Question

one number is 7 more than another and its square is 77 more than the square of the smaller number what are the number? ​

Answer 1

Answer:

$\huge{\boxed{\mathfrak{Answer=9\:and\:2}}}$

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Let's consider the bigger number as 'X' and the smaller number as 'Y' .

According to question now:-

$\tt X = Y + 7 \: \: \: \:(eq.1) \\\\ \tt {X}^{2} = {Y}^{2} + 77 \: \: \: \:(eq.2)$

Now putting the value of (eq.1) into the (eq.2) we get,

$\tt {(Y+7)} ^{2} = {Y}^{2} + 77 \\ \\ \leadsto\tt {Y}^{2} + 2 \times Y \times 7 +{7}^{2} = {Y}^{2} + 77 \\ \\ \leadsto\tt {Y}^{2} + 14Y + 49 = {Y}^{2} + 77 \\ \\ \leadsto\tt {Y}^{2} + 14Y - {Y}^{2} = 77 - 49 \\ \\ \leadsto\tt 14Y = 28 \\ \\ \leadsto\tt Y =\frac{28}{14} \\ \\ \leadsto\tt Y = 2$

$\therefore\bold {\underline{Smaller\:number(Y)=2}}$

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Putting the value of Y into the (eq.1) we get,

$\tt X = 2 + 7 \\ \\ \leadsto\tt X = 9$

$\therefore\bold {\underline{Bigger\:number(X)=9}}$

Verification:-

Putting the value of X and Y in both the equations,we get

$\tt 9 = 2 + 7 \\ \\ \leadsto\tt 9 = 9 \: \:[Verified]$

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$\tt {9}^{2} = {2}^{2} + 77 \\ \\ \leadsto\tt 81 = 4 + 77 \\ \\ \leadsto\tt 81 = 81 \: \: \:[Verified]$

$\therefore\bold{\underline{ Numbers\:are\:9 \: \:and \: \: 2}}$

Answer 2

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: SOLUTION \: \: } \mid}}}}}$

$\text{Suppose,} \\ \\ \: \: \: \: \: \: \: \: \: \: \text{The \: bigger \: number = x} \\ \\ \: \: \: \: \: \: \: \: \: \: \text{The \: smaller \: number = y}</p><p>$

$\underline{ \bold{ \: \: Given \: : \mapsto}} \\ \\ \text{ \: \: \: One \: number \: is \: 7 \: more \: than \: another.} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \bold{x - y = 7 \: \: \: - - - - - - - > (1)} \\ \\ \text{ \: \: \: \: The \: \: square \: \: of \: \: the \: \: bigger \: \: number \:} \\ \text{is \: \: 77 \: \: more \: \: than \: \: the \: \: square \: \: of \: \: the \: } \\ \text{ smaller \: \: number.} \\ \\ \bold{x {}^{2} - y {}^{2} = 77 \: \: \: - - - - - - - - >( 2) }</p><p></p><p>$

$\bold{(2) \Rightarrow \: x {}^{2} - y {}^{2} = 77} \\ \\ \bold{\Longrightarrow \: (x + y)(x - y) = 77} \\ \\ \bold{\Longrightarrow \: (x + y) \times 7 = 77 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: [From \: (1)]} \\ \\ \bold{\Longrightarrow \: x + y = \frac{77}{7} } \\ \\ \: \: \: \: \: \bold{ \therefore \: \: x + y = 11 \: \: \: - - - - - - - > (3)}$

$\underline{ \text{ \: Now, \: }} \\ \\ \: \: \: \: \: \: \bold{ (1) +(3)} \\ \\ \bold{ \Longrightarrow \: (x - y) +( x + y) = 7 + 11} \\ \\ \bold{\Longrightarrow \: x + x = 18} \\ \\ \bold{\Longrightarrow \: 2x = 18} \\ \\ \bold{\Longrightarrow \: x = \frac{18}{2} } \\ \\ \: \: \: \: \: \bold{ \therefore \: \: \underline {\: x = 9 \:} }$

$\bold{ \therefore \: \: (1) \Rightarrow \: x - y = 7} \\ \\ \bold{ \Longrightarrow \: 9 - y = 7} \\ \\ \bold{\Longrightarrow \: - y = 7 - 9} \\ \\ \bold{ \Longrightarrow \: - y = - 2} \\ \\ \bold{ \: \: \: \: \: \therefore \: \: \underline{ \: y = 2 \: }}$

$\mathsf{Hence,} \\ \\ \mathsf{The \: numbers \: \: are :- \: \: \underline {\: \: 9 \: \: and \: \: 2\: \: }}$

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: VERIFICATION \: \: } \mid}}}}}$

$\underline{\text{ \: Putting \: \: the \: \: value \: \: of \: \: \red{ x } \: \: and \: }}\\ \underline{\text{ \: \red{ y }\: \: in \: \: both \: \: the \: \: equations \: :- }}$

From eq. (1) , we get,

$\bold{L.H.S. = x - y }\\ \\ \bold{= 9 - 2 } \\ \\ \bold{= 7 = R.H.S.}$

From eq. (2), we get,

$\bold{L.H.S. = x {}^{2} - y {}^{2} } \\ \\ \bold{ = 9{}^{2} - 2{}^{2}} \\ \\ \bold{ = 81 - 4} \\ \\ \bold{= 77 = R.H.S.}$

Hence, Verified.