Question

Реши систему уравнений методом подстановки.

Answer 1

$\left \{ {{ \frac{z + y}{6} - \frac{z - y}{3} = 1 } \atop { \frac{2z - y}{6} - \frac{3z + 2y}{ 3} = - 13}} \right.{} \\ \left \{ {{z = - 6 + 3y} \atop { \frac{2z - y}{6} - \frac{3z + 2y}{3} = - 13 }} \right.{} \\ \frac{2( - 6 + 3y) - y}{6} - \frac{3( - 6 + 3y) + 2y}{3} = - 13 \\ \boldsymbol{y = 6} \\ z = - 6 + 3 \times 6 \\ \boldsymbol{z = 12} \\$

Вот ответ

Answer 2

z = 12; y =6

Объяснение:

(z+y)/6 - (z-y)/3 = 1

((z+y) - 2*(z-y))/6 = 1

(z + y - 2z + 2y)/6 = 1

(3y-z)/6 = 1

3y-z = 6

z = 3 y - 6

z = 3(y-2)

Подставляем во 2 уравнение:

$\frac{6(y-2)-y}{6} -\frac{9(y-2)+2y}{3} = -13$

$\frac{6y-12-y}{6} -\frac{9y-18+2y}{3} = -13$

$\frac{5y-12}{6} -\frac{11y-18}{3} = -13$    |* 6

5y-12 - 22y + 36 = -78

-17y = -78+12-36

-17y = 102 |: ( -17)

y= 6

z = 3*6 - 6

z = 18-6

z = 12

$\left \{ {{z=12} \atop {y=6}} \right.$

ответ: z = 12; y = 6