1) решите систему методом исключения переменной а) 2x+3y=4 и 3x-2y=5; б) 3/4x-5/6y=1 и 5/6x-3/4y=2

а) Умножим первое уравнение на 2, второе - на 3, и сложим их:

{2x + 3y = 4;

{3x - 2y = 5; {4x + 6y = 8;

{9x - 6y = 15; {13x = 23;

{2y = 3x - 5; {x = 23/13;

{y = (3x - 5) / 2; {x = 23/13;

{y = (3 * 23/13 - 5) / 2; {x = 23/13;

{y = (69/13 - 65/13) / 2; {x = 23/13;

{y = 4/13 : 2 = 2/13.

б) Умножим оба уравнения на 12:

{3/4 * x - 5/6 * y = 1;

{5/6 * x - 3/4 * y = 2; {9x - 10y = 12;

{10x - 9y = 24; {9x - 10y = 12;

{10x = 24 + 9y; {9 (2,4 + 0,9y) - 10y = 12;

{x = 2,4 + 0,9y; {21,6 + 8,1y - 10y = 12;

{x = 2,4 + 0,9y; {8,1y - 10y = 12 - 21,6;

{x = 2,4 + 0,9y; {-1,9y = - 9,6;

{x = 2,4 + 0,9y; {y = 9,6/1,9 = 96/19;

{x = 2,4 + 0,9 * 96/19; {y = 96/19;

{x = 45,6/19 + 86,4/19; {y = 96/19;

{x = 132/19.