System with Non-Linear Equations
So in today's class we learned about non-linear equations and this question was given to solve for:1) y=x-1
y=x2-6x-9
^ There are TWO variables so that means you should have TWO equations.
To solve it, substitution is the best method to use since y=x-1 already.
Now plug in x-1 as the value for the second equation:
x-1=x2-6x-9
Put everything on one side of the equation:
0=x2-7x+10
Factor ! :
0=(x-2)(x-5)
There are TWO values so plug in the values to one of the original equations:
x=2 or 5
ORIGINAL EQUATION: y=x-1
x=2
y=2-1
y=1 --> (2,1)
or
x=5
y=5-1
y=4 --> (5,4)
Solution: (2,1) or (5,4)
NOTE: You will have two points where it intersects.
Like so..
The second example was:
2) x2-y2=5
xy-6=0
Again substitution is going to be needed in order to solve the equation.
Rewrite the second equation so that it equals to y:
xy-6=0
y=-6/x
Now substitute that into the first equation:
x2-y2=5
x2-(-6/x)2=5
Then eliminate the denominator:
x2(x2-36/x2=5)
x4-36=5x2
Factor it:
x4-5x2-36=0
(x2-9)(x2+4)=0
(x+3)(x-3)(x+4)=0
x=3 or -3
x+4 = imaginary root (4 will be a negative number)
Now plug in the values of x to find the points of y:
y=-6/3
y=-2 --> (3,-2)
y=-6/-3
y=2 --> (-3,2)
Solution: (3,-2) or (-3,2)
NOTE: Once again two points where it intersects:
YES ! I'm done. I was gonna pick Richard to get him back but then he chose ME again soo now I have to pick the LUCKY person that wins to be the next scribe.. and that person is PAULO ! Congrats and have fun. I heard and saw on the shoutbox that you haven't blogged sooo there you go! =D
OHHHH yeaaah..
Don't forget to do your homework:
Exercise 26 questions 1-8 & 14
posted by Rojuane @ 4:56 PM
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