Question

pomoze ktos?

układ rownan

rozwiazac metoda algebraiczną bez graficznej

1 { x do kwadratu - 4x + 3 - y = 0

{ 1/2 y + x= 3

2 { y = 2x + 3

{ 2x do kwadr - 4x - y + 3 = 0

3 { 2y + x do kwadr - 2 x + 5 = 0

{ 2y + 2 - x = 0

4 { y - x do kwadr - 5x - 8 = 0

{ y - x = 3

5 { y - x do kwadr + x + 2 = 0

{ y - 2x - 2 = 0

{

{ klamra

/ ułamek

Answer 1

1.

x^2 - 4 x + 3  - y = 0

(1/2) y + x = 3 / * 2

--------------------------------

y = 6 - 2 x

x^2 - 4 x + 3 - ( 6 -2 x) = 0

-------------------------------

x^2 - 2 x - 3 = 0

delta = ( -2)^2 - 4*1*(-3) = 4 + 12 = 16

p (delty) = 4

x1 = [ 2 - 4]/2 = - 1

x2 = [ 2 + 4]/2 = 3

zatem

y1 = 6 - 2*( -1) = 6 + 2 = 8

y2 = 6 - 2*3 = 6 - 6 = 0

Odp.  x = - 1 ,  y = 8     lub  x = 3 , y = 0

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2.

y = 2 x + 3

2 x^2 - 4 x - y + 3 = 0

-----------------------------------

2 x^2 - 4 x - ( 2 x + 3) + 3 = 0

y = 2 x + 3

-----------------

2 x^2 - 4 x -2 x - 3 + 3 = 0

2 x^2 - 6 x = 0

2x*( x -3) = 0

x = 0  lub  x - 3 = 0

x = 0   lub  x = 3

y = 2*0 + 3 = 3  lub  y = 2*3 + 3 = 9

Odp.  x = 0, y = 3    lub  x = 3,  y = 9

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3.

2 y + x^2 -2 x + 5 = 0

2 y + 2 - x = 0

----------------------

2 y = x - 2

x - 2 + x^2 -2 x + 5 = 0

---------------------------------

y = 0,5  x - 1

x^2 - x + 3 = 0

---------------------------

delta = ( -1)^2 - 4*1*3 = 1 - 12 = - 11 < 0  - brak rozwiązań

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4.

y - x^2 - 5 x - 8 = 0

y - x = 3

------------------------

y = x + 3

x + 3 - x^2 - 5 x - 8 = 0

--------------------------------

- x^2 - 4 x - 5 = 0

x^2 + 4 x + 5 = 0

delta = 4^2 - 4*1*5 = 16 - 20 = - 4 < 0   - brak rozwiązań w liczbach rzeczywistych

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5.

y - x^2 + x + 2 = 0

y -2 x - 2 = 0

--------------------------

y = 2 x + 2

2 x + 2 - x^2 + x + 2 = 0

---------------------------------

- x^2 +3 x + 4 = 0

x^2 -3 x - 4 = 0

delta = ( -3)^2 - 4*1*( -4) = 9 + 16 = 25

p( delty) = 5

x = [ 3 - 5]/2 = - 1      lub  x = [ 3 + 5 ]/2 = 4

y = 2*( -1) + 2 = -2 + 2 = 0    lub  y = 2*4 + 2 = 10

Odp. x = - 1,  y = 0   lub  x = 4,  y = 10

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