Question

Решите системы уравнений методом подстановки

Answer 1

$\left. \begin{cases} { x + 4y - 4 = 0 } \\ { x {}^{2} + 5xy = 0 } \end{cases} \right.\left. \begin{cases} {x = - 4y + 4 } \\ {x {}^{2} + 5xy = 0 } \end{cases} \right.$

$( - 4y + 4) {}^{2} + 5( - 4y + 4)y = 0 \\ ( - 4y + 4) \times ( - 4y + 4 + 5y) = 0 \\ ( - 4(y - 1)) \times (y + 4) = 0 \\ - 4(y - 1) \times (y + 4) = 0 \\ (y - 1) \times (y + 4) = 0 \\y - 1 = 0~~~~~~~~~~~~y + 4 = 0 \\ y = 1~~~~~~~~~~~~~~~y = - 4$

Раставим в место y:

$x = - 4 \times 1 + 4 = - 4 + 4 = 0 \\ x = - 4 \times ( - 4) + 4 = 16 + 4 = 20$

$ответ:(x1;y1) = (0;1) \\ ~~~~~~~~~~~(x2;y2) = (20; - 4)$

Answer 2

$\displaystyle \left \{ {{x+4y-4=0} \atop {x^2+5xy=0}} \right.\\\\ \left \{ {{x=-4y+4} \atop {x^2+5xy=0}} \right.\\\\ (-4y+4)^2+5(-4y+4)y=0 \\y=1,y=-4\\x=-4*1+4,x=-4*(-4)+4\\x=0,x=20\\(x_{1},y_{1})=(0,1),(x_{2},y_{2} )=(20,-4)$