Question

Прошу пожалуйста помогите​

Answer 1

$\displaystyle\bf\\1)\\\\\left \{ {{x-y=3} \ |\cdot 2 \atop {3x+2y=1}} \right.\\\\\\+\left \{ {{2x-2y=6} \atop {3x+2y=1}} \right. \\---------\\5x=7\\\\x=1,4\\\\y=x-3=1,4-3=-1,6\\\\Otvet \ : \ \Big(1,4 \ ; \ -1,6\Big)\\\\\\2)\\\\\left \{ {{3x+4y=10} \ |\cdot(-3)\atop {4x+3y=5} \ |\cdot 4} \right. \\\\\\+\left \{ {{-9x-12y=-30} \atop {16x+12y=20}} \right. \\------------\\7x=-10\\\\x=-1\frac{3}{7}$

$\displaystyle\bf\\3y=5-4x=5-4\cdot\Big(-1\frac{3}{7} \Big)=5+5\frac{5}{7} =10\frac{5}{7} \\\\y=10\frac{5}{7} :3=1\frac{19}{21} \\\\\\Otvet \ : \ \Big(-1\frac{3}{7} \ ; \ 10\frac{5}{7} \Big)\\\\\\3)\\\\\left \{ {{3p-c=2} \ |\cdot 2\atop {3p+2c=6}} \right. \\\\\\+\left \{ {{6p-2c=4} \atop {3p+2c=6}} \right. \\---------\\9p=10\\\\p=1\frac{1}{9} \\\\c=3p-2=3\cdot 1\frac{1}{9} -2=3\frac{1}{3} -2=1\frac{1}{3} \\\\\\Otvet \ : \ \Big(1\frac{1}{9} \ ; \ 1\frac{1}{3} \Big)$

Answer 2

Ответ:

a ) x - y = 3 | × 2

3x + 2y = 1

2x - 2y = 6

3x + 2y = 1

5x = 7

x - y = 3

x = 7 ÷ 5 = 1,4

1,4 - y = 3

x = 1,4

y = 1,4 - 3 = - 1,6

ответ: ( 1.4 , - 1,6 )

б ) 3x + 4y = 10 | × ( - 3 )

4x + 3y = 5 | × ( 4 )

- 9x - 12y = - 30

16x + 12y = 20

7x = - 10

3x + 4y = 10

3

x = - 10 ÷ 7 = 1 _

7

3 × 1 3/7 + 4y = 10