Question

(x + y = 6, y = x² ;

Answer 1

$\displaystyle\bf\\\left \{ {{x+y=6} \atop {y=x^{2} }} \right. \\\\\\\left \{ {{x^{2} +x-6=0} \atop {y=x^{2} }} \right. \\\\\\x^{2} +x-6=0\\\\D=1^{2} -4\cdot(-6)=1+24=25=5^{2} \\\\\\x_{1} =\frac{-1-5}{2} =\frac{-6}{2} =-3\\\\\\x_{2} =\frac{-1+5}{2} =\frac{4}{2} =2\\\\\\1)\\\\x=-3 \ \ \ \Rightarrow \ \ \ y=(-3)^{2} =9\\\\2)\\\\x=2 \ \ \ \Rightarrow \ \ \ y=2^{2} =4\\\\\\Otvet \ . \ (-3 \ ; \ 9) \ , \ (2 \ ; \ 4)$

Answer 2

To solve the system of equations:

1. x + y = 6
2. y = x²

You can substitute the expression for y from equation 2 into equation 1, which will allow you to solve for the values of x and y:

1. x + (x²) = 6

Now, you have a quadratic equation in terms of x. Rearrange it to isolate x:

x² + x - 6 = 0

To solve this quadratic equation, you can use the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

In this case, a = 1, b = 1, and c = -6. Plug these values into the formula:

x = (-1 ± √(1² - 4(1)(-6))) / (2(1)) x = (-1 ± √(1 + 24)) / 2 x = (-1 ± √25) / 2

Now, you have two possible values for x:

1. x = (-1 + 5) / 2 = 4/2 = 2
2. x = (-1 - 5) / 2 = -6/2 = -3

So, you have two potential solutions for the values of x: x = 2 and x = -3.

Now, you can find the corresponding values of y using equation 2:

For x = 2: y = (2)² = 4

For x = -3: y = (-3)² = 9

So, the solutions to the system of equations are:

1. When x = 2, y = 4
2. When x = -3, y = 9