# Week 12 – precalculus 11

Solving Systems of Equations Algebraically

In this case, we will be using liner-quadratic systems. A system is a series of two equations in which you can substitute one equation and put it in the other and use this to find the values for x, usually by factoring at one point. You can also find Y if you please. You can plug the variables you found back in to get points, some linear-quadratic systems have no answers, sometimes just one point, and can have two points. The answers or points are where the two shapes intersect. If there is only one answer then they only intersect once if there are two then they intersect twice.

We will show this process with the system
1: $y=3x -2$

2: $x^2+ 4x -2 =y$

We will plug the first one into the second one like so $x^2+ 4x -2 = 3x -2$

This will give us

We will now bring over 3x-2 to the left side. We do this because it is quadratic and this is the only way we can find x by factoring. By using the zero product law we can isolate x $x^2 + x =0$

This  looks a little different and harder to factor but it is not, we will just bring out x. $x(x+1) =0$

and so to make this zero we know that x is 0 and -1

We are not done here, however. We must find the y values and this will give us the points of intersection. We will plug 0 and -1 back into the equation to find y. I will use the first equation and start with 0=x

y=3x -2

insert x $y=3(0) -2$

then $y=-2$

So one of the points is (0,-2)

We will now find the next one using -1 $y=3(-1) -2$

then $y=-3 -2$

so $y=-5$

and the other point is (-1, -5)

So the two solutions or points of intersection are  (0,-2) and (-1,-5). This is how to solve a Linear-Quadratic system. We can also check our answers using desmos. 