00:01
What are the point or points that satisfy this given nonlinear system of equations? well, to solve this, we are going to use substitution.
00:10
I'm going to take my second equation and solve for either x or y.
00:16
In this case, i'm going to solve it for x.
00:19
Either one works.
00:20
It doesn't matter which one.
00:22
You just have to pick.
00:22
So i've picked x.
00:24
And i'm going to substitute that back into my first equation.
00:28
3x squared plus 2y squared equals 5.
00:31
And i'm using a value of x now of y minus 2.
00:36
So let's get rid of parentheses and set it all equal to 0.
00:41
Let's see if i can factor this equation.
00:43
Well, y minus 2 squared is y squared minus 4y plus 4.
00:53
And i will be setting that equal to 0, so i'm going to bring that 5 over.
00:57
Okay, 3y squared.
00:59
Well, i already have a 2y squared, so that's 5 y squared, minus 12y, y, and my my third one here is a plus 12, and i already have a minus 5, so that's a plus 7 equals 0.
01:13
This is a factorable trinomial.
01:17
It factors into 5y minus 7 and y minus 1.
01:23
I'm going to move it to the top where i've got a little more room.
01:26
If i set the first factor equal to 0, i get a value of y of 7 .5ths.
01:32
For my second factor, i get a value of y equaling 1.
01:37
Okay, let's take those values, and i'm going to plug them into this second equation.
01:43
It's the easier of the two.
01:45
I'm not squaring anything.
01:46
That's a good place to substitute back to find out what our x is.
01:51
So x equals y, in this case 7 fifths, minus 2.
01:56
Well, we do need a common denominator.
01:59
So this is 7 .5ths minus 10 fifths.
02:03
I get negative 3 fifths...