What are the solutions of the following system?

The solution set is (0, -5)

Step-by-step explanation:

In order to solve the following system, you need to solve the second equation for either variable.

2x + y = -5

y = -2x - 5

Now use that answer in for y in the previous equation.

x^2 + y^2 = 25

x^2 + (-2x - 5)^2 = 25

x^2 + 4x^2 + 20x + 25 = 25

5x^2 + 20x = 0

x = 0

Now we can use that to solve for y.

2x + y = -5

2(0) + y = -5

y = -5

(0,-5) and (-4,3)

Step-by-step explanation:

We have given two equations. We have to find their solution.

x²+y² = 25                                   eq(1)

2x+y = -5                                     eq(2)

From eq(2), separate y

y = -5-2x

Putting above value of y in eq(1), we have

x²+(-5-2x)² = 25

x²+25+4x²+20x = 25

adding like terms, we have

5x²+20x+25 = 25

Adding -25 to both sides of above equation, we have

5x²+20x = 0

Taking 5x common , we have

5x(x+4) = 0

Applying Zero-Product Property , we have

5x = 0 or x+4 = 0

x =0 or x = -4

Putting above values in eq (3) , we have

y = -5-2(0)   or y = -5-2(-4)

y = -5-0   or y = -5+8

y = -5 or y = 3  

The solution sets are (0,-5) and (-4,3).

-8/2y-8=5/y+4-7y+8/y^2-16

we move all terms to the left:

-8/2y-8-(5/y+4-7y+8/y^2-16)=0

we can not solve this equation

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step-by-step explanation: