Question

X2+3xy=544y2+xy=115​

Answer 1

x²+3xy=54

4y²+xy=115​

тупо складываем два уравнения

x²+3xy +4y²+xy=54 +115​

x²+4xy +4y²= 169​

(x + 2y)²= 13²

|x + 2y| = 13

1. x + 2y = 13

x = 13 - 2y​

4y² + (13 - 2y)y = 115​

2y² + 13y - 115 = 0​

D=13² + 4*2*115 = 1089 = 33²

y12 = (-13 +-33)/4 = 5   -23/2

x1 = 13 - 2*5 = 13 - 10 = 3

x2 = 13 - 2*(-23/2) = 13 + 23 = 36

2. x + 2y = -13

x = -13 - 2y​

4y² + (-13 - 2y)y = 115​

2y² - 13y - 115 = 0​

D=13² + 4*2*115 = 1089 = 33²

y12 = (13 +-33)/4 = -5   23/2

x1 = 13 + 2*5 = 13 + 10 = 23

x2 = 13 - 2*(23/2) = 13 - 23 = 10

Ответ {3, 5} {36, -23/2} {23, -5} {10, 23/2}

Answer 2

To solve the system of equations:

Equation 1: x^2 + 3xy = 544 Equation 2: y^2 + xy = 115

We can solve this system by using substitution or elimination method. Let's use the substitution method:

1. Solve Equation 2 for x: xy = 115 - y^2 (rearrange Equation 2) x = (115 - y^2) / y

2. Substitute the value of x in Equation 1: (115 - y^2)^2 / y^2 + 3(115 - y^2) = 544

Now, we can simplify and solve this equation to find the values of y.