What is x+2y=1,2x^2+2y=42 ?

The solution to x+2y=1,2x^2+2y=42 is

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

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

(x = - \frac{- 1 + 329}{4}, x = \frac{1 + 329}{4}, y = \frac{3 + 329}{8} y = \frac{3 - 329}{8})

Solution steps

[x + 2 y = 1 2 x^{2} + 2 y = 42]

Subtract the equations

x + 2 y = 1

x + 2 y - (2 x^{2} + 2 y) = 1 - 42

Simplify

- 2 x^{2} + x = - 41

Solve

- 2 x^{2} + x = - 41 : x = - \frac{- 1 + 329}{4}, x = \frac{1 + 329}{4}

Plug the solutions

x = - \frac{- 1 + 329}{4}, x = \frac{1 + 329}{4}

into

x + 2 y = 1

For

x + 2 y = 1

, subsitute

x

with

- \frac{- 1 + 329}{4} : y = \frac{3 + 329}{8}

For

x + 2 y = 1

, subsitute

x

with

\frac{1 + 329}{4} : y = \frac{3 - 329}{8}

Verify solutions by plugging them into the original equations

Therefore, the final solutions for

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

are

(x = - \frac{- 1 + 329}{4}, x = \frac{1 + 329}{4}, y = \frac{3 + 329}{8} y = \frac{3 - 329}{8})

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

x + y = 1, y - 15 = - x^{2} + 4 x

y^{2} = 9 x, x - y + 2 = 0