To find the solutions to each equation, we'll first have to set Y to 0. We have:[tex]-3x-3 &=0\\ x^2-3x+5=0[/tex]
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:
[tex]x= \frac{-b\pm\sqrt{b^2-4ac}}{2a} [/tex]
For the sake of this question, the only part we're concerned with is the [tex]b^2-4ac[/tex] bit, which is referred to as the descriminant of the quadratic. If [tex]b^2-4ac \geq 1[/tex], 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 [tex]b^2-4ac=0[/tex], we only have one real solution, since adding and subtracting 0 from the [tex]-b[/tex] term have the same effect. If [tex]b^2-4ac \ \textless \ 0[/tex], 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 [tex]x^2-3x+5[/tex].
Looking at the coefficients and constant, we have a=1, b=-3, and c=5, which makes our discriminant
[tex](-3)^2-4(1)(5) = 9-20=-11[/tex]
-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.
