Question

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Answer 1

Answer:

45

Step-by-step explanation:

We know that:

(a+b)^2 = a^2+2ab+b^2

In question, a = x & b = 11y

xy = 2

x+11y = 7

(x+11y)^2 = 7^2

x^2+2xy+(11y)^2 = 49

x^2+121(y^2)+2(2) = 49

x^2+121(y^2) +4 = 49

x^2+121(y^2) = 45

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Answer 2

Given,
x + 11y = 7 —-—--——------———>1
xy = 2 —---————---———>2

Using
${(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab$

$\: \: \: \: \: {(x + 11y)}^{2} = {x}^{2} + {(11y)}^{2} + 2 \times x \times 11y \\ = > {(x + 11y)}^{2} = {x}^{2} + 121 {y}^{2} + 22xy$
From eq. 1&2,
$= > {x}^{2} + 121 {y}^{2} = {(x + 11y)}^{2} + 22xy \\ = > {x}^{2} + 121 {y}^{2} = {7}^{2} + 22 \times 2 \\ = > {x}^{2} + 121 {y}^{2} = 49 + 44 \\ = > {x}^{2} + 121 {y}^{2} = 93.$



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