Quadratic systems: a line and a circle

What are the solutions to the system of equations y is equal to x plus 1, and x squared plus y squared is equal to 25? So let's first just visualize what we're trying to do. So let me try to roughly graph these two equations. So my y-axis, this is my x-axis. This right over here, x squared plus y squared is equal to 25, that's going to be a circle centered at 0 with radius 5. You don't have to know that to solve this problem, but it helps to visualize it. So if this is 5, this is 5, 5, 5. This right over here is negative 5. This right over here is negative 5. This equation would be represented by this set of points, or this is a set of points that satisfy this equation. So let me-- there you go. Trying to draw it as close to a perfect circle as I can. And then y equals x plus 1 is a line of slope 1 with a 1 y-intercept. So this is 1, 2, 3, 4. y-intercept is there and has a slope of 1 so it looks something like this. So when we're looking for the solutions, we're looking for the points that satisfy both. The points that satisfy both are the points that sit on both. So it's that point-- let me do it in green-- It's this point and it's this point right over here. So how do we actually figure that out? Well, the easiest way is to-- well, sometimes the easiest way is to substitute one of these constraints into the other constraint. And since they've already solved for y here, we can substitute y in the blue equation with x plus 1 with this constraint right over here. So instead of saying x squared plus y squared equals 25, we can say x squared plus, and instead of writing a y, we're adding the constraint that y must be x plus 1. So x squared plus x plus 1 squared must be equal to 25. And now, we can attempt to solve for x. So we get x squared plus-- now we square this. We'll get-- let me write in magenta-- we'll get x squared plus 2x plus 1, and that must be equal to 25. We have 2x squared-- now, I'm just combining these two terms-- 2x squared plus 2x plus 1 is equal to 25. Now, we could just use the quadratic formula to find the-- well, we have to be careful. We have to set this equal to 0, and then use the quadratic formula. So let's subtract 25 from both sides, and you could get 2x squared plus 2x minus 24 is equal to 0. And actually, let's-- just to simplify this-- let's divide both sides by 2, and you get x squared plus x minus 12 is equal to 0. And actually, we don't even have to use the quadratic formula, we can factor this right over here. What are two numbers that when we take their product we get negative 12, and when we add them, we get positive 1? Well, positive 4 and negative 3 would do the trick. So we have x plus 4 times x minus 3 is equal to 0. So x could be equal to-- well, if x plus 4 is 0 then that would make this whole thing true. So x could be equal to negative 4 or x could be equal to positive 3. So this right over here is a situation where x is negative 4. This right over here is a situation where x is 3. So we're almost done, we just have to find the corresponding y's. And for that, we can just resort to the simplest equation right over here, y is x plus 1. So in this situation when x is negative 4, y is going to be that plus 1. So y is going to be negative 3. This is the point negative 4 comma negative 3. Likewise, when x is 3, y is going to be equal to 4. So this is the point 3 comma 4. These are the two solutions to this non-linear system of equations.