Question

Решите методом подстановки систему уравнений: 1) х-у=3
ху=28
2) у в квадрате-х=14
х-у=-2
3) у-2х в квадрате=2
3х+у=1
4) х в квадрате-2у в квадрате=8
х+у=6

Answer 1

1) (-4;-7) ; (4;7)

2) (2;4) ; (-5;-3)

3) (-0,5;2,5) ; (-1;4)

4) (20;-14) ; (4;2)

Объяснение:

1)

$x - y = 3 \\ xy = 28 \\ \\ x = y + 3 \\ y(y + 3) = 28 \\ \\ x = y + 3 \\ {y}^{2} + 3y - 28 = 0$

${y}^{2} + 3y - 28 = 0$

По теореме Виета:

у1 = -7

у2 = 4

$x = y + 3 \\ y = - 7 \\ \\ x = y + 3 \\ y = 4\\ \\ x = - 7 + 3 \\ y = - 7 \\ \\ x = 4 + 3 \\ y = 3 \\ \\ x = - 4 \\ y = - 7 \\ \\ x= 7 \\ y = 4$

2)

${y}^{2} - x = 14 \\ x - y = - 2 \\ \\ x = y - 2 \\ {y}^{2} - (y - 2) = 14 \\ \\ x = y - 2 \\ {y}^{2} - y + 2 - 14 = 0 \\ \\ x = y - 2 \\ {y}^{2} - y - 12 = 0$

${y}^{2} - y - 12 = 0$

По теореме Виета:

у1=4

у2=-3

$x = y - 2 \\ y = 4 \\ \\ x = y - 2 \\ y = - 3 \\ \\ x = 4 - 2 \\ y = 4 \\ \\ x = - 3 - 2 \\ y = - 3 \\ \\ x = 2 \\ y = 4 \\ \\ x = - 5 \\ y = - 3$

3)

$y - 2 {x}^{2} = 2 \\ 3x + y = 1 \\ \\ y = 1 - 3x \\ 1 - 3x - 2 {x}^{2} = 2 \\ \\ 2 {x}^{2} + 3x + 1 = 0 \\ y = 1 - 3x$

$2 {x}^{2} + 3x + 1 = 0 \\ d = {3}^{2} - 4 \times 2 \times 1 = 9 - 8 = 1 \\ x1 = \frac{ - 3 + 1}{4} = \frac{ - 2}{4} = - \frac{ 1}{2} = - 0.5\\ x2 = \frac{ - 3 - 1}{4} = \frac{ - 4}{4} = - 1$

$x = - 0.5 \\ y= 1 - 3x \\ \\ x = - 1 \\ y = 1 - 3x \\ \\ x = - 0.5 \\ y = 1 + 3 \times 0.5 \\ \\ x = - 1 \\ y = 1 + 3 \times 1 \\ \\ x = - 0.5 \\ y = 1 + 1.5 \\ \\ x = - 1 \\ y = 1 + 3 \\ \\ x = - 0.5 \\ y = 2.5 \\ \\ x = - 1 \\ y = 4$

4)

${x}^{2} - 2 {y}^{2} = 8 \\ x + y = 6 \\ \\ x = 6 - y \\ {(6 - y)}^{2} - 2 {y}^{2} = 8 \\ \\ x = 6 - y \\ 36 - 12y + {y}^{2} - 2 {y}^{2} - 8 = 0 \\ \\ x = 6 - y \\ - {y}^{2} - 12y + 28 = 0 \\ \\ x = 6 - y \\ {y}^{2} + 12y - 28 = 0$

${y}^{2} + 12y - 28 = 0$

По теореме Виета:

у1 = -14

у2 = 2

$x = 6 - y \\ y = - 14 \\ \\ x = 6 - y \\ y = 2 \\ \\ x = 6 + 14 \\ y = - 14 \\ \\ x = 6 - 2 \\ y = 2 \\ \\ x = 20 \\ y = - 14 \\ \\ x = 4 \\ y = 2$