Question

4х - y= 3 2x^2+y^2=3 система

Answer 1

{4х - y= 3
{2x^2+y^2=3

{y=4x-3
{2x²+(4x-3)²=3

{y=4x-3
{2x²+16x²-24x+9-3=0

{y=4x-3
{18x²-24x+6=0

{y=4x-3
{3x²-4x+1=0        D=16-12=4      x₁=(4+2)/6=1    x₂=(4-2)/6=1/3
                                                    y₁=4-3=1          y₂=4/3-3=-5/3

ответ   (1;1)   (1/3;-5/3)

Answer 2

To solve the system of equations, we'll use the substitution method or elimination method. Let's use the substitution method:

Step 1: Solve one of the equations for one variable in terms of the other. Step 2: Substitute the expression from Step 1 into the other equation. Step 3: Solve the resulting equation for the variable found in Step 1. Step 4: Substitute the value of the variable found in Step 3 back into one of the original equations and solve for the other variable. Step 5: Check the solution by verifying if it satisfies both equations.

Let's start solving the system:

1. 4x - y = 3 ........... (Equation 1)
2. 2x^2 + y^2 = 3 ....... (Equation 2)

Step 1: Solve Equation 1 for y: y = 4x - 3

Step 2: Substitute the value of y from Step 1 into Equation 2: 2x^2 + (4x - 3)^2 = 3

Step 3: Expand and simplify Equation 2: 2x^2 + (16x^2 - 24x + 9) = 3 2x^2 + 16x^2 - 24x + 9 = 3 18x^2 - 24x + 9 - 3 = 0 18x^2 - 24x + 6 = 0

Step 4: Solve the quadratic equation for x. We can divide the entire equation by 6 to simplify it: 3x^2 - 4x + 1 = 0

Now, let's factor this quadratic equation: (3x - 1)(x - 1) = 0

Setting each factor to zero and solving for x:

1. 3x - 1 = 0 3x = 1 x = 1/3

2. x - 1 = 0 x = 1

So, we have two potential values for x: x = 1/3 and x = 1.

Step 5: Now that we have potential values for x, let's find the corresponding y-values using Equation 1 (y = 4x - 3):

For x = 1/3: y = 4(1/3) - 3 y = 4/3 - 3 y = 4/3 - 9/3 y = -5/3

For x = 1: y = 4(1) - 3 y = 4 - 3 y = 1

So, we have two potential solutions to the system of equations: (x = 1/3, y = -5/3) and (x = 1, y = 1).

To verify the solutions, we can substitute them back into both original equations:

For (x = 1/3, y = -5/3): 4(1/3) - (-5/3) = 3 4/3 + 5/3 = 3 9/3 = 3 (True)

2(1/3)^2 + (-5/3)^2 = 3 2/9 + 25/9 = 3 27/9 = 3 (True)

For (x = 1, y = 1): 4(1) - 1 = 3 4 - 1 = 3 3 = 3 (True)

2(1)^2 + 1^2 = 3 2 + 1 = 3 3 = 3 (True)

Both solutions satisfy the system of equations, so the final answer is: x = 1/3, y = -5/3, and x = 1, y = 1.