The degrees of the given systems will be different. Means one equation may be quadratic, cubic or polynomial.
To solve this system of equation, we may use substitution.
Solve the nonlinear system. Justify your answer with a graph.
Problem 1 :
y^{2} = x − 3 and y = x - 3
Solution :
y^{2} = x − 3 -------(1)
and
y = x - 3 -------(2)
Apply (2) in (1)
(x - 3)^{2} = x - 3
x^{2} - 2x(3) + 3^{2} = x - 3
x^{2} - 6x + 9 = x - 3
x^{2} - 6x -x + 9 + 3 = 0
x^{2} - 7x+ 12 = 0
(x - 4) (x - 3) = 0
Equating each factor to 0, we get
x = 4 and x = 3
When x = 4 y = 4 - 3 y = 1 |
When x = 3 y = 3 - 3 y = 0 |
So, the point of intersections are (4, 1) and (3, 0).
Check with graph :
Problem 2 :
y^{2} = 4x + 17 and y = x + 5
Solution :
y^{2} = 4x + 17 -----(1)
and
y = x + 5 -----(2)
Applying (2) in (1), we get
(x + 5)^{2} = 4x + 17
x^{2} + 2x(5) + 5^{2} = 4x + 17
x^{2} + 10x +25 = 4x + 17
x^{2} + 10x - 4x +25 - 17 = 0
x^{2} + 6x + 8 = 0
(x + 2)(x + 4) = 0
Equating each factor to zero, we get
x = -2 and x = -4
When x = -2 y = -2 + 5 y = 3 |
When x = -4 y = -4 + 5 y = 1 |
So, the point of intersections are (-2, 3) and (-4, 1)
Check with graph :
Problem 3 :
x^{2} + y^{2} = 4
y = x - 2
Solution :
x^{2} + y^{2} = 4 -------(1)
y = x - 2 -------(2)
Applying (2) in (1), we get
x^{2} + (x - 2)^{2} = 4
x^{2} + x^{2} - 2x(2) + 2^{2} = 4
2x^{2} - 4x + 4 = 4
Subtracting 4 on both sides, we get
2x^{2} - 4x = 0
2x(x - 2) = 0
2x = 0 and x - 2 = 0
x = 0 and x = 2
When x = 0, then y = 0 - 2 ==> -2
When x = 2, then y = 2 - 2 ==> 0
So, the points of intersections are (0, -2) and (2, 0).
Problem 4 :
x^{2} + y^{2} = 25
y = (-3/4)x + (25/4)
Solution :
x^{2} + y^{2} = 25 ----(1)
y = (-3/4)x + (25/4) -----(2)
Applying (2) in (1), we get
(x - 3) (x - 3) = 0
x = 3 and x = 3
When x = 3, then y = (-3/4)(3) + (25/4)
y = -9/4 + 25/4
y= 16/4
y = 4
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM