−y
2
=(x+y)(x−y)
★(x+y)
2
−(x−y)
2
=4xy
★(x+y)
2
+(x−y)
2
=2(x
2
+y
2
)
★(x+y)
3
=x
3
+y
3
+3xy(x+y)
★(x−y)
3
=x
3
−y
3
−3xy(x−y)
★x
3
+y
3
=(x+y)(x
2
−xy+y
2
)