[tex] \frac{5 + \sqrt{11} }{3 – 2 \sqrt{11} } = x + y \sqrt{11} [/tex]Find the value of ‘x’ and ‘y’ About the author Anna
Answer: 3−2 11 5+ 11 =x+y 11 To Find: Find the value of x and y Solution: \begin{gathered} \implies \tt \bold{ \frac{5 + \sqrt{11} }{3 – 2 \sqrt{11} } = x + y \sqrt{11} } \\ \end{gathered} ⟹ 3−2 11 5+ 11 =x+y 11 Rewrite the LHS and RHS . \begin{gathered}\implies \tt \bold{ x + y \sqrt{11} = \frac{5 + \sqrt{11} }{3 – 2 \sqrt{11} } } \\ \end{gathered} ⟹x+y 11 = 3−2 11 5+ 11 Now , Rationalize the denominator by 3+2√11 \begin{gathered}\implies \tt \bold{ x + y \sqrt{11} = \frac{5 + \sqrt{11} }{3 – 2 \sqrt{11}} } \bold{ \times \frac{3 + 2 \sqrt{11} }{3 + 2 \sqrt{11} }} \\ \end{gathered} ⟹x+y 11 = 3−2 11 5+ 11 × 3+2 11 3+2 11 Simplify the RHS By using the formula a²-b²=(a+b)(a-b) \begin{gathered}\implies \tt \bold{ x + y \sqrt{11} = \frac{(5 + \sqrt{11})(3 + 2 \sqrt{11}) }{ {3}^{2} – 4(11)} } \\ \end{gathered} ⟹x+y 11 = 3 2 −4(11) (5+ 11 )(3+2 11 ) \begin{gathered}\implies \tt \bold{ x + y \sqrt{11} = \frac{15 + 13 \sqrt{11} + 22 }{ – 35} } \\ \end{gathered} ⟹x+y 11 = −35 15+13 11 +22 Adding the numbers 15 and 22 is 37 . \begin{gathered}\implies \tt \bold{ x + y \sqrt{11} = \frac{37 + 13 \sqrt{11} }{ – 35}} \\ \end{gathered} ⟹x+y 11 = −35 37+13 11 Now , the comparing the LHS AND RHS. \begin{gathered}\implies \tt \bold{ x + y \sqrt{11} = \frac{ – 37}{35} – \frac{13}{35} \sqrt{11}} \\ \end{gathered} ⟹x+y 11 = 35 −37 − 35 13 11 \therefore \boxed{ \bf{ \red{x = \frac{ – 37}{35}}} }∴ x= 35 −37 \therefore \boxed{ \red{ \bf{y = \frac{ – 13}{35}}}}∴ y= 35 −13 H Step-by-step explanation: please mark me as brainlist Reply
Step-by-step explanation: [tex] \huge \textbf{ \underline{Answer :-}}[/tex] Given : [tex] \frac{5 + \sqrt{11} }{3 – 2 \sqrt{11} } = x + y \sqrt{11} [/tex] To Find: Find the value of x and y Solution: [tex] \implies \tt \bold{ \frac{5 + \sqrt{11} }{3 – 2 \sqrt{11} } = x + y \sqrt{11} } \\ [/tex] Rewrite the LHS and RHS . [tex]\implies \tt \bold{ x + y \sqrt{11} = \frac{5 + \sqrt{11} }{3 – 2 \sqrt{11} } } \\ [/tex] Now , Rationalize the denominator by 3+2√11 [tex]\implies \tt \bold{ x + y \sqrt{11} = \frac{5 + \sqrt{11} }{3 – 2 \sqrt{11}} } \bold{ \times \frac{3 + 2 \sqrt{11} }{3 + 2 \sqrt{11} }} \\ [/tex] Simplify the RHS By using the formula a²-b²=(a+b)(a-b) [tex]\implies \tt \bold{ x + y \sqrt{11} = \frac{(5 + \sqrt{11})(3 + 2 \sqrt{11}) }{ {3}^{2} – 4(11)} } \\ [/tex] [tex]\implies \tt \bold{ x + y \sqrt{11} = \frac{15 + 13 \sqrt{11} + 22 }{ – 35} } \\ [/tex] Adding the numbers 15 and 22 is 37 . [tex]\implies \tt \bold{ x + y \sqrt{11} = \frac{37 + 13 \sqrt{11} }{ – 35}} \\ [/tex] Now , the comparing the LHS AND RHS. [tex]\implies \tt \bold{ x + y \sqrt{11} = \frac{ – 37}{35} – \frac{13}{35} \sqrt{11}} \\ [/tex] [tex] \therefore \boxed{ \bf{ \red{x = \frac{ – 37}{35}}} }[/tex] [tex] \therefore \boxed{ \red{ \bf{y = \frac{ – 13}{35}}}} [/tex] Hope it will help you mate !! Mark as Brainlist !! Reply
Answer:
3−2
11
5+
11
=x+y
11
To Find:
Find the value of x and y
Solution:
\begin{gathered} \implies \tt \bold{ \frac{5 + \sqrt{11} }{3 – 2 \sqrt{11} } = x + y \sqrt{11} } \\ \end{gathered}
⟹
3−2
11
5+
11
=x+y
11
Rewrite the LHS and RHS .
