Question

AJUTOOOR VA ROOG MUULT RAPIIID!!

Rezolvati in Ζ:
a) 4(3x + 2) + 3(2x + 3) - 28 = 7(2x +5) + 6 ;
b) Ι2x + 7Ι = 9 ;
c) 4x - 5y = 25 si ΙxΙ < 12 .

AJUTOOOR PLSSSS RAPIIID!!

Answer 1

a)

4(3x+2)+3(2x+3)-28=7(2x+5)+6

12x+8+6x+9-28=14x+35+6

18x-11=14x+41

18x=14x+41+11=14x+52

18x-14x=52

4x=52

$x = \frac{52}{4} = 13$

b)

|2x+7|=9

2x+7=9

-(2x+7)=9

2x=9-7=2

x=1

-2x-7=9

-2x=9+7=16

$x = - \frac{16}{2} = - 8$

Answer 2

b)

$\it |2x+7| =9 \Rightarrow 2x+7=\pm9 \Rightarrow 2x+7\in\{-9,\ 9\}|_{-7} \Rightarrow2x \in \{-16, \ 2 \}|_{:2} \Rightarrow\\ \\\Rightarrow x\in \{-8,\ \ 1\}$

c)

$\it 4x-5y=25 \Rightarrow 4x = 25 +5y \ \ \ \ \ (1)\\ \\ 25+5y\in M_{5} \stackrel{(1)}{\Longrightarrow} 4x \in M_{5}\ \ \ \ (2)\\ \\ |x| <12 \stackrel{(2)}{\Longrightarrow} x\in\{\pm10,\ \pm5,\ 0\} \Rightarrow x\in \{-10,\ -5,\ 0,\ 5,\ 10\} \ \ \ \ \ \ (3)$

$\it (3) \Rightarrow 4x \in\{-40,\ -20,\ 0,\ 20,\ 40\} \stackrel{(1)}{\Longrightarrow} 25+5y\in\{-40,\ -20,\ 0,\ 20,\ 40\}|_{-25}\Rightarrow\\ \\ \Rightarrow 5y \in \{-65,\ -45,\ -25,\ -5,\ 15\}|_{:5}\Rightarrow y\in\{-13,\ -9,\ -5,\ -1,\ 3\} \ \ \ \ \ (4)$

$\it (3),\ (4) \Rightarrow (x,\ y) \in\{(-10,\ -13),\ (-5,\ -9),\ (0,\ -5),\ (5,\ -1),\ (10,\ 3)\}$