What is 2y+x=1,x^2-2x+y^2-4y=8 ?

The solution to 2y+x=1,x^2-2x+y^2-4y=8 is

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2y+x=1,x^2-2x+y^2-4y=8

2 y + x = 1, x^{2} - 2 x + y^{2} - 4 y = 8

(x = 3, x = - \frac{13}{5}, y = - 1 y = \frac{9}{5})

Solution steps

[2 y + x = 1 x^{2} - 2 x + y^{2} - 4 y = 8]

Isolate

y

for

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

Plug the solutions

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

into

x^{2} - 2 x + y^{2} - 4 y = 8

For

x^{2} - 2 x + y^{2} - 4 y = 8

, subsitute

y

with

\frac{1 - x}{2} : x = 3, x = - \frac{13}{5}

Plug the solutions

x = 3, x = - \frac{13}{5}

into

2 y + x = 1

For

2 y + x = 1

, subsitute

x

with

3 : y = - 1

For

2 y + x = 1

, subsitute

x

with

- \frac{13}{5} : y = \frac{9}{5}

Verify solutions by plugging them into the original equations

Therefore, the final solutions for

2 y + x = 1, x^{2} - 2 x + y^{2} - 4 y = 8

are

(x = 3, x = - \frac{13}{5}, y = - 1 y = \frac{9}{5})

x + y = 10, x y = 40

a + b = 30, a - b = 20 \frac{a}{b}

f (x) = 2, x^{2} - 7 x + 3 = 12

x + y - z = 5, - x + y + z = 1

x + y = 7, x + z = 4 x + y + z