Question

X+y=4x^2-2xy+y^2=64​

Answer 1

1)y=4-x

2)x²-2xy+y²=64

x²-2x(4-x)+(4-x)²=64

x²-8x+2x²+16-8x+x²=64

4x²-16x+16=64

x²-4x+4=16

x²-4x-12=0

D=b²-4ac=16+48=64

$x1 = \frac{4 + 8}{2 } = 6$

$x2 = \frac{4 - 8}{2} = - 2$

Если х1=6, то y1=4-6=-2

Если х2=-2, то y2=4-(-2)=6

Ответ : (6;-2),(-2;6)

Answer 2

To solve the system of equations, we need to find the values of 'x' and 'y' that satisfy both equations simultaneously. There are a few ways to approach this problem, but one common method is to use substitution or elimination.

Let's use the substitution method:

Step 1: Solve one equation for one variable in terms of the other variable. Step 2: Substitute the expression obtained in Step 1 into the other equation. Step 3: Solve the resulting equation for one variable. Step 4: Substitute the value found in Step 3 back into one of the original equations to find the corresponding value of the other variable.

Given equations:

1. x + y = 4
2. 4x^2 - 2xy + y^2 = 64

Step 1: Solve the first equation for x in terms of y. x = 4 - y

Step 2: Substitute the expression for x into the second equation. 4(4 - y)^2 - 2(4 - y)y + y^2 = 64

Step 3: Simplify and solve for y: 16 - 8y + y^2 - 8y + 2y^2 + y^2 = 64 4y^2 - 16y - 48 = 0

Step 4: Solve for y by factoring the quadratic equation: 4(y^2 - 4y - 12) = 0 (y^2 - 4y - 12) = 0

Now, we can use the quadratic formula to find the values of y:

y = [4 ± √(4^2 - 4 * 1 * (-12))]/(2 * 1) y = [4 ± √(16 + 48)]/2 y = [4 ± √64]/2 y = [4 ± 8]/2

This gives us two possible values for y:

1. y = (4 + 8)/2 = 12/2 = 6
2. y = (4 - 8)/2 = -4/2 = -2

Now that we have the values of y, we can find the corresponding values of x using the first equation (x + y = 4):

1. For y = 6: x + 6 = 4 x = 4 - 6 x = -2

2. For y = -2: x + (-2) = 4 x = 4 + 2 x = 6

So, the solutions to the system of equations are:

1. x = -2, y = 6
2. x = 6, y = -2