Question

X^2+y^2=4 x-y=2 x+y=7xy=12​

Answer 1

Объяснение:

1.x^2+y^2=4

x=2+y

(2+y)^2+y^2=4

4+4y+y^2+y^2=4

2y^2+4y+4=4

2y^2+4y=0 /2

y^2+2y=0

y(y+2)=0

y1=0 y+2=0

y2=-2

x1=2+0=2

x2=2-2=0

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x+y=7

xy=12

x=7-y

(7-y)y=12

7y-y^2=12 /-1

y^2-7y+12=0

D= 49-4*1*12=1

y1=3

y2=4

x1=7-3=4

x2=7-4=3

Answer 2

To solve the system of equations:

1. X^2 + y^2 = 4
2. x - y = 2
3. x + y = 7
4. xy = 12

We can use a combination of substitution and elimination methods. Let's go step by step:

From equation 3, we have x + y = 7. We can rearrange it to express y in terms of x: y = 7 - x.

Now, substitute this value of y in equation 2: x - (7 - x) = 2. Simplifying, we get 2x - 7 = 2, or 2x = 9. Solving for x, we find x = 9/2 or 4.5.

Next, substitute the value of x into equation 3 to find the corresponding y-value: y = 7 - x = 7 - 4.5 = 2.5.

So, one solution to the system of equations is x = 4.5 and y = 2.5.

Now, let's substitute these values into equation 4 to check if it holds: xy = 4.5 * 2.5 = 11.25, which is not equal to 12. Therefore, the solution we found does not satisfy equation 4.

This means that there is no solution that satisfies all four equations simultaneously. The system of equations is inconsistent.