What is x^2+y^2=1,8y=-x+6 ?

The solution to x^2+y^2=1,8y=-x+6 is

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x^2+y^2=1,8y=-x+6

x^{2} + y^{2} = 1, 8 y = - x + 6

x = 6 - \frac{8 ( 48 + 29 )}{65}, x = 6 - \frac{8 ( 48 - 29 )}{65}, y = \frac{48 + 29}{65} y = \frac{48 - 29}{65}

Solution steps

[x^{2} + y^{2} = 1 8 y = - x + 6]

Isolate

x

for

8 y = - x + 6 : x = 6 - 8 y

Plug the solutions

x = 6 - 8 y

into

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

For

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

, subsitute

x

with

6 - 8 y : y = \frac{48 + 29}{65}, y = \frac{48 - 29}{65}

Plug the solutions

y = \frac{48 + 29}{65}, y = \frac{48 - 29}{65}

into

8 y = - x + 6

For

8 y = - x + 6

, subsitute

y

with

\frac{48 + 29}{65} : x = 6 - \frac{8 ( 48 + 29 )}{65}

For

8 y = - x + 6

, subsitute

y

with

\frac{48 - 29}{65} : x = 6 - \frac{8 ( 48 - 29 )}{65}

Verify solutions by plugging them into the original equations

Therefore, the final solutions for

x^{2} + y^{2} = 1, 8 y = - x + 6

are

x = 6 - \frac{8 ( 48 + 29 )}{65}, x = 6 - \frac{8 ( 48 - 29 )}{65}, y = \frac{48 + 29}{65} y = \frac{48 - 29}{65}

x + y = 1, 7 x^{2} + y^{2} = 1 3^{2}

a + b + z = 5, 3 z - x - y = 81

f (x) = x^{2} - 2 x + 3, f (x) = - 2 x \cdot 2

7 y = 2, y^{3} = 2

(x^{y} - y)^{2} + x^{y} - y = 2, x^{2 y} + 3 x^{y} = 10