Question

rootx +root y=5 and x +y =13 find x and y​

Answer 1

Answer:

In the attachment I have answered this problem.

The equations have been solved by Substitution method.

The first equation is modified and substituted in second equation.

Simplifications are done step by step to find the value of x.

See the attachment for detailed solution.

Answer 2

${ \large{ \star}} \: \: \: \: \blue{ \underline{ Given :}}$

$\tt { \large{ {\rightsquigarrow}}} \: \sqrt{x} + \sqrt{y} = 5$

$\tt { \large{ {\rightsquigarrow}}} \: x + y = 13$

${ \large{ \star}} \: \: \: \: \blue{ \underline{Solution :}}$

$\tt \sqrt{x} + \sqrt{y} = 5 \\ \\ \implies \tt \: {( \sqrt{x} + \sqrt{y} )}^{2} = 25 \\ \\ \tt \implies \: x + y + 2 \sqrt{xy} = 25 \\ \\ \implies \tt 13 + 2 \sqrt{xy} = 25 \\ \\ \implies \tt \sqrt{xy} = 6 \\ \\ \implies \tt x = \frac{36}{y} \underline{ \: \: \: \: \: \: \: \: \: \: \: \: }(1) \\ \\ \tt \: from \: \: given \: {eq}^{n} \\ \\ \tt \: x + y = 13 \\ \\ \implies \tt \: x = 13 - y \\ \\ \tt \natural \: \: putting \: \: value \: \: of \: \: x \: \: from \: \: {eq}^{n} \: \: (1) \\ \\ \implies \tt \frac{36}{y} = 13 - y \\ \\ \implies \tt \: {y}^{2} - 13y + 36 = 0 \\ \\ \implies \tt {y}^{2} - 9y - 4y + 36 = 0 \\ \\ \implies \tt \: (y - 9)(y - 4) = 0 \\ \\ \therefore{ \green{ \boxed{ \tt \: y = 9 \: \: \: \: \: \: \mid \: y = 4 \: \: \: \: \: \: \: }}} \\ \\ \\ \tt \natural \: \: \: putting \: \: the \: \: value \: \: of \: \: y \: \: in \: \: {eq}^{n} (1) \\ \\ \begin{array}{c|c} \tt \: x + y = 13 & \: \tt \: x + y = 13 \\ \\ \implies \tt x +9 = 13 \: & \: \implies \tt \: x + 4 = 13 \\ \\ \implies \tt \: x = 4& \: \tt \implies x =9 \end{array} \\ \\ \therefore{ \green{ \boxed{ \tt{ x = 4 \: \: \: \: \: \: \mid \: x = 9 \: \: \: \: \: \: }}}} \\ \\ \\ \huge\mathfrak \red{Merry \: \: Christmas}$

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$\: \: \: \: \: \: \: \: \: \: \: \: \tt \colorbox{aqua}{@StayHigh}$