Question

Решить систему уравнений: 3x-y=2 . x^2+xy+y^2=3 подробно

Answer 1

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Answer 2

(1) 3x-y=2

(2) x^2+xy+y^2=3

Из (1) y=3x-2

Значит, x^2+x(3x-2)+(3x-2)^2=3

x^2+3x^2-2x+9x^2-12x+4=3

13x^2-14x+1=0

D=14^2-4*13=196-52=144=12^2

$x_{1}=(14+12)/26=1, x_{2}=(14-12)/26=1/13$

$y_{1}=3-2=1; y_{2}= \frac{3}{13}-2=-\frac{23}{13}$

Answer 3

To solve the system of equations, we'll use the method of substitution. We'll first solve one equation for one variable and then substitute that expression into the other equation. Let's start:

Step 1: Solve the first equation for y in terms of x: 3x - y = 2 y = 3x - 2

Step 2: Substitute the expression for y from Step 1 into the second equation: x^2 + xy + y^2 = 3 x^2 + x(3x - 2) + (3x - 2)^2 = 3

Step 3: Simplify and solve for x: Expand the square in the last term: x^2 + x(3x - 2) + (3x - 2)(3x - 2) = 3 x^2 + 3x^2 - 2x + 9x^2 - 12x + 4 = 3 Combine like terms: 13x^2 - 14x + 4 = 3 Subtract 3 from both sides: 13x^2 - 14x + 1 = 0

Step 4: Solve the quadratic equation for x using the quadratic formula: The quadratic formula is given by: x = (-b ± √(b^2 - 4ac)) / 2a

In our equation, a = 13, b = -14, and c = 1. x = [14 ± √((-14)^2 - 4 * 13 * 1)] / 2 * 13 x = [14 ± √(196 - 52)] / 26 x = [14 ± √144] / 26 x = (14 ± 12) / 26

So, we have two potential solutions for x:

1. x = (14 + 12) / 26 = 26 / 26 = 1
2. x = (14 - 12) / 26 = 2 / 26 = 1 / 13

Step 5: Find the corresponding values of y for each x value: For x = 1: y = 3x - 2 y = 3(1) - 2 y = 3 - 2 y = 1

For x = 1/13: y = 3x - 2 y = 3(1/13) - 2 y = 3/13 - 26/13 y = (3 - 26) / 13 y = -23/13

So, the solutions to the system of equations are:

1. x = 1, y = 1
2. x = 1/13, y = -23/13