Quadratic systems

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

the first part I got this x^2+(4-2x)^2=10
and I don't know what steps comes next.
could you show work on how to do it I'm so confuse on this problem.

Hmmm. By algebra 2 you should surely know how to solve quadratic equations...

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

I didn't learn this in class so could you tell me which step goes next. I want to learn how to do it and be able to do my homework by myself.

OK. But this is Algebra I, which you have presumably already passed. But, moving right along,

5x^2 - 16x + 6 = 0

The quadratic formula says that the roots of

ax^2+bx+c are

x = (-b±√(b^2-4ac)/2a

So, for this problem, that means

x = (16±√(16^2-4*5*6))/10
= (16±√136)/10
= (16±2√34)/10
= (8±√34)/5

Or, you can always solve by completing the square. Recall that

(x-a)^2 = x^2+2ax+a^2

So, we have

5x^2 - 16x + 6 = 0
5x^2 - 16x = -6
5(x^2 - 16/5) = -6
5(x^2 - 16/5 + (8/5)^2) = -6 + 5(8/5)^2
5(x - 8/5)^2 = 34/5
x - 8/5 = ±√34/5
x = (8±√34)/5

Sure! Let's go through the steps to solve this quadratic system of equations.

Step 1: Rewrite the second equation.
You correctly rewrote the second equation as 2y + y = 4. However, there is a typo in your equation as 2y + y should equal 3y, not 4. So the correct equation should be 3y = 4.

Step 2: Choose a method to solve the system.
There are several methods to solve a system of equations, such as substitution, elimination, or graphing. In this case, since the first equation is already solved for x^2, we can use substitution to solve the system.

Step 3: Solve one equation for one variable.
We'll start by solving the second equation, 3y = 4, for y. Dividing both sides of the equation by 3 gives us y = 4/3.

Step 4: Substitute the value of the variable into the other equation.
Now that we have the value of y, we can substitute it into the first equation: x^2 + y^2 = 10. Plugging in y = 4/3, we get x^2 + (4/3)^2 = 10.

Step 5: Solve for x.
Now we can solve for x. We'll simplify the equation by squaring the fraction (4/3)^2. This gives us x^2 + 16/9 = 10. To get rid of the fraction, we can multiply both sides of the equation by 9, resulting in 9x^2 + 16 = 90.

Step 6: Rearrange the equation and solve for x^2.
To solve for x, we'll rearrange the equation by moving the constant term to the right side: 9x^2 = 90 - 16, which simplifies to 9x^2 = 74.

Step 7: Solve for x.
To solve for x, we'll take the square root of both sides of the equation to eliminate the square. This gives us x = ± √(74/9). This is the exact solution for x.

Step 8: Calculate the approximate values of x and y.
To find the approximate values of x and y, we can use a calculator to evaluate the square root and divide it by 3: x ≈ ± 2.716. Plugging this approximate value of x into the second equation, we get y ≈ 4/3 ≈ 1.333.

So the solutions to the quadratic system are approximately (x, y) ≈ (2.716, 1.333) and (x, y) ≈ (-2.716, 1.333).

Remember to always check your solutions by substituting them back into the original equations to ensure they satisfy both equations.