Q:

Use the Newton-Raphson method to determine the solution of the simultaneous nonlinear equations: y=βˆ’x2+x+0.75 y+5xy=x2 Use the initial guesses of x = y = 1.2, and iterate until the 4th iteration. (Round the final answers to five decimal places.) The values of x and y are as follows: iterationxy01.21.21 0.0290321.39412 3 0.239294

Accepted Solution

A:
Answer:Step-by-step explanation:Let's solve for y. βˆ’x2+x+0.75y+5xy=x2 Step 1: Add x^2 to both sides. βˆ’x2+5xy+x+0.75y+x2=x2+x2 5xy+x+0.75y=2x2 Step 2: Add -x to both sides. 5xy+x+0.75y+βˆ’x=2x2+βˆ’x 5xy+0.75y=2x2βˆ’x Step 3: Factor out variable y. y(5x+0.75)=2x2βˆ’x Step 4: Divide both sides by 5x+0.75. y(5x+0.75) 5x+0.75 = 2x2βˆ’x 5x+0.75 y= 2x2βˆ’x 5x+0.75 Answer: y= 2x2βˆ’x 5x+0.75