|
Question 135089: solve the nonlinear system
y-x^2 =-1
2y-3x =0
Answer by algebrapro18(249)   (Show Source):
You can put this solution on YOUR website! well here we can use a substitution method similar to the one used in linear system of equations. The steps of this substitution will be:
1) Solve the first equation for y
2) Plug in the expression found in 1 into equation two and find the values of x
3) Plug in the value's found step 2 into the expression found in step 1 and find your values of y.
y-x^2 =-1
2y-3x =0
1)Solving equation 1 for y you get:
y = x^2-1
2)Now plugging that into equation two and solving you get:
2y-3x =0 --> plug in the expression for y
2(x^2-1)-3x=0 --> distribute the 2
2x^2-2-3x=0 --> reorder in standard form
2x^2-3x-2=0 --> Multiply 2 by 2 to get 4 and see what factors of 4 add to -3
2x^2-4x+x-2=0 --> Factor by grouping
2x(x-2)+1(x-2)=0 --> Factor by grouping
(2x+1)(x-2)=0 --> Apply the zero product property
2x+1 = 0 and x-2 = 0
2x = -1 and x = 2
x = -1/2 and x = 2
3)Now plugging x = -1/2 and x = 2 into the expression found in step one we find the values for y
y = x^2-1 and y = x^2-1
y = (-1/2)^2-1 and y = (2)^2-1
y = 1/4 -1 and y = 4-1
y = -3/4 and y = 3
So our answer is x = -1/2 y = -3/4 and x = 2 and y = 3 or in ordered pair notation (-1/2,-3/4) and (2,3).
Now lets check our solutions(note: you have to check both equations with each solution):
(-1/2,-3/4)
y-x^2 =-1
-3/4 - (-1/2)^2 = -1
-3/4 - 1/4 = -1
-1 = -1
2y-3x =0
2(-3/4)-3(-1/2)=0
-3/2 + 3/2 = 0
0 = 0
this checks out now lets check the other solution
(2,3)
y-x^2 =-1
3-(2)^2=-1
3-4=-1
-1 = -1
2y-3x =0
2(3)-3(2)=0
6 - 6 = 0
0 = 0
both of our solutions checked out so we know that we did the problem right. If this was to fast for you let me know just drop me an e-mail at dettmanb@uwplatt.edu
|
|
|
| |
