Question

If a1 x +biy +c1=0 and a2x +b2y+c2+0 are such that a1 b1 c1 a2 b2 c2 are consecutive integers in same order then find the value of x+y

Answer 1

$\textbf{Given:}$

$a_1x+b_1y+c_1=0$

$a_2x+b_2y+c_2=0$

$\textbf{To find:}$

$\text{The value of x+y}$

$\textbf{Solution:}$

$\text{Since a_1,b_1,c_1,a_2,b_2,c_2 are consecutive integers, we can write}$

$b_1=a_1+1$

$c_1=a_1+2$

$a_2=a_1+3$

$b_2=a_1+4$

$c_2=a_1+5$

$\text{The given equations become}$

$a_1x+(a_1+1)y+(a_1+2)=0$

$(a_1+3)x+(a_1+4)y+(a_1+5)=0}$

$\text{By cross multiplication rule, we get}$

$\dfrac{x}{(a_1+1)(a_1+5)-(a_1+2)(a_1+4)}=\dfrac{y}{(a_1+2)(a_1+3)-a_1(a_1+5)}$

$=\dfrac{1}{a_1(a_1+4)-(a_1+1)(a_1+3)}$

$\dfrac{x}{{a_1}^2+6\,a_1+5-{a_1}^2-6\,a_1-8}=\dfrac{y}{{a_1}^2+5\,a_1+6-{a_1}^2-5\,a_1}=\dfrac{1}{{a_1}^2+4\,a_1-{a_1}^2-4\,a_1-3}$

$\dfrac{x}{-3}=\dfrac{y}{6}=\dfrac{1}{-3}$

$\implies\dfrac{x}{-3}=\dfrac{1}{-3}$

$\implies\,x=1$

$\text{and}$

$\dfrac{y}{6}=\dfrac{1}{-3}$

$\implies\,y=-2$

$x+y=1+(-2)=-1$

$\therefore\textbf{The value of \bf\,x+y is -1}$