How many real solutions does the system have y=-3x-3 y=x^2-3x+5

To find the solutions to each equation, we'll first have to set Y to 0. We have:

To obtain the solutions for both, you'll have to solve them for x. The first equation is linear, so obtaining a solution there is fairly straightforward, and you're guaranteed to get one real solution there. The second equation is a little more involved, and we'll need the quadratic formula to settle that one out. You might remember from earlier math classes that the quadratic formula gives you a way to quickly find the roots of a particular quadratic equation. Here's the full thing:



For the sake of this question, the only part we're concerned with is the  bit, which is referred to as the descriminant of the quadratic. If , we'll have two real solutions, since we'll need to evaluate the square root of the term for both positive and negative outcomes. If , we only have one real solution, since adding and subtracting 0 from the  term have the same effect. If , we have no real solutions; the discriminant is nested inside a square root, and taking the square root of a negative number only produces imaginary results.

With that in mind, let's look at the discriminant of .

Looking at the coefficients and constant, we have a=1, b=-3, and c=5, which makes our discriminant



-11 is less than 0, so we have no real solutions to the second equation. This means that our first equation is the only one with a real solution, so the total number of real solutions for the system is 1.