Question

Система 1)х-у=5 2у+5х*=-3 2)х*-у=-2 2х+у=2 3)х+у=6 5х-2у=9 4)х-6у=-2 2х+3у=11 5)4х-3у=-1 х-5у=4

Answer 1

1) 
$\left \{ {{x-y=5} \atop {2y+5x^2 = -3}} \right.$
$x=5+y$
$2y+5*(5+y)^{2}+3=0$
$2y+5*(5^2+10y+y^2)+3=0$
$2y+125+50y+5y^2+3=0$
$5y^2+52y+128=0$
$D=b^2-4ac=2704-4*5*128=144$
$y_{1} = \frac{-b- \sqrt{D}}{2a} = \frac{-52-12}{10}=- 6,4$
$y_{1} = \frac{-b+ \sqrt{D}}{2a} = \frac{-52+12}{10}=-4$
$x _{1} =5-6,4=-1,4$
$x_{2} =5-4=1$

3)
$\left \{ {{x+y=6} \atop {5x-2y=9}} \right.$
x = 6 - y
5*(6 - y) - 2y = 9
30 - 5y - 2y = 9
-7y = -21
y = 3
x = 6 - 3 = 3

4)
$\left \{ {{x - 6y=-2} \atop {2x+3y=11}} \right.$
x = 6y - 2
2*(6y - 2) + 3y = 11
12y - 4 + 3y = 11
15y = 15
y = 1
x = 6*1-2 = 4

5)
$\left \{ {{4x-3y=-1} \atop {x-5y=4}} \right.$
x = 5y + 4
4*(5y+4) - 3y = -1
20y + 16 - 3y = -1
17y = -17
y = -1
x = 5 * (-1) + 4 = -1

2)
$\left \{ {{x^2-y=-2} \atop {2x+y=2}} \right.$
$y=2-2x$
$x^2-(2-2x)=-2$
$x^2-2+2x+2=0$
$x^2+2x=0$
$x*(x+2)=0$
$x _{1} = 0$
$x_{2} +2=0$
$x_{2} =-2$