Question

Решите системы:^x^2+y^2=74x+y=12ещеx^2+y^2=10x+y=4С решением, пожалуйста. Очень нужно

Answer 1

см.рис.

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Answer 2

System of Equations 1:

The given system of equations is:

Equation 1: x^2 + y^2 = 74 Equation 2: x + y = 12

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

From Equation 2, we can express x as x = 12 - y.

Substituting this value of x into Equation 1, we get:

(12 - y)^2 + y^2 = 74

Expanding and simplifying the equation, we have:

144 - 24y + y^2 + y^2 = 74

Combining like terms, we get:

2y^2 - 24y + 70 = 0

Dividing the equation by 2, we have:

y^2 - 12y + 35 = 0

Now, we can solve this quadratic equation for y. Factoring the equation, we get:

(y - 5)(y - 7) = 0

Setting each factor equal to zero, we have:

y - 5 = 0 --> y = 5 y - 7 = 0 --> y = 7

So, we have two possible values for y: y = 5 and y = 7.

Now, let's substitute these values back into Equation 2 to find the corresponding values of x.

For y = 5: x + 5 = 12 x = 12 - 5 x = 7

For y = 7: x + 7 = 12 x = 12 - 7 x = 5

Therefore, the solutions to the first system of equations are: (x, y) = (7, 5) and (x, y) = (5, 7).

System of Equations 2:

The given system of equations is:

Equation 1: x^2 + y^2 = 10 Equation 2: x + y = 4

Again, we can use the method of substitution or elimination to solve this system of equations. Let's use the substitution method.

From Equation 2, we can express x as x = 4 - y.

Substituting this value of x into Equation 1, we get:

(4 - y)^2 + y^2 = 10

Expanding and simplifying the equation, we have:

16 - 8y + y^2 + y^2 = 10

Combining like terms, we get:

2y^2 - 8y + 6 = 0

Dividing the equation by 2, we have:

y^2 - 4y + 3 = 0

Now, we can solve this quadratic equation for y. Factoring the equation, we get:

(y - 1)(y - 3) = 0

Setting each factor equal to zero, we have:

y - 1 = 0 --> y = 1 y - 3 = 0 --> y = 3

So, we have two possible values for y: y = 1 and y = 3.

Now, let's substitute these values back into Equation 2 to find the corresponding values of x.

For y = 1: x + 1 = 4 x = 4 - 1 x = 3

For y = 3: x + 3 = 4 x = 4 - 3 x = 1

Therefore, the solutions to the second system of equations are: (x, y) = (3, 1) and (x, y) = (1, 3).

I hope this helps! Let me know if you have any further questions.