Question

If 9x² + y² = 13 and xy = 2, find the value of
(i) 3x + y, (ii) 3x - y.​

Answer 1

$\large\bf\underline{Given:-}$

• 9x² + y² = 13
• xy = 2

$\large\bf\underline {To \: find:-}$

• Value of 3x + y
• and 3x - y

$\huge\bf\underline{Solution:-}$

We know that,

$\leadsto \tt \: (a + b {)}^{2} = a {}^{2} + {b}^{2} + 2ab \\ \tt \leadsto \: (a + b) = \sqrt{ {a}^{2} + {b}^{2} + 2ab }$

$\underbrace { \rm \: (i) \: value \: of \: \: 3x + y : - }$

• 9x² + y² = 13
• xy = 2

$\tt \rightarrowtail \: (3x + y) {}^{2} = (3 {x})^{2} + y {}^{2} + 2 \times 3x \times y \\ \\ \tt \rightarrowtail \: (3x + {y)}^{2} = 9x {}^{2} + {y}^{2} + 6 \times xy \\ \\ \tt \rightarrowtail \:(3x + y) {}^{2} = 13 + 6 \times 2 \\ \\ \tt \rightarrowtail \: (3x + y {)}^{2} = 13 + 12 \\ \\ \tt \rightarrowtail \: 3x + y = \sqrt{25} \\ \\ \rightarrowtail \boxed{ \bf \: 3x + y = 5}$

$\underbrace { \rm \: (ii) \: value \: of \: \: 3x - y : - }$

we know that,

$\leadsto \tt \: (a - b {)}^{2} = a {}^{2} + {b}^{2} - 2ab \\ \tt \leadsto \: (a -b) = \sqrt{ {a}^{2} + {b}^{2} -2ab }$

$\tt \rightarrowtail \: (3x - y) {}^{2} = (3 {x})^{2} + y {}^{2} - 2 \times 3x \times y \\ \\ \tt \rightarrowtail \: (3x - {y)}^{2} = 9x {}^{2} + {y}^{2} - 6 \times xy \\ \\ \tt \rightarrowtail \:(3x - y) {}^{2} = 13 - 6 \times 2 \\ \\ \tt \rightarrowtail \: (3x - y {)}^{2} = 13 - 12 \\ \\ \tt \rightarrowtail \: 3x - y = \sqrt{1} \\ \\ \rightarrowtail \boxed{ \bf \: 3x - y = 1}$

Hence ,

• »★ Value of 3x + y = 5
• »★ value of 3x - y = 1

Answer 2

$\sf\large{\underline{\purple{\underline{\red{Question:-}}}}}$

If 9x² + y² = 13 and xy = 2, find the value of

(i) 3x + y, (ii) 3x - y.

$\sf\large{\underline{\purple{\underline{\red{To\:find:-}}}}}$

• 3x + y=?
• 3x - y.=?

$\sf\large{\underline{\purple{\underline{\red{Given:-}}}}}$

• 9x² + y² = 13 and xy = 2,

$\sf\large{\underline{\purple{\underline{\red{Solution:-}}}}}$

$\sf→{\purple{\underline{\red{Using\: (a+b)^2}}}}\\\sf→ (a+b)^2 = a^2+b^2+2ab$

$\sf→ (3x+y)^2= 3x^2+y^2+2×3x×y\\\sf→ (3x+y)^2=13+6×xy\\\sf→ 3x+y=13+6×2\\\sf→(3x+y)^2=13+12\\\sf→ (3x+y)^2=25 \\\sf→ 3x+y=\sqrt25\\\sf→ {\fbox{\red{\underline{\purple{3x+y=5}}}}}$

$\rule{210}2$

• ¡¡ nd part

$\sf\large{\underline{\purple{\underline{\red{Now:-}}}}}$

$\sf\large{\underline{\purple{\underline{\red{Finding\: 3x-y =?:-}}}}}$

$\sf\large{\underline{\purple{\underline{\red{we write (a-b)={\sqrt a^2+b^2-2ab}:-}}}}}$

$\sf\large{\underline{\purple{\underline{\red{Solution:-}}}}}$

$\sf\large{\underline{\purple{\underline{\red{By \:the\: formula:-}}}}}$

$\sf→ (3x-y)^2= 3x^2+y^2-2×3x×y\\\sf→ (3x-y)^2=9x^2+y^2-6×xy\\\sf→ (3x-y)^2=13-6×2\\\sf→(3x-y)^2=13-12\\\sf→(3x-y)^2=1\\\sf→ 3x-y=\sqrt1\\\sf→{\fbox{\red{\underline{\purple{ 3x-y=1}}}}}$