What is (x)2x-y+1=3,x+7=-2y ?

The solution to (x)2x-y+1=3,x+7=-2y is

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(x)2x-y+1=3,x+7=-2y

(x) 2 x - y + 1 = 3, x + 7 = - 2 y

(x = - \frac{1}{8} - i \frac{47}{8}, x = - \frac{1}{8} + i \frac{47}{8}, y = - \frac{55}{16} + i \frac{47}{16} y = - \frac{55}{16} - i \frac{47}{16})

Solution steps

[(x) \cdot 2 x - y + 1 = 3 x + 7 = - 2 y]

Isolate

y

for

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

Plug the solutions

y = 2 x^{2} - 2

into

x + 7 = - 2 y

For

x + 7 = - 2 y

, subsitute

y

with

2 x^{2} - 2 : x = - \frac{1}{8} - i \frac{47}{8}, x = - \frac{1}{8} + i \frac{47}{8}

Plug the solutions

x = - \frac{1}{8} - i \frac{47}{8}, x = - \frac{1}{8} + i \frac{47}{8}

into

(x) 2 x - y + 1 = 3

For

(x) 2 x - y + 1 = 3

, subsitute

x

with

- \frac{1}{8} - i \frac{47}{8} : y = - \frac{55}{16} + i \frac{47}{16}

For

(x) 2 x - y + 1 = 3

, subsitute

x

with

- \frac{1}{8} + i \frac{47}{8} : y = - \frac{55}{16} - i \frac{47}{16}

Verify solutions by plugging them into the original equations

Therefore, the final solutions for

(x) 2 x - y + 1 = 3, x + 7 = - 2 y

are

(x = - \frac{1}{8} - i \frac{47}{8}, x = - \frac{1}{8} + i \frac{47}{8}, y = - \frac{55}{16} + i \frac{47}{16} y = - \frac{55}{16} - i \frac{47}{16})

x + y - z = 17, 0 x + z - y = 130 x

x - y = 5, x^{2} + y^{2} = 13

2 x + y = 5, x^{2} + y^{2} = 3

2 x - y = 3, x^{2} - x y = - 4

x + 3 y^{2} = 5, 3 x - y = 5 y