Systes of Nonlinear Equations and Their Solutions非线性方程及其解.pptVIP

Systems of Nonlinear Equations and Their Solutions Example: Solving a Nonlinear System by the Substitution Method Example: Solving a Nonlinear System by the Addition Method Example: Solving a Nonlinear System by the Addition Method Examples Solve: 4. Find the length and width of a rectangle whose perimeter is 20 ft. an whose area is 21 sq.ft. * A system of two nonlinear equations in two variables contains at least one equation that cannot be expressed in the form Ax + By = C. Here are two examples: A solution to a nonlinear system in two variables is an ordered pair of real numbers that satisfies all equations in the system. The solution set to the system is the set of all such ordered pairs. x2 = 2y + 10 3x – y = 9 y = x2 + 3 x2 + y2 = 9 Solve by the substitution method: The graph is a line. The graph is a circle. x – y = 3 (x – 2)2 + (y + 3)2 = 4 Solution Graphically, we are finding the intersection of a line and a circle whose center is at (2, -3) and whose radius measures 2. Step 1 Solve one of the equations for one variable in terms of the other. We will solve for x in the linear equation - that is, the first equation. (We could also solve for y.) x – y = 3 This is the first equation in the given system. x = y + 3 Add y to both sides. Solution Step 2 Substitute the expression from step 1 into the other equation. We substitute y + 3 for x in the second equation. x = y + 3 ( x – 2)2 + (y + 3)2 = 4 This gives an equation in one variable, namely (y + 3 – 2)2 + (y + 3)2 = 4. The variable x has been eliminated. Step 3 Solve the resulting equation containing one variable. (y + 3 – 2)2 + (y + 3)2 = 4 This is the equation containing one variable. (y + 1)2 + (y + 3 )2 = 4 Combine numerical terms in the first parentheses. y2 + 2y + 1 + y2 + 6y + 9 = 4 Square each binomial. 2y2 + 8y + 10 = 4 Combine like terms on the left. 2y2 + 8y + 6 = 0 Subtract 4 from both sides and set the quadratic equation equal to 0