Question

пожалуйста,методом подстановки(сколько сможете)​

Answer 1

$1)\; \; \left \{ {{\frac{x}{5}=1-\frac{y}{15}\; |\cdot 15} \atop {2x-5y=0\quad }}\right.\; \; \left \{ {{3x=15-y} \atop {2x=5y}} \right. \; \left \{ {{y=15-3x\quad } \atop {2x=5\cdot (15-3x)}} \right. \; \left \{ {{y=15-3x} \atop {2x=75-15x}} \right. \; \left \{ {{y=15-3x} \atop {17x=75}} \right. \\\\\left \{ {{y=15-3\cdot \frac{75}{17}} \atop {x=\frac{75}{17}}} \right. \; \left \{ {{y=\frac{30}{17}} \atop {x=\frac{75}{17}}} \right. \; \; \to \; \; (\frac{75}{17}\, ,\, \frac{30}{17})$

$2)\; \; \left \{ {{4x-3y=1} \atop {\frac{2x+1}{6}=\frac{9-5y}{8} }} \right.\; \left \{ {{4x=1+3y\qquad \; } \atop {8(2x+1)=6(9-5y)}} \right.\; \left \{ {{4x=1+3y\qquad } \atop {16x+8=54-30y}} \right.\; \left \{ {{4x=1+3y\; |\cdot 2\quad } \atop {16x=46-30y\, |:2}} \right. \\\\\left \{ {{8x=2+6y} \atop {8x=23-15y}} \right. \opminus \left \{ {{4x=1+3y} \atop {0=21-21y}} \right. \; \left \{ {{4x=1+3\cdot 1} \atop {y=1}} \right. \; \left \{ {{x=1} \atop {y=1}} \right. \; \; \to \; \; \; (1,1)$