What is x^2+(y-1)2=13,2x+3y=3 ?

The solution to x^2+(y-1)2=13,2x+3y=3 is

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x^2+(y-1)2=13,2x+3y=3

x^{2} + (y - 1) 2 = 13, 2 x + 3 y = 3

(x = - 3, x = \frac{13}{3}, y = 3 y = - \frac{17}{9})

Solution steps

[x^{2} + (y - 1) \cdot 2 = 13 2 x + 3 y = 3]

Isolate

x

for

2 x + 3 y = 3 : x = \frac{3 - 3 y}{2}

Plug the solutions

x = \frac{3 - 3 y}{2}

into

x^{2} + (y - 1) 2 = 13

For

x^{2} + (y - 1) 2 = 13

, subsitute

x

with

\frac{3 - 3 y}{2} : y = 3, y = - \frac{17}{9}

Plug the solutions

y = 3, y = - \frac{17}{9}

into

2 x + 3 y = 3

For

2 x + 3 y = 3

, subsitute

y

with

3 : x = - 3

For

2 x + 3 y = 3

, subsitute

y

with

- \frac{17}{9} : x = \frac{13}{3}

Verify solutions by plugging them into the original equations

Therefore, the final solutions for

x^{2} + (y - 1) 2 = 13, 2 x + 3 y = 3

are

(x = - 3, x = \frac{13}{3}, y = 3 y = - \frac{17}{9})

x + y = 1, x^{2} - 2 x + 3 y = - 1

x + 2 y = 1, 2 x^{2} + 2 y = 42

x + 2 y = 2, x + 2 y = 6

8 a = 4, b = 3 c

x^{2} + y^{2} + 4 x - 6 y - 12 = 0, x^{2} + y^{2} - 8 x + 6 y + 12 = 0