What is 2x+y=1,x^2+y^2=1 ?

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

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2x+y=1,x^2+y^2=1

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

(x = 0, x = \frac{4}{5}, y = 1 y = - \frac{3}{5})

Solution steps

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

Isolate

x

for

2 x + y = 1 : x = \frac{1 - y}{2}

Plug the solutions

x = \frac{1 - y}{2}

into

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

For

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

, subsitute

x

with

\frac{1 - y}{2} : y = 1, y = - \frac{3}{5}

Plug the solutions

y = 1, y = - \frac{3}{5}

into

2 x + y = 1

For

2 x + y = 1

, subsitute

y

with

1 : x = 0

For

2 x + y = 1

, subsitute

y

with

- \frac{3}{5} : x = \frac{4}{5}

Verify solutions by plugging them into the original equations

Therefore, the final solutions for

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

are

(x = 0, x = \frac{4}{5}, y = 1 y = - \frac{3}{5})

x - y = 5, x^{2} + y = 53

f (x) = x^{2} - 2 x + 3, f (x) = - 2 x + 12

x^{2} + y^{2} = 2, 5 x + y = 5

4 x^{2} + y = 10, 0 yy + 9 = 9 (x + 1)

2 x + 3 y = 7, (a - b) x + (a + b) y = 3 a + b - 2