Question

1426. 1) [4x + 3y - 5 = 0,Ix-y-3= 0;
2) (x – y - 7 = 0,15x – 3y - 1 = 0;
3) (6x + 5y - 6'= 0,(2x + y + 2 = 0.

Answer 1

Пошаговое объяснение:

1)

$4x + 3y = 5 \\( x - y = 3) \times 4 \\ \\ 4x + 3y = 5 \\ 4x - 4y = 12 \\ \\ 7y = - 7 \\ y = - 1 \\ \\ x - ( - 1) = 3 \\ x = 2$

$(x – y - 7 = 0) \times 3 \\ 15x – 3y - 1 = 0 \\ \\ 3x - 3y = 21 \\ 15x - 3y = 1 \\ \\ 18x = 22 \\ x = \frac{11}{9} \\ \\ \frac{11}{9} - y = 7 \\ y = - 5 \frac{8}{9}$

$6x + 5y = 6 \\ (2x + y = - 2) \times 3 \\ \\6x + 5y = 6 \\ 6x + 3y = - 6 \\ \\ 2y = 12 \\ y = 6 \\ \\ 2x + 6 = - 2 \\ 2x = - 8 \\ x = - 4$