Rozwiąż układ równań:

a) 2(3x + 4y) - 7 (y-x) = -37
4(x +2) - (y-4) = -2

b) 2/3 x + 1/2 y = 5
1/3 x - 1/3 y = -1

c) 1/4 x +1/6 y =2
1/2 x - 1/3 y =0

d) 1/3 x + 2/5 y = 4
1/2 x - 1/10 y = -1

e) 1/2(x-4) - 2 (y-3)= 11
3(x -1)- 1/3 (y+6) = 2


Prosze, błagam rozwiązcie ... pomocy ;)

1

Odpowiedzi

2010-02-21T15:05:28+01:00
A) 2(3x + 4y) - 7 (y-x) = -37
4(x +2) - (y-4) = -2

6x + 8y - 7y + 7x = -37
4x + 8 - y + 4 = -2

13x + y = -37
4x - y = -2 - 4 - 8

13x + y = -37
4x - y = -14

17x = -51 |:17
x = -3

4x - y = -14
-12 - y = -14
-y = -14 + 12
-y = -2
y = 2

b) 2/3 x + 1/2 y = 5
1/3 x - 1/3 y = -1

⅔x + ½y = 5
⅓x - ⅓y = -1 |*(-2)

⅔x + ½y = 5
-⅔x + ⅔y = 2

½y + ⅔y = 7 |*6
3y + 4y = 42
7y = 42 |:7
y = 6

⅔x + ½y = 5
⅔x = 5 - 3½
⅔x = 1½ |*3
2x = 4½ |:2
x = 2¼


c) 1/4 x +1/6 y =2
1/2 x - 1/3 y =0

¼x + ⅙y = 2 |*(-2)
½x - ⅓y = 0

-½x - ⅓y = -4
½x - ⅓y = 0

-⅓y - ⅓y = -4
-⅔y = -4 |*3
-2y = -12 |:(-2)
y = 6

½x - ⅓y = 0
½x = ⅓y
½x = ⅓*6
½x = 2 |*2
x = 4

d) 1/3 x + 2/5 y = 4
1/2 x - 1/10 y = -1

⅓x + ⅖y = 4 |*(-6)
½x - 1/10y = -1 |*4

-2x - 12/5y = -24
2x - 4/10y = -4

-12/5y - 2/5y = -28
-14/5y = -28 |*5
-14y = -140 |:(-14)
y = 10

⅓x + ⅖y = 4
⅓x = 4 - ⅖*10
⅓x = 4 - 20/5
⅓x = 4 - 4
⅓x = 0
x = 0 (sprawdź wynik w odpowiedziach...)

e) 1/2(x-4) - 2 (y-3)= 11
3(x -1)- 1/3 (y+6) = 2

½x - 2 - 2y + 6 = 11
3x -3 - ⅓y - 2 = 2

½x - 2y = 11 + 2 - 6
3x - ⅓y = 2 + 2 + 3

½x - 2y = 7
3x - ⅓y = 7 |*(-6)

½x - 2y = 7
-18x + 2y = -42

-17½x = -35 |:(-17½)
x = 2

½x - 2y = 7
-2y = 7 - ½x
-2y = 7 - 1
-2y = 6 |:(-2)
y = -3