What is x^2+y^2=2,5y-x+7=0 ?

The solution to x^2+y^2=2,5y-x+7=0 is

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x^2+y^2=2,5y-x+7=0

x^{2} + y^{2} = 2, 5 y - x + 7 = 0

(x = \frac{5 3 + 7}{26}, x = - \frac{5 3 - 7}{26}, y = \frac{- 35 + 3}{26} y = - \frac{35 + 3}{26})

Solution steps

[x^{2} + y^{2} = 2 5 y - x + 7 = 0]

Isolate

x

for

5 y - x + 7 = 0 : x = 5 y + 7

Plug the solutions

x = 5 y + 7

into

x^{2} + y^{2} = 2

For

x^{2} + y^{2} = 2

, subsitute

x

with

5 y + 7 : y = \frac{- 35 + 3}{26}, y = - \frac{35 + 3}{26}

Plug the solutions

y = \frac{- 35 + 3}{26}, y = - \frac{35 + 3}{26}

into

5 y - x + 7 = 0

For

5 y - x + 7 = 0

, subsitute

y

with

\frac{- 35 + 3}{26} : x = \frac{5 3 + 7}{26}

For

5 y - x + 7 = 0

, subsitute

y

with

- \frac{35 + 3}{26} : x = - \frac{5 3 - 7}{26}

Verify solutions by plugging them into the original equations

Therefore, the final solutions for

x^{2} + y^{2} = 2, 5 y - x + 7 = 0

are

(x = \frac{5 3 + 7}{26}, x = - \frac{5 3 - 7}{26}, y = \frac{- 35 + 3}{26} y = - \frac{35 + 3}{26})

x + y = 3, \neg 65 x + 119 y = 249

5 x - 6 y = 20, 4 x - 3 \frac{y}{4} x - 3 y = 23

x + y = 8, x y = \frac{25}{9}

p = - 20, 0 + \frac{1}{4} q^{2} p = 1000 - \frac{1}{2} q^{2}

2^{x} + 2^{y} = 40, 2^{x + y} = 256