What is x^2+y^2-12x+26=0,((x^2))/2-y^2= 7/2 ?

The solution to x^2+y^2-12x+26=0,((x^2))/2-y^2= 7/2 is

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x^2+y^2-12x+26=0,((x^2))/2-y^2= 7/2

x^{2} + y^{2} - 12 x + 26 = 0, \frac{( x ^{2} )}{2} - y^{2} = \frac{7}{2}

x = 5, x = 5, x = 3, x = 3, y = 3 y = - 3 y = 1 y = - 1

Solution steps

[x^{2} + y^{2} - 12 x + 26 = 0 \frac{( x ^{2} )}{2} - y^{2} = \frac{7}{2}]

Add the equations

x^{2} + y^{2} - 12 x + 26 = 0

x^{2} + y^{2} - 12 x + 26 + (\frac{( x ^{2} )}{2} - y^{2}) = 0 + \frac{7}{2}

Simplify

\frac{3 x ^{2} - 24 x}{2} + 26 = \frac{7}{2}

Solve

\frac{3 x ^{2} - 24 x}{2} + 26 = \frac{7}{2} : x = 5, x = 3

Plug the solutions

x = 5, x = 3

into

x^{2} + y^{2} - 12 x + 26 = 0

For

x^{2} + y^{2} - 12 x + 26 = 0

, subsitute

x

with

5 : y = 3, y = - 3

For

x^{2} + y^{2} - 12 x + 26 = 0

, subsitute

x

with

3 : y = 1, y = - 1

Verify solutions by plugging them into the original equations

Therefore, the final solutions for

x^{2} + y^{2} - 12 x + 26 = 0, \frac{( x ^{2} )}{2} - y^{2} = \frac{7}{2}

are

x = 5, x = 5, x = 3, x = 3, y = 3 y = - 3 y = 1 y = - 1

x + y + x y = - 11, x^{2} + y^{2} + x y = 13

x - y - 4 = 0, x y - 12 = 0

35 x^{2} - 12 x + 1 = 0, 20 y^{2} - 14 y + 2 = 0

y^{2} x - 7 y = 10, x = 6

x y = 4, 2 x + 3 y = 14

What is x^2+y^2-12x+26=0,((x^2))/2-y^2= 7/2 ?
The solution to x^2+y^2-12x+26=0,((x^2))/2-y^2= 7/2 is