Question

X-y=8 x ^2+y^2=90 X x Y=?

Answer 1

х=8+y

x ^2+y^2=90

(8+y)^2+y^2=90

y=-4+(корень)29

y=-4-(корень)29

х=8-4+(корень)29

х=8-4-(корень)29

х=4+(корень)29

х=4-(корень)29

(х1 , y1)=(4+(корень)29,-4-(корень)29)

(х2 , y2)=(4-(корень)29,-4-(корень)29)

4+(корень)29-(-4-(корень)29)=8

(4+(корень)29)^2+(-4+(корень)29)^2=90

4-(корень)29-(-4-(корень)29)=8

(4-(корень)29)^2+(-4-(корень)29)^2=90

8=8

90=90

8=8

90=90

Ответ:

(х1 , y1)=(4+(корень)29,-4+(корень)29)

(х2 , y2)=(4-(корень)29,-4-(корень)29)

Answer 2

To find the value of x and y, we can use the given equations:

1. x - y = 8 ...(Equation 1)
2. x^2 + y^2 = 90 ...(Equation 2)

Let's solve this system of equations:

Step 1: Solve Equation 1 for x in terms of y: x = 8 + y

Step 2: Substitute the value of x from Step 1 into Equation 2: (8 + y)^2 + y^2 = 90

Step 3: Expand and simplify the equation: 64 + 16y + y^2 + y^2 = 90 2y^2 + 16y + 64 - 90 = 0 2y^2 + 16y - 26 = 0

Step 4: Divide the entire equation by 2 to simplify: y^2 + 8y - 13 = 0

Step 5: Use the quadratic formula to solve for y: The quadratic formula is given by: y = (-b ± √(b^2 - 4ac)) / 2a

For our equation, a = 1, b = 8, and c = -13. y = (-8 ± √(8^2 - 4 * 1 * -13)) / 2 * 1 y = (-8 ± √(64 + 52)) / 2 y = (-8 ± √116) / 2 y = (-8 ± 2√29) / 2 y = -4 ± √29

So, we have two possible values for y: y₁ = -4 + √29 y₂ = -4 - √29

Step 6: Now that we have the possible values for y, we can find the corresponding values of x using Equation 1 (x = 8 + y):

For y₁ = -4 + √29: x₁ = 8 + (-4 + √29) x₁ = 4 + √29

For y₂ = -4 - √29: x₂ = 8 + (-4 - √29) x₂ = 4 - √29

So the two sets of solutions for x and y are:

1. x = 4 + √29, y = -4 + √29
2. x = 4 - √29, y = -4 - √29

Note: The values of x and y are irrational numbers (due to the presence of √29), so the solutions are in decimal form.