Question

5х+3у=7 3х-2у=12 4х-2у=2 2х+3у=11 решить 2 системы подстановки

Answer 1

$1)\; \; \left \{ {{5x+3y=7} \atop {3x-2y=12}} \right. \; \; \left \{ {{20+\frac{10}{3}y+3y=7} \atop {x=\frac{12+2y}{3}}} \right. \; \; \left \{ {{\frac{19}{3}y=7-20} \atop {x=4+\frac{2}{3}y}} \right. \; \; \left \{ {{y=-\frac{39}{19}} \atop {x=4-\frac{2}{3}\cdot \frac{39}{19}}} \right. \\\\ \left \{ {{y=-\frac{39}{19}} \atop {x=\frac{50}{19}}} \right. \; \; \left \{ {{y=-2\frac{1}{19}} \atop {x=2\frac{12}{19}}} \right.$

$2)\; \; \left \{ {{4x-2y=2} \atop {2x+3y=11}} \right. \; \; \left \{ {{y=2x-1} \atop {2x+3(2x-1)=11}} \right. \; \; \left \{ {{y=2x-1} \atop {8x=11+3}} \right. \; \; \left \{ {{y=2x-1} \atop {x=\frac{7}{4}}} \right. \\\\ \left \{ {{y=2\cdot \frac{7}{4}-1} \atop {x=\frac{7}{4}}} \right. \; \; \left \{ {{y=2,5} \atop {x=1,75}} \right.$