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 <em>descriminant </em>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 <em>no real solutions</em>; 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 <em>no real solutions</em> 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.