Question

Y=x(квадрат) x-3y=-3

Answer 1

1)xy+x-3y-3=0

x-3+y(x-3)=0

y(x-3)=3-x

y=(3-x):(x-3)

 

(2) x^2 + ((3-x):(x-3))^2=10

(x^2(x-3)^2+(3-x)^2):(x-3)^2=0

((x^2-1)(x-3)^2):(x-3)^2 =0

x^2-1=0

x^2=1

x=-1, x=1

 

y^2=10-1

y^2=9

y= -3, y=3

 

Ответ - (-1;3) (-1;-3) (1;3) (1;-3)

Answer 2

To solve the system of equations, you can use the method of substitution or elimination. Let's use the substitution method.

Given the system of equations:

1. $Y = x^2$
2. $x - 3y = -3$

Step 1: Substitute the value of $Y$ from the first equation into the second equation: $x - 3(x^2) = -3$

Step 2: Simplify and rearrange the equation to a quadratic equation form: $x - 3x^2 = -3$ $3x^2 - x + 3 = 0$

Now, we have a quadratic equation in terms of $x$. To solve this quadratic equation, we can use the quadratic formula:

The quadratic formula is given by: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 3$, $b = -1$, and $c = 3$.

Step 3: Calculate the discriminant ($b^2 - 4ac$): $D = (-1)^2 - 4(3)(3) = 1 - 36 = -35$

Since the discriminant is negative ($-35$), the quadratic equation has no real solutions. This means that the system of equations has no real solutions as well. Geometrically, this means that the two graphs (a parabola and a straight line) do not intersect, so there is no common point that satisfies both equations simultaneously.