Question

Решите систему уравнений: 4x+y=-3 -y-x^2=6 Также: 6x-y=2 -x+y=-1 Иии: 3x+y=3 y-x^2=-7 Срочно, пжжж

Answer 1

) 4x+y=-3
    -y-x^2=6

y=-3-4x
Подставляем во второе и решаем его: 3+4х-х^2=6
-x^2+4x-3=0  |умножаем на -1
x^2-4x+3=0
D=16-12 = 4
sqrtD = 2
x1 = (4+2)/2= 3
x2= (4-2)/2 = 1
х1 = 3, отсюда у1 = -15
х2= 1, отсюда у2 = -7
Ответ: (3,-15) (1,-7)

2)  6х-у=2
     -х+у=-1

у=х-1
6х-(х-1)=2
6х-х+1=2
5х=1
х=1/5, отсюда у= -4/5
Ответ (1/5, -4/5)

3) 3х+у=3
    у-х^2=-7

у=3-3х
3-3х-х^2=-7
-x^2-3x+10=0
x^2+3x-10=0
D=9+40=49
sqrtD=7
x1=(-3+7)/2=2 
x2=(-3-7)/2= -5

x1=2, отсюда у1=-3
х2=-5, отсюда у2= 18
Ответ: (2,-3) (-5,18)

Answer 2

1) 4x+y=-3
    -y-x^2=6

y=-3-4x
Подставляем во второе и решаем его: 3+4х-х^2=6
-x^2+4x-3=0  |умножаем на -1
x^2-4x+3=0
D=16-12 = 4
sqrtD = 2
x1 = (4+2)/2= 3
x2= (4-2)/2 = 1
х1 = 3, отсюда у1 = -15
х2= 1, отсюда у2 = -7
Ответ: (3,-15) (1,-7)

2)  6х-у=2
     -х+у=-1

у=х-1
6х-(х-1)=2
6х-х+1=2
5х=1
х=1/5, отсюда у= -4/5
Ответ (1/5, -4/5)

3) 3х+у=3
    у-х^2=-7

у=3-3х
3-3х-х^2=-7
-x^2-3x+10=0
x^2+3x-10=0
D=9+40=49
sqrtD=7
x1=(-3+7)/2=2 
x2=(-3-7)/2= -5

x1=2, отсюда у1=-3
х2=-5, отсюда у2= 18
Ответ: (2,-3) (-5,18)

Answer 3

System of Equations

The given system of equations is:

Equation 1: 4x + y = -3 Equation 2: -y - x^2 = 6

Equation 3: 6x - y = 2 Equation 4: -x + y = -1

Equation 5: 3x + y = 3 Equation 6: -y - x^2 = -7

To solve this system of equations, we can use various methods such as substitution, elimination, or matrix methods. Let's solve it using the substitution method.

Solving the System of Equations

We'll start by solving Equation 1 and Equation 2:

From Equation 1, we can express y in terms of x as: y = -3 - 4x. (Equation 7) Substituting Equation 7 into Equation 2, we get: -(-3 - 4x) - x^2 = 6.

Simplifying the equation, we have: 3 + 4x - x^2 = 6. Rearranging the terms, we get: x^2 - 4x + 3 = 0.

Now, we can solve this quadratic equation. Factoring it, we have: (x - 1)(x - 3) = 0. So, the solutions for x are: x = 1 and x = 3.

Substituting these values back into Equation 7, we can find the corresponding values for y:

For x = 1: y = -3 - 4(1) = -7. For x = 3: y = -3 - 4(3) = -15.

Therefore, the first solution to the system of equations is: (x, y) = (1, -7). And the second solution is: (x, y) = (3, -15).

Now, let's solve the remaining set of equations:

For Equation 3 and Equation 4, we can use the substitution method again:

From Equation 3, we can express y in terms of x as: y = 6x - 2. (Equation 8) Substituting Equation 8 into Equation 4, we get: -x + (6x - 2) = -1.

Simplifying the equation, we have: 5x - 2 = -1. Rearranging the terms, we get: 5x = 1.

So, the solution for x is: x = 1/5. Substituting this value back into Equation 8, we can find the corresponding value for y:

For x = 1/5: y = 6(1/5) - 2 = -1/5.

Therefore, the third solution to the system of equations is: (x, y) = (1/5, -1/5).

Lastly, let's solve Equation 5 and Equation 6:

From Equation 5, we can express y in terms of x as: y = 3 - 3x. (Equation 9) Substituting Equation 9 into Equation 6, we get: - (3 - 3x) - x^2 = -7.

Simplifying the equation, we have: -3 + 3x - x^2 = -7. Rearranging the terms, we get: x^2 - 3x + 4 = 0.

This quadratic equation does not have real solutions. Therefore, there are no real solutions for Equation 5 and Equation 6.

Summary of Solutions

The system of equations has the following solutions:

1. (x, y) = (1, -7) 2. (x, y) = (3, -15) 3. (x, y) = (1/5, -1/5)

Please note that the solutions are approximate values and may be rounded for simplicity.

I hope this helps! Let me know if you have any further questions.