system of equations

"); grf.document.close(); } function expcls(){ grf.window.close(); } function tip(){ popup=window.open('','','toolbar=no,width=350,height=100,location=top'); popup.document.open(); popup.document.writeln("the intermediate value theorem for continuous functions "); popup.document.writeln("For any continuous function f (x) in the interval [a,b] which satisfies f (a) * f (b) < 0 must have a zero of 'f ' in the interval [a,b] ."); popup.document.writeln(" "); popup.document.close(); } function tipclose(){ popup.window.close(); } function hlts(){ hlt=window.open('','','toolbar=no,width=350,height=250,location=top'); hlt.document.open(); hlt.document.writeln("highlights"); hlt.document.writeln("Advantages

Iteration scheme is simple to implement.
Iterations always converge for any continuous function f(x).

Disadvantages

Convergence process is slow.
Always one has to find the interval [a,b] which brackets the zero of f(x)= 0 more importantly f(a) * f(b) must be less than zero.

."); hlt.document.writeln(" "); hlt.document.close(); } function hltsclose(){ hlt.window.close(); }

System Of Non-Linear Equations - Examples

**Solve for x and y if**
Example 1	1 + y²- 4x²= 0 and 3 + 2x - x²- y²= 0	Solution
Example 2	y - x³ = 0 and 36 - 4x² - 9y² = 0	Solution
Example 3	3 + y - x² = 0 and 3 - xy = 0	Solution
Example 4	y - x³ + 3x² - 4x = 0 and y² - x - 2 = 0	Solution
Example 5	x² -2x - y + 0.5 = 0 and x² + 4y² - y = 0	Solution
Problems to workout

1. Solve for x and y if 1 + y²- 4x²= 0 and 3 + 2x - x²- y²= 0.

Let the initial approximation be (1, 2)

By Fixed-Point Iteration Method

Let x_i+1 = g₁(x_i, y_i) = (8x_i - 4x²_i + y²_i + 1)/8
y_i+1 = g₂(x_i, y_i) = (2x_i - x²_i + 4y_i + y²_i + 3)/4

i	0	1	2	3	4
x_i	1	1.125	1.117	1.116	1.116
y_i	2	2	1.996	1.997	1.997

So by Fixed-Point Iteration Method x = 1.116 & y = 1.997

[Solved Examples]

By Newton's Method

i	0	1	2	3
x_i	1	1.125	1.117	1.117
y_i	2	2	1.997	1.997

So by Newton's Method x = 1.117 & y = 1.997

[Solved Examples]

2. Solve for x and y if y - x³ = 0 and 36 - 4x² - 9y² = 0

Let the initial approximation is (1, 2)

By Fixed-Point Iteration Method

Let x_i+1 = g₁(x_i, y_i) = (4x_i - x³_i + y_i) /4
y_i+1 = g₂(x_i, y_i) = (-x2_i / 9) + (4y_i -y2_i)/4 + 1

i	0	1	2	3	4
x_i	1	1.125	1.234	1.22	1.222
y_i	2	1.889	1.823	1.823	1.823

So by Fixed-Point Iteration Method x = 1.222 & y = 1.823

[Solved Examples]

By Newton's Method

i	0	1	2	3	4
x_i	1	1.276	1.225	1.222	1.222
y_i	2	1.828	1.826	1.826	1.826

So by Newton's Method x = 1.222 & y = 1.826

[Solved Examples]

3.Solve for x and y if 3 + y - x² = 0 and 3 - xy = 0.

Let the initial approximation is (2, 1.5)

By Fixed-Point Iteration Method

Let x_i+1 = g₁(x_i, y_i) = (4x_i - x²_i + y_i+ 3)/4
y_i+1 = g₂(x_i, y_i) = (3 - x_iy_i + 2y_i)/2

i	0	1	2	3	4	5	6	7	8
x_i	2	2.125	2.121	2.098	2.101	2.105	2.104	2.104	2.104
y_i	1.5	1.5	1.406	1.415	1.431	1.431	1.425	1.426	1.426

So by Fixed-Point Iteration Method x = 2.104 & y = 1.426

[Solved Examples]

By Newton's Method

i	0	1	2	3	4
x_i	2	2.105	2.104	2.104	2.104
y_i	1.5	1.421	1.426	1.426	1.426

So by Newton's Method x = 2.104 & y = 1.426

[Solved Examples]

4. Solve for x and y if y - x³ + 3x² - 4x = 0 and y² - x - 2 = 0.

Let the initial approximation is (-0.5, -1.5)

By Fixed-Point Iteration Method

Let x_i+1 = g₁(x_i, y_i) = (y_i - x³_i + 3x²_i - 3x_i)/7
y_i+1 = g₂(x_i, y_i) = (y²_i + 2y_i - x_i - 2)/2

i	0	1	2	3	4	5
x_i	-0.5	-0.304	-0.247	-0.269	-0.27	-0.27
y_i	-1.5	-1.125	-1.340	-1.319	-1.315	-1.315

So by Fixed-Point Iteration Method x = -0.27 & y = -1.315

[Solved Examples]

By Newton's Method

i	0	1	2	3
x_i	-0.5	-0.299	-0.27	-0.27
y_i	-1.5	-1.317	-1.315	-1.315

So by Newton's Method x = -0.27 & y = -1.315

[Solved Examples]

5. Solve for x and y if x² -2x - y + 0.5 = 0 and x² + 4y² - y = 0

Let the initial approximation is (0, 1)

By Fixed-Point Iteration Method
Let x_i+1 = g₁(x_i, y_i) = (x²_i -y_i +0.5)/2
y_i+1 = g₂(x_i, y_i) = (-x²_i -4y²_i +8y_i +4)/8

i	0	1	2	3	4
x_i	0	-0.25	-0.219	-0.222	-0.222
y_i	1	1	0.992	0.994	0.994

So by Fixed-Point Iteration Method x = -0.222 & y = 0.994

[Solved Examples]

By Newton's Method

i	0	1	2	3	4	5	6
x_i	0	-0.036	0.063	0.12	0.143	0.148	0.148
y	1	0.571	0.367	0.271	0.235	0.226	0.226

So by Newton's Method x = 0.148 & y = 0.226

[Solved Examples]

Problems to Work-Out: