Question

X+2=5 x-11=-7 12+x=17 x+7=7 -13+x=-5 2x=10 -5x=0 -2x=-4 -6x=18 3x+1=7 5-2x=1 5x-2=1 -5x-2=-12 7-3(x+1)=6 5-2(3-x)=11 2x-(7+x)=2 -3-3(3-2x)=1

Answer 1

Ответ:

Пошаговое объяснение:

1) 3

2)-18

3)5

4)0

5)8

6)5

7)0

8)2

9)-3

10)2

11)2

12)2

13)?

14)6

15)9

16)?

в 13 и 16 не решается нацело

Answer 2

Let's solve each equation step by step:

1. x + 2 = 5 Subtract 2 from both sides: x = 5 - 2 x = 3

2. x - 11 = -7 Add 11 to both sides: x = -7 + 11 x = 4

3. 12 + x = 17 Subtract 12 from both sides: x = 17 - 12 x = 5

4. x + 7 = 7 Subtract 7 from both sides: x = 7 - 7 x = 0

5. -13 + x = -5 Add 13 to both sides: x = -5 + 13 x = 8

6. 2x = 10 Divide both sides by 2: x = 10 / 2 x = 5

7. -5x = 0 Divide both sides by -5: x = 0

8. -2x = -4 Divide both sides by -2: x = -4 / -2 x = 2

9. -6x = 18 Divide both sides by -6: x = 18 / -6 x = -3

10. 3x + 1 = 7 Subtract 1 from both sides: 3x = 7 - 1 3x = 6 Divide both sides by 3: x = 6 / 3 x = 2

11. 5 - 2x = 1 Subtract 5 from both sides: -2x = 1 - 5 -2x = -4 Divide both sides by -2: x = -4 / -2 x = 2

12. 5x - 2 = 1 Add 2 to both sides: 5x = 1 + 2 5x = 3 Divide both sides by 5: x = 3 / 5

13. -5x - 2 = -12 Add 2 to both sides: -5x = -12 + 2 -5x = -10 Divide both sides by -5: x = -10 / -5 x = 2

14. 7 - 3(x + 1) = 6 Distribute the -3 on the left side: 7 - 3x - 3 = 6 Subtract 7 from both sides: -3x - 3 = 6 - 7 -3x - 3 = -1 Add 3 to both sides: -3x = -1 + 3 -3x = 2 Divide both sides by -3: x = 2 / -3 x = -2/3 or approximately -0.6667

15. 5 - 2(3 - x) = 11 Distribute the -2 on the left side: 5 - 6 + 2x = 11 Subtract 5 from both sides: 2x = 11 - 5 - 6 2x = 0 Divide both sides by 2: x = 0

16. 2x - (7 + x) = 2 Remove parentheses: 2x - 7 - x = 2 Combine like terms: x - 7 = 2 Add 7 to both sides: x = 2 + 7 x = 9

17. -3 - 3(3 - 2x) = 1 Distribute the -3 on the left side: -3 - 9 + 6x = 1 Combine like terms: 6x - 12 = 1 Add 12 to both sides: 6x = 1 + 12 6x = 13 Divide both sides by 6: x = 13 / 6

I have now solved all the given equations for x. If you have any more questions or need further assistance, feel free to ask!