Question

x^2+y(x+1)=17,y^2+x(y+1)=13
​brainly 360 plz help meeee!​ this is important and not for free point if u answer something i will REPORT YOUR QUESTION

Answer 1

Answer:

$\sf{The \ values \ of \ x \ and \ y \ are \ \dfrac{-23}{7}}$

$\sf{and \ \dfrac{-19}{7} \ respectively \ or \ values}$

$\sf{of \ x \ and \ y \ are \ 3 \ and \ 2 \ respectively. }$

Given:

$\sf{The \ given \ equations \ are \ x^{2}+y(x+1)=17}$

$\sf{and \ y^{2}+x(y+1)=13}$

To find:

$\sf{The \ value \ of \ x \ and \ y.}$

Solution:

$\sf{x^{2}+y(x+1)=17...(1)}$

$\sf{y^{2}+x(y+1)=13...(2)}$

$\sf{Add \ equations \ (1) \ and \ (2), \ we \ get}$

$\sf{x^{2}+y(x+1)=17}$

$\sf{+}$

$\sf{y^{2}+x(y+1)=13}$

____________________

$\sf{x^{2}+y^{2}+y(x+1)+x(y+1)=30}$

$\sf{\therefore{x^{2}+y^{2}+xy+y+xy+x=30}}$

$\sf{\therefore{x^{2}+y^{2}+2xy+x+y=30}}$

$\boxed{\sf{a^{2}+b^{2}+2ab=(a+b)^{2}...using \ identity}}$

$\sf{\therefore{(x+y)^{2}+(x+y)=30}}$

$\sf{Substitute \ x+y=p, \ we \ get}$

$\sf{p^{2}+p=30}$

$\sf{\therefore{p^{2}+p-30=0}}$

$\sf{\therefore{p^{2}+6p-5p-30=0}}$

$\sf{\therefore{p(p+6)-5(p+6)=0}}$

$\sf{\therefore{(p+6)(p-5)=0}}$

$\sf{\therefore{p=-6 \ or \ p=5}}$

$\sf{When \ p=-6}$

$\sf{x+y=-6}$

$\sf{\therefore{y=-6-x...(3)}}$

$\sf{Substituting \ y=-6-x \ in \ equation(2), \ we \ get}$

$\sf{(-6-x)^{2}+x(-6-x+1)=13}$

$\sf{\therefore{36+12x+x^{2}-6x-x^{2}+x=13}}$

$\sf{\therefore{7x=13-36}}$

$\sf{\therefore{7x=-23}}$

$\boxed{\sf{\therefore{x=\dfrac{-23}{7}}}}$

$\sf{Substitute \ x=\dfrac{-23}{7} \ in \ equation(3), \ we \ get}$

$\sf{y=-6-(\dfrac{-23}{7})}$

$\sf{\therefore{y=-6+\dfrac{23}{7}}}$

$\sf{\therefore{y=\dfrac{-42+23}{7}}}$

$\boxed{\sf{\therefore{y=\dfrac{-19}{7}}}}$

$\sf{When \ p=5, \ we \ get}$

$\sf{x+y=5}$

$\sf{\therefore{y=5-x...(4)}}$

$\sf{Substitute \ y=5-x \ in \ equation(2), \ we \ get}$

$\sf{(5-x)^{2}+x(5-x+1)=13}$

$\sf{\therefore{25-10x+x^{2}+5^{2}-x^{2}+x=13}}$

$\sf{\therefore{-4x=13-25}}$

$\sf{\therefore{-4x=-12}}$

$\sf{\therefore{x=\dfrac{-12}{-4}}}$

$\boxed{\sf{\therefore{x=3}}}$

$\sf{Substitute \ x=3 \ in \ equation(4), \ we \ get}$

$\sf{y=5-3}$

$\boxed{\sf{\therefore{y=2}}}$

$\sf\purple{\tt{\therefore{The \ values \ of \ x \ and \ y \ are \ \dfrac{-23}{7}}}}$

$\sf\purple{\tt{and \ \dfrac{-19}{7} \ respectively \ or \ values}}$

$\sf\purple{\tt{of \ x \ and \ y \ are \ 3 \ and \ 2 \ respectively. }}$

Answer 2

Answer is y=2 approximately