Question

Решить систему уравнений: x^2+3y=1 |x|-y=3 уравнения в скобке находятся*

Answer 1

{x²+3y=1
{|x|-y=3⇒y=|x|-3
подставим в 1
x²+3|x|-9=1
x²+3|x|-10=0
1)x<0
x²-3x-10=0
x1+x2=3 U x1*x2=-10
x1=-2⇒y=|-2|-3=2-3=-1
x2=5 не удов усл
2)x≥0
x²+3x-10=0
x1+x2=-3 U x1*x2=-10
x1=-5 не удов усл
x2=2⇒y=|2|-3=-1
Ответ (-2;-1);(2;-1)

Answer 2

Solving the System of Equations

To solve the given system of equations:

1. Equation 1: x^2 + 3y = 1 2. Equation 2: |x| - y = 3

We can use various methods, such as substitution or elimination, to find the values of x and y that satisfy both equations.

Let's solve the system step by step:

1. Start with Equation 2, which involves an absolute value. We can split it into two cases: one when x is positive and another when x is negative.

- Case 1: x is positive (x > 0) In this case, the equation becomes x - y = 3.

- Case 2: x is negative (x < 0) In this case, the equation becomes -x - y = 3.

2. Now, let's solve each case separately:

- Case 1: x - y = 3 Rearrange the equation to solve for y: y = x - 3

- Case 2: -x - y = 3 Rearrange the equation to solve for y: y = -x - 3

3. Now, let's substitute the values of y from each case into Equation 1 (x^2 + 3y = 1) and solve for x:

- Case 1: Substitute y = x - 3 into Equation 1: x^2 + 3(x - 3) = 1 Simplify the equation: x^2 + 3x - 9 = 1 Rearrange the equation: x^2 + 3x - 10 = 0

- Case 2: Substitute y = -x - 3 into Equation 1: x^2 + 3(-x - 3)