Question

1) решите систему уравнений x-3y=7 xy=-22)решите систему уравнений x+2y=7 xy=63) решите систему уравнений x+y=7 x*y=6помогите пожалуйста очень срочно нужно

Answer 1

$1) \left \{ {{x-3y=7} \atop {xy=-2}} \right. \\ \left \{ {{x=3y+7} \atop {(3y+7)y=-2}} \right.$
решаем второе уравнение 3у²+7у+2=0,  D=b²-4ac=49-4·3·2=25
 у₁=(-7-5)/6=-2  или у₂=(-7+5)/6=-1/2=-0,5  тогда х₁=3у₁+7=-6+7=1    у₂=3у₂+7=-1,5+7=5,5
Ответ (1;2) (5,5; -0,5)

$2) \left \{ {{x+2y=7} \atop {xy=6}} \right.$
$\left \{ {{x=-2y+7} \atop {(-2y+7)y=6}} \right.$
-2y²+7y-6=0,
2y²-7y+6=0  D=(-7)²-4·2·6=49-48=1
y₁=(7-1)/4=3/2=1,5    y₂=(7+1)/4=2
x₁=-2y₁+7=-3+7=4      x₂=-2y₂+7=-2·2+7=3

Ответ (4; 1,5)  (3;2)

$\left \{ {{x+y=7} \atop {xy=6}} \right.$

$\left \{ {{y=7-x} \atop {x(7-x)=6}} \right.$

7x-x²-6=0,
x²-7x+6=0
D=49-24=25
x₁=(7+5)/2=6     x₂=(7-5)/2=1
y₁=7-x₁=7-6=1    y₂=7-x₂=7-1=6
ответ (6;1)(1;6)

Answer 2

System of Equations 1: x - 3y = 7, xy = -22

To solve the system of equations: 1. We can start by solving the first equation for x in terms of y. - Rearranging the equation x - 3y = 7, we get x = 7 + 3y. 2. Substituting this expression for x into the second equation, we have (7 + 3y)y = -22. - Expanding the equation, we get 7y + 3y^2 = -22. 3. Rearranging the equation, we have 3y^2 + 7y + 22 = 0. - This is a quadratic equation in terms of y. 4. We can solve this quadratic equation using the quadratic formula: y = (-b ± √(b^2 - 4ac)) / (2a), where a = 3, b = 7, and c = 22. - Plugging in the values, we get y = (-7 ± √(7^2 - 4 * 3 * 22)) / (2 * 3). - Simplifying further, we have y = (-7 ± √(49 - 264)) / 6. - Since the discriminant (b^2 - 4ac) is negative, the quadratic equation has no real solutions. - Therefore, there are no real values for y that satisfy the system of equations.

Conclusion: The system of equations x - 3y = 7 and xy = -22 has no real solutions.