Question

решите систему уравнений:
(2х-1)²-(2х+3)²= 10у,
(у+2)²-(у-4)²= -30х.
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Answer 1

Відповідь:

х=2
у=-4

Пояснення:

-16х-10у=8
30х+12у=12

96х+60у=-48

-150х-60у=-60

-54х=-108

х=2

-16*2-10у=8

у=-4

Answer 2

Ответ:

$\displaystyle \left \{ {{x=2} \atop {y=-4}} \right.$

Объяснение:

$\displaystyle\\\left \{ {{(2x-1)^2-(2x+3)^2=10y} \atop {(y+2)^2-(y-4)^2=-30x}} \right. \\\\\\\left \{ {{(4x^2-4x+1)-(4x^2+12x+9)=10y} \atop {(y^2+4y+4)-(y^2-8y+16)=-30x}} \right. \\\\\left \{ {{4x^2-4x+1-4x^2-12x-9=10y} \atop {y^2+4y+4-y^2+8y-16=-30x}} \right.\\\\\left \{ {{-16x-8=10y} \atop {12y-12=-30x}} \right.\\\\\left \{ {{-8x-4=5y} \atop {6y-6=-15x}} \right.\\\\\left \{ {{-8x-4=5y} \atop {6y-6=-15x}} \right.$

$\displaystyle \left \{ {{x=-\dfrac58y-\dfrac12} \atop {6y-6=-15\cdot\Big(-\dfrac58y-\dfrac12\Big)}} \right.\\\\\left \{ {{x=-\dfrac58y-\dfrac12} \atop {6y-6=\dfrac{75}{8}y+\dfrac{15}{2}}} \right.\\\\\left \{ {{x=-\dfrac58y-\dfrac12} \atop {-\dfrac{27}{8}y-\dfrac{27}{2}=0}} \right.\\\\\left \{ {{x=-\dfrac58y-\dfrac12} \atop {y=-4}} \right.\\\\\left \{ {{x=\dfrac{5}{2}-\dfrac12} \atop {y=4}} \right.\\\\\left \{ {{x=2} \atop {y=-4}} \right.$