Find the equation of the straight line passing through the point of intersection of 2x+3y+1=0 and 3x-5y-5=0 and equally inclined to the axes.

$The\;equations,\;2x+3y+1=0\;and\;3x-5y-5=0\\*The\;equation\;of\;the\; straight\;line\;passing\;through\;the\;points\;of\;intersection\\*of\;2x+3y+1=0\;and\;3x-5y-5=0\;is\\*2x+3y+1+\lambda\;(3x-5y-5)=0\\*(2+3\lambda)x+(3-5\lambda)y+1-5\lambda=0\\*y=-\frac{(2+3\lambda)}{(3-5\lambda)}-\frac{(1-5\lambda)}{(3-5\lambda)}\\* The\;required\;line\;is\;equally\; inclined\;to\;the\;axes.\;So,\;the\;\;slope\;of\;the\;required\\*line\;is\;either\;1\;or\;1.\;So,\\*-\frac{(2+3\lambda)}{(3-5\lambda)}=1\;and-\frac{(2+3\lambda)}{(3-5\lambda)}=-1\\*-2-3\lambda=3-5\lambda\; and\; 2+3\lambda=3-5\lambda\\*\lambda=\frac{5}{2}\;and\;\frac{1}{8}\\*Now,\; substitute\;the\;values\;of\;\lambda\;in\;(2+3\lambda)x+(3-5\lambda)y+1-5\lambda=0,\;we\;get\;the\;equations\;of\;the\;required\;lines\;as:\\*\Rightarrow\;(2+\frac{15}{2})x+(3-\frac{25}{2})y+(1-\frac{25}{2})=0\;and\;(2+\frac{3}{8})x+(3-\frac{5}{8})y+(1-\frac{5}{8})=0\\*19x-19y-23=0\;and\; 19x+19y+3=0\\*\therefore The\;required\;equation\;is\;19x-19y-23=0\; and\;19x+19y+3=0$

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