# Verify if the system has solutions? y= square root (25-x^2) x=square root (400-25y^2)/4

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### 2 Answers

To verify if the system has solutions:

y= square root (25-x^2).(1)

x=square root (400-25y^2)/4...(2).

From (1): we get y^2 = 25 - x^2

So x^2 +y^2 = 25......(3)

From (2) x^2 = (400 - 25y^2)/16.

So 16 x^2 = 400 -25 y^2.

16x^2 +25 y^2 = 400...(4).

16*(3) - (4) : -9y^2 = 25*16- 400. So -9y^2 = 0, Or y = 0.

Therefore put y ^2 = 0 in (3): x^2 = 25. So x= 5, or -5.

**Therefore the solution is x = 5 or x = -5 and y = 0**.

We'll raise to square the first equation, to eliminate the square root:

y^2 = 25 - x^2 (1)

We'll raise to square the 2nd equation, to eliminate the square root:

x^2 = (400-25y^2)/16

16x^2 = 400-25y^2 (2)

We'll substitute y^2 in the second equation:

16x^2 + 25(25 - x^2) = 400

We'll remove the brackets:

16x^2 + 625 - 25x^2 - 400 = 0

We'll combine like terms:

-9x^2 = -225

We'll divide by -9:

x^2 = 25

x1 = 5

x2 = -5

We'll substitute x1 and x2 in the first equation:

y1^2 = 25 - 25

y1^2 = 0

y1 = y2 = 0

**The system has 2 solutions and they are {5 ; 0} and {-5 ; 0}.**