Question

1257. 1) (8x + 7y + 53 = 0, 4x - 5y + 1 = 0;
х нужно пожалуйста дам45 бал​

Answer 1

$\displaystyle\\\left \{ {{8x+7y+53=0} \atop {4x-5y+1=0|\times-2}} \right. \\\\\left \{ {{8x+7y+53=0} \atop {-8x+10y-2=0}} \right. \\(1)+(2)=8x-8x+7y+10y=-53+2\\17y=-51\\y=-3= > x=-4$