Решите систему уравнений способом подстановки -x+y=4 4x+y=-1
{-x+y=4
{ 4x+y=-1
{y=4+x
{4x+4+x=-1
{y=4+x
{5x=-1-4
{y=4+x
{5x=-5
{y=4+x
{x=-1
{y=4-1
{x=-1
{y=3
{x=-1
ответ: (-1; 3)
a)
b)
Пошаговое объяснение:
a) [3+2x]
5
раскрываем модуль.
1) 3+2x5 2) 3+2x-5
2x5-3 2x-5-3
2x2. 2x-8
x1 x-4.
< -4 < 1 (они идет влево)
[---.---------------.---------------->
xэ [-бесконечность;1]
точка должна быть закрашенной.
b) [10-x]>11
раскрываем модуль.
1) 10-x>11 2) 10-x>-11
-x>11-10 -x>-11-10
-x>1 -x>-21
теперь меняем знак на <, потому что перед x стоит минус -
x<-1 x<21
< -1 21 >
[-----,-------------,------------->
xэ нет решения.
так как они не пересекаются .
=8
=87a + 5b = 8a
=87a + 5b = 8a5b = 8a - 7a
=87a + 5b = 8a5b = 8a - 7aa = 5b
=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32
=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b
=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b
=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b
=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b =
=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b = b
=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b = b5∗5b+7b
=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b = b5∗5b+7b
=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b = b5∗5b+7b =
=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b = b5∗5b+7b = b
=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b = b5∗5b+7b = b25b+7b
=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b = b5∗5b+7b = b25b+7b
=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b = b5∗5b+7b = b25b+7b =
=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b = b5∗5b+7b = b25b+7b = b
=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b = b5∗5b+7b = b25b+7b = b32b
=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b = b5∗5b+7b = b25b+7b = b32b
=87a + 5b = 8a5b = 8a - 7aa = 5b\frac{5a+7b}{b}= \frac{5*5b+7b}{b} = \frac{25b+7b}{b}= \frac{32b}{b} =32 b5a+7b = b5∗5b+7b = b25b+7b = b32b =32
