[x+2y=1 [x+y²=4

To solve this system of equations, we can use the substitution method.

First, let's solve the first equation for x:

x + 2y = 1

x = 1 - 2y

Now, substitute this value of x into the second equation:

1 - 2y + y^2 = 4

Rearrange this equation to obtain a quadratic equation:

y^2 - 2y + 3 = 0

Now we can solve this quadratic equation by factoring or using the quadratic formula. However, it cannot be factored easily, so we will use the quadratic formula:

y = (-(-2) ± sqrt((-2)^2 - 4(1)(3))) / (2(1))

Simplifying, we have:

y = (2 ± sqrt(4 - 12)) / 2

y = (2 ± sqrt(-8)) / 2

Since the square root of a negative number is not defined in the real number system, there are no real solutions for y in this system of equations. Therefore, the system has no solution.