Question

How many solutions does this system of linear equations have? Explain. 1= 2 =1+3 2

Answer 1

Answer: There are no real solutions.

Step-by-step explanation:

We have the equation:

x = y^2

y = x + 3/2

To solve this, we can start by noticing that x is already isolated in the first equation, then we can replace it in the second equation to get:

y = y^2 + 3/2

And now we can rewrite this as:

y^2 - y + 3/2 = 0

This is a quadratic equation, and the solutions can be calculated with Bhaskara's equation, the solutions are:

$y = \frac{-(-1) +- \sqrt{(-1)^2 - 4*3/2*1} }{2*1} = \frac{1 + \sqrt{-5} }{2}$

We have a negative number inside the square root, this means that the solutions are complex numbers.

Then we can conclude that the system of equations has no real solutions.