Question

$if(x + y ):(x - y) = 4:1 \: then \: ( {x}^{2} + {y}^{2} ):( {x}^{2} - {y}^{2} ) =$
?​

Answer 1

$\underline{\texttt{Solution :}}$

$\mathrm{Given,\:(x+y):(x-y)=4:1}$

$\to \mathrm{\frac{x+y}{x-y}=\frac{4}{1}}$

$\to \mathrm{x+y=4x-4y}$

$\to \mathrm{4x-x=y+4y}$

$\to \mathrm{3x=5y}$

$\to \mathrm{\frac{x}{5}=\frac{y}{3}=k\:(\neq 0), \:say}$

$\implies \mathrm{x=5k,\:y=3k}$

$\mathrm{Now,\:\frac{x^{2}+y^{2}}{x^{2}-y^{2}}}$

$\mathrm{=\frac{(5k)^{2}+(3k)^{2}}{(5k)^{2}-(3k)^{2}}}$

$\mathrm{=\frac{25k^{2}+9k^{2}}{25k^{2}-9k^{2}}}$

$\mathrm{=\frac{34k^{2}}{16k^{2}}}$

$\mathrm{=\frac{17}{8},\:since\:k\neq 0}$

$\to \boxed{\mathrm{(x^{2}+y^{2}):(x^{2}-y^{2})=17:8}}$

Answer 2

Question:

$if(x + y ):(x - y) = 4:1 \: then \: ( {x}^{2} + {y}^{2} ):( {x}^{2} - {y}^{2} ) = ?$

Answer:

$(x^{2} +y^{2} ):(x^{2} -y^{2} ) = 17:8$

Step-by-step Explanation:

According to the question

$=> (x + y):(x - y) = 4:1\\\\=> \frac{(x+y)}{(x-y)} = \frac{4}{1} \\\\=> x+y = 4x - 4y\\\\=> y + 4y = 4x - x\\\\=> 5y = 3x\\\\=> y = \frac{3}{5} x$

We need to find

$(x^{2} +y^{2} ):(x^{2} -y^{2} )$

By putting the value of $y = \frac{3}{5} x$ we get;

$=>(x^{2} +y^{2} ):(x^{2} -y^{2} )\\\\=> \frac{(x^{2}+y^{2}) }{(x^{2} -y^{2}) } \\\\=> \frac{[x^{2} + (\frac{3}{5}x) ^{2} ]}{[x^{2} - (\frac{3}{5}x) ^{2} ]}\\\\=> \frac{[x^{2} + \frac{9}{25}x ^{2} ]}{[x^{2} - \frac{9}{25}x^{2} ]}\\\\=> \frac{\frac{25x^{2}+9x^{2} }{25} }{\frac{25x^{2}-9x^{2} }{25} } \\\\=> \frac{34x^{2} }{16x^{2} } \\\\=> \frac{17}{8} \\\\=> 17:8$