Solving Quadratic Systems

x^2+y^2=10
2x+y=4

Please show work I need to know how to do it

since we know y = 4-2x, use that:

x^2+(4-2x)^2 = 10
5x^2 - 16x + 6 = 0

Now it's just the normal routine for solving a quadratic equation.

how do I solve the answer

from the type of problem that you posted, you surely MUST know how to solve quadratic equations.

Steve took you as far as 5x^2 - 16x + 6 = 0 and made the same assumption I made.
I suggest using the quadratic formula.
(you will get two irrational answers)

To solve the quadratic system of equations:

1. Begin with the first equation x^2 + y^2 = 10, which represents a circle with a radius of √10 centered at the origin (0, 0) on a Cartesian plane.

2. Rearrange the second equation 2x + y = 4 to isolate one of the variables. Subtract 2x from both sides of the equation to get y = 4 - 2x.

3. Substitute the expression for y from the second equation into the first equation by replacing y with (4 - 2x):

x^2 + (4 - 2x)^2 = 10

4. Expand and simplify the equation:

x^2 + 16 - 16x + 4x^2 = 10
5x^2 - 16x + 6 = 0

5. Now, to solve the quadratic equation, use the quadratic formula:

x = (-b ± √(b^2 - 4ac))/(2a)

In this case, a = 5, b = -16, and c = 6.

x = (-(-16) ± √((-16)^2 - 4(5)(6)))/(2(5))
x = (16 ± √(256 - 120))/10
x = (16 ± √136)/10

6. Simplify further if possible:

x = (16 ± 2√34)/10
x = (8 ± √34)/5

7. Therefore, the x-values for the solutions of the quadratic system are:

x₁ = (8 + √34)/5
x₂ = (8 - √34)/5

8. Substituting these values of x into the second equation 2x + y = 4 to find the corresponding y-values:

For x₁:
2((8 + √34)/5) + y = 4
(16 + 2√34)/5 + y = 4
y = 4 - (16 + 2√34)/5
y = 20/5 - (16 + 2√34)/5
y = (20 - 16 - 2√34)/5
y = (4 - 2√34)/5

For x₂:
2((8 - √34)/5) + y = 4
(16 - 2√34)/5 + y = 4
y = 4 - (16 - 2√34)/5
y = 20/5 - (16 - 2√34)/5
y = (20 - 16 + 2√34)/5
y = (4 + 2√34)/5

9. The solutions to the quadratic system are:

(x₁, y₁) = ( (8 + √34)/5 , (4 - 2√34)/5 )
(x₂, y₂) = ( (8 - √34)/5 , (4 + 2√34)/5 )