Question

кто сделает ето очень или мне поставлять два нужна буду очень благодарен​

Answer 1

№1

$+ \left \{ {{x - y = 1} \atop {x + y = 7}} \right. \\x - y + x + y = 1 + 7\\2x = 8\\x = 4\\4 + y = 7 \implies y = 3$

№2

$+ \left \{ {{3x - 8y = 18} \atop {-3x + 4y = -6}} \right. \\3x - 8y - 3x + 4y = 18 - 6\\-4y = 12\\y = -3\\-3x + 4*(-3) = -6 \implies x = -2$

№3

$+ \left \{ {{2x + 5y = 6} \atop {8x - 5y = -1}} \right. \\2x + 5y + 8x - 5y = 6 - 1\\10x = 5\\x = \frac{1}{2}\\2*\frac{1}{2} + 5y = 6 \implies y = 1$

№4

$- \left \{ {{4x + 5y = 6} \atop {4x + 3y = 2}} \right. \\4x + 5y - 4x - 3y = 6 - 2\\2y = 4 \implies y = 2\\4x + 3*2=2 \implies x = -1$

№5

$- \left \{ {{5m + 2n = -2} \atop {3m + 2n = 2}} \right. \\5m + 2n - 3m - 2n = -2 - 2\\2m = -4 \implies m = -2\\3*(-2) + 2n = 2 \implies n = 4$

№6

$- \left \{ {{3x - 3y = 12} \atop {3x + 2y = 17}} \right. \\3x - 3y - 3x - 2y = 12 - 17\\-5y = -5 \implies y = 1\\3x + 2*1 = 17 \implies x = 5$

Решение методом подстановки:

№1

$\left \{ {{x - y = 1} \atop {x + y = 7}} \right. \\\left \{ {{x = 1 + y} \atop {(1 + y) + y = 7}} \right. \\\left \{ {{x = 1 + y} \atop {2y = 6}} \right. \\\left \{ {x = 1 + y} \atop {y = 3}} \right. \\\left \{ {{x = 1 + 3} \atop {y = 3}} \right. \\\left \{ {{x = 4} \atop {y = 3}} \right.$

№2

$\left \{ {{3x - 8y = 18} \atop {-3x + 4y = -6}} \right. \\\left \{ {{3x = 18 + 8y} \atop {-(18+8y) + 4y = -6}} \right. \\\left \{ {{3x = 18 + 8y} \atop {-18 - 8y + 4y = -6}} \right. \\\left \{ {{3x = 18 + 8y} \atop {-4y = 12}} \right. \\\left \{ {{3x = 18 + 8y} \atop {y = -3}} \right. \\\left \{ {{3x = 18 + 8*(-3)} \atop {y = -3}} \right. \\\left \{ {{3x = 18-24} \atop {y = -3}} \right. \\\left \{ {{x = -2} \atop {y = -3}} \right.$

№3

$\left \{ {{2x + 5y = 6} \atop {8x - 5y = -1}} \right. \\\left \{ {{5y = 6 - 2x} \atop {8x - (6 - 2x) = -1}} \right. \\\left \{ {{5y=6 - 2x} \atop {8x - 6 + 2x = -1}} \right. \\\left \{ {{5y = 6 - 2x} \atop {10x = 5}} \right. \\\left \{ {{5y = 6 - 2x} \atop {x = \frac{1}{2}}} \right. \\\left \{ {{5y = 6 - 2*\frac{1}{2}} \atop {x=\frac{1}{2}}} \right. \\\left \{ {{5y=5} \atop {x=\frac{1}{2}}} \right. \\\left \{ {{y=1} \atop {x=\frac{1}{2}}} \right.$

№4

$\left \{ {{4x + 5y = 6} \atop {4x + 3y = 2}} \right. \\\left \{ {{4x = 6 - 5y} \atop {(6 - 5y) + 3y = 2}} \right. \\\left \{ {{4x = 6 - 5y} \atop {-2y=-4}} \right. \\\left \{ {{4x = 6 - 5y} \atop {y=2}} \right.\\\left \{ {{4x = 6 - 5*2} \atop {y=2}} \right.\\\left \{ {{4x = -4} \atop {y=2}} \right.\\\left \{ {{x = -1} \atop {y=2}} \right.$

№5

$\left \{ {{5m + 2n = -2} \atop {3m + 2n = 2}} \right. \\\left \{ {{2n = -2 - 5m} \atop {3m + (-2-5m) = 2}} \right. \\\left \{ {{2n = -2 - 5m} \atop {-2m = 4}} \right. \\\left \{ {{2n = -2 - 5m} \atop {m = -2}} \right. \\\left \{ {{2n = -2 - 5*(-2)} \atop {m = -2}} \right. \\\left \{ {{2n = 8} \atop {m = -2}} \right. \\\left \{ {{n = 4} \atop {m = -2}} \right.$

№6

$\left \{ {{3x - 3y = 12} \atop {3x + 2y = 17}} \right. \\\left \{ {{3x = 12 + 3y} \atop {12 + 3y + 2y = 17}} \right. \\\left \{ {{3x = 12 + 3y} \atop {5y = 5}} \right. \\\left \{ {{3x = 12 + 3y} \atop {y = 1}} \right. \\\left \{ {{3x = 15} \atop {y = 1}} \right. \\\left \{ {{x = 5} \atop {y = 1}} \right.$