Question

if (2x+y=3 and xy=1 then find the value of (x+y)^x-y.... plz help me brainly
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Answer 1

Answer:

1

Step-by-step explanation:

2x+y=3

y=3-2x

then in 2nd equation put the value of y and solve

x=1 y=1

then put the value of xand y and u will get ans

Answer 2

Answer:

$\sf{The \ value \ of \ (x+y)^{x-y} \ is \ (\frac{5}{2})^{\frac{-3}{2}}}$

Given:

$\sf{The \ given \ equations \ are \ 2x+y=3}$

$\sf{and xy=1}$

To find:

$\sf{The \ value \ of \ (x+y)^{x-y}.}$

Solution:

$\sf{2x+y=3...(1)}$

$\sf{xy=1...(2)}$

$\sf{From \ equation(2), \ we \ get}$

$\sf{xy=1}$

$\sf{\therefore{y=\dfrac{1}{x}}}$

$\sf{Substitute \ y=\dfrac{1}{x} \ in \ equation(1), \ we \ get}$

$\sf{2x+\dfrac{1}{x}=3}$

$\sf{Multiply \ throughout \ by \ x, \ we \ get}$

$\sf{2x^{2}+1=3x}$

$\sf{\therefore{2x^{2}-3x+1=0}}$

$\sf{\therefore{2x^{2}-2x-x+1=0}}$

$\sf{\therefore{2x(x-1)-1(x-1)=0}}$

$\sf{\therefore{(2x-1)(x-1)=0}}$

$\sf{\therefore{x=\dfrac{1}{2} \ or \ x=1}}$

$\sf{When \ x=\dfrac{1}{2}}$

$\sf{Substitute \ x=\dfrac{1}{2} \ in \ equation(2), \ we \ get}$

$\sf{(\frac{1}{2})y=1}$

$\sf{\implies{y=2}}$

$\sf{When \ x=1}$

$\sf{Substitute \ x=1 \ in \ equation(1), \ we \ get}$

$\sf{(1)y=1}$

$\sf{\implies{y=1}}$

$\sf{Considering \ x=\dfrac{1}{2} \ and \ y=2}$

$\sf{\leadsto{(\dfrac{1}{2}+2)^{\frac{1}{2}-2}}}$

$\sf{\leadsto{(\frac{5}{2})^{\frac{-3}{2}}}}$

$\sf{Considering \ x=1 \ and \ y=1}$

$\sf{\leadsto{(1+1)^{1-1}}}$

$\sf{\leadsto{2^{0}}}$

$\sf{\leadsto{1}}$

$\sf\purple{\tt{\therefore{The \ value \ of \ (x+y)^{x-y} \ is \ (\frac{5}{2})^{\frac{-3}{2}}}}}$

$\sf\purple{\tt{or \ 1.}}$