Question

Решите систему уравнений способом подстановки: 1){x/4+y/8=2, 2x/9+y/6=1; 2){7x/3-y/9=4, 8x/5-y/30=3

Answer 1

первая система: $\left \{ {{ \frac{x}{4}+ \frac{y}{8} =2 } \atop { \frac{2x}{9}+ \frac{y}{6}= 1}} \right. \\ \left \{ {{2x+y=16} \atop {4x+3y=18}} \right. \\ \left \{ {{y=16-2x} \atop {4x+3(16-2x)=18}} \right. \\ 4x+3(16-2x)=18 \\ 4x+48-8x=18 \\ 4x-8x=18-48 \\ -4x=-30 \\ x=7.5 \\ \left \{ {{y=16-2x} \atop {x=7.5}} \right. \\ \left \{ {{y=16-2*7.5} \atop {x=7.5}} \right. \\ \left \{ {{y=1} \atop {x=7.5}} \right.$
вторая система: $\left \{ {{ \frac{7x}{3}- \frac{y}{9}=4 } \atop { \frac{8x}{5}- \frac{y}{30}=3 }} \right. \\ \left \{ {{21x-y=36} \atop {48x-y=90}} \right. \\ \left \{ {{y=21x-36} \atop {48x-(21x-36)=90}} \right. \\ 48x-(21x-36)=90 \\ 48x-21x+36=90 \\ 27x=90-36 \\ 27x=54 \\ x=2 \\ \left \{ {{y=21x-36} \atop {x=2}} \right. \\ \left \{ {{y=21*2-36} \atop {x=2}} \right. \\ \left \{ {{y=42-36} \atop {x=2}} \right. \\ \left \{ {{y=6} \atop {x=2}} \right.$