\begin{gathered}\implies \tt \bold{ x + y \sqrt{11} = \frac{5 + \sqrt{11} }{3 – 2 \sqrt{11} } } \\ \end{gathered}
⟹x+y
11
=
3−2
11
5+
11
Now , Rationalize the denominator by 3+2√11
\begin{gathered}\implies \tt \bold{ x + y \sqrt{11} = \frac{5 + \sqrt{11} }{3 – 2 \sqrt{11}} } \bold{ \times \frac{3 + 2 \sqrt{11} }{3 + 2 \sqrt{11} }} \\ \end{gathered}
⟹x+y
11
=
3−2
11
5+
11
×
3+2
11
3+2
11
Simplify the RHS
By using the formula a²-b²=(a+b)(a-b)
\begin{gathered}\implies \tt \bold{ x + y \sqrt{11} = \frac{(5 + \sqrt{11})(3 + 2 \sqrt{11}) }{ {3}^{2} – 4(11)} } \\ \end{gathered}
⟹x+y
11
=
3
2
−4(11)
(5+
11
)(3+2
11
)
\begin{gathered}\implies \tt \bold{ x + y \sqrt{11} = \frac{15 + 13 \sqrt{11} + 22 }{ – 35} } \\ \end{gathered}
⟹x+y
11
=
−35
15+13
11
+22
Adding the numbers 15 and 22 is 37 .
\begin{gathered}\implies \tt \bold{ x + y \sqrt{11} = \frac{37 + 13 \sqrt{11} }{ – 35}} \\ \end{gathered}
⟹x+y
11
=
−35
37+13
11
Now , the comparing the LHS AND RHS.
\begin{gathered}\implies \tt \bold{ x + y \sqrt{11} = \frac{ – 37}{35} – \frac{13}{35} \sqrt{11}} \\ \end{gathered}
⟹x+y
11
=
35
−37
−
35
13
11
\therefore \boxed{ \bf{ \red{x = \frac{ – 37}{35}}} }∴
x=
35
−37
\therefore \boxed{ \red{ \bf{y = \frac{ – 13}{35}}}}∴
y=
35
−13
H
Step-by-step explanation:
please mark me as brainlist
Step-by-step explanation:
[tex] \huge \textbf{ \underline{Answer :-}}[/tex]
Given :
To Find:
Solution:
[tex] \implies \tt \bold{ \frac{5 + \sqrt{11} }{3 – 2 \sqrt{11} } = x + y \sqrt{11} } \\ [/tex]
Rewrite the LHS and RHS .
[tex]\implies \tt \bold{ x + y \sqrt{11} = \frac{5 + \sqrt{11} }{3 – 2 \sqrt{11} } } \\ [/tex]
Now , Rationalize the denominator by 3+2√11
[tex]\implies \tt \bold{ x + y \sqrt{11} = \frac{5 + \sqrt{11} }{3 – 2 \sqrt{11}} } \bold{ \times \frac{3 + 2 \sqrt{11} }{3 + 2 \sqrt{11} }} \\ [/tex]
Simplify the RHS
By using the formula a²-b²=(a+b)(a-b)
[tex]\implies \tt \bold{ x + y \sqrt{11} = \frac{(5 + \sqrt{11})(3 + 2 \sqrt{11}) }{ {3}^{2} – 4(11)} } \\ [/tex]
[tex]\implies \tt \bold{ x + y \sqrt{11} = \frac{15 + 13 \sqrt{11} + 22 }{ – 35} } \\ [/tex]
Adding the numbers 15 and 22 is 37 .
[tex]\implies \tt \bold{ x + y \sqrt{11} = \frac{37 + 13 \sqrt{11} }{ – 35}} \\ [/tex]
Now , the comparing the LHS AND RHS.
[tex]\implies \tt \bold{ x + y \sqrt{11} = \frac{ – 37}{35} – \frac{13}{35} \sqrt{11}} \\ [/tex]
[tex] \therefore \boxed{ \bf{ \red{x = \frac{ – 37}{35}}} }[/tex]
[tex] \therefore \boxed{ \red{ \bf{y = \frac{ – 13}{35}}}} [/tex]
Hope it will help you mate !!
Mark as Brainlist !!