Question

Решите сисемы уравнений: 3x+2y=5 2x^2+3y=12 2(x+y)^2-7(x+y)+3=0 2x-3y=-1

Answer 1

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Answer 2

System of Equations

The given system of equations is:

1. 3x + 2y = 5 2. 2x^2 + 3y = 12 3. 2(x+y)^2 - 7(x+y) + 3 = 0 4. 2x - 3y = -1

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

Solving the System of Equations

We will use the substitution method to solve this system of equations.

From equation 1, we can express x in terms of y: 3x = 5 - 2y x = (5 - 2y)/3 Substituting this value of x into equation 2, we get: 2((5 - 2y)/3)^2 + 3y = 12 Simplifying this equation, we have: 2(25 - 20y + 4y^2)/9 + 3y = 12 Multiplying both sides by 9 to eliminate the denominator, we get: 2(25 - 20y + 4y^2) + 27y = 108 Expanding and rearranging the equation, we have: 50 - 40y + 8y^2 + 27y = 108 8y^2 - 13y + 58 = 0

Now, let's solve this quadratic equation for y. We can use the quadratic formula: y = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = 8, b = -13, and c = 58. Substituting these values into the quadratic formula, we get: y = (-(-13) ± √((-13)^2 - 4(8)(58))) / (2(8)) y = (13 ± √(169 - 1856)) / 16 y = (13 ± √(-1687)) / 16

Since the discriminant is negative, the solutions for y will be complex numbers. However, the given system of equations does not have complex solutions. Therefore, there is no solution to this system of equations.

Conclusion

The given system of equations 3x + 2y = 5, 2x^2 + 3y = 12, 2(x+y)^2 - 7(x+y) + 3 = 0, 2x - 3y = -1 does not have a solution.