Question

{х+у=3,{х2+у2=29 решите систему уравнения

Answer 1

{х+у=3,

{х2+у2=29

y = 3-x

x²+(3-x)² = 29

x² + 9 - 6x + x² = 29

2x² - 6x - 20 = 0

x² - 3x - 10 = 0

(x+2)(x-5) = 0

x¹ = -2, x² = 5

Answer 2

To solve the system of equations:

1. х + у = 3
2. х^2 + у^2 = 29

We can use the method of substitution or elimination. I'll use the substitution method:

Step 1: Solve the first equation (1) for one variable in terms of the other: From equation (1), we have: х = 3 - у.

Step 2: Substitute this value of х into equation (2): (3 - у)^2 + у^2 = 29.

Step 3: Expand and simplify the equation: 9 - 6у + у^2 + у^2 = 29, 2у^2 - 6у - 20 = 0.

Step 4: Solve the quadratic equation for у. Let's factor it, if possible: 2(y^2 - 3y - 10) = 0.

Step 5: Factor the quadratic equation: 2(y - 5)(y + 2) = 0.

Step 6: Set each factor to zero and solve for у: y - 5 = 0 => y = 5, y + 2 = 0 => y = -2.

Step 7: Now that we have two potential values of у, we can find the corresponding values of х using equation (1):

For y = 5: х = 3 - 5 = -2.

For y = -2: х = 3 - (-2) = 5.

So, the solution to the system of equations is: х = 5, у = -2, х = -2, у = 5.