§ 10.5 Systems of Nonlinear Equations in Two Variables

Solving Systems of Nonlinear Equations We looked at solving systems of linear equations in an earlier section. In this section, we will look at solving systems of nonlinear equations. A system of nonlinear equations contains at least one equation that cannot be expressed in the form Ax +By = C. As with systems of linear equations in two variables, the solution of a nonlinear system (if there is one) corresponds to the intersection point(s) of the graphs of the equations. Unlike linear systems, the graphs can be circles, ellipses, parabolas, hyperbolas, or anything other than two lines. We will solve nonlinear systems using the substitution method and the addition method. Blitzer, Intermediate Algebra, 5e – Slide #2 Section 10.5

Solving Systems of Nonlinear Equations EXAMPLE Solve by the substitution method: SOLUTION 1) Solve one of the equations for one variable in terms of the other. We begin by isolating one of the variables raised to the first power in either of the equations. By solving for x in the second equation, which has a coefficient of 1, we can avoid fractions. This is the second equation in the given system. Add 2y - 14 to both sides. Blitzer, Intermediate Algebra, 5e – Slide #3 Section 10.5

Solving Systems of Nonlinear Equations CONTINUED 2) Substitute the expression from step 1 into the other equation. We substitute 2y – 14 for x in the first equation. This gives us an equation in one variable, namely The variable x has been eliminated. 3) Solve the resulting equation containing one variable. This is the given equation containing one variable. Blitzer, Intermediate Algebra, 5e – Slide #4 Section 10.5

Solving Systems of Nonlinear Equations CONTINUED Use the distributive property. Add 12 to both sides. Factor 2 out of the left side. Divide both sides by 2. Factor. Set each factor equal to zero. Solve for y. 4) Back-substitute the obtained values into the equation from step 1. Now that we have the y-coordinates of the solutions, we back-substitute 6 for y and 1 for y into the equation x = 2y - 14. Blitzer, Intermediate Algebra, 5e – Slide #5 Section 10.5

Solving Systems of Nonlinear Equations CONTINUED If y is 6, so (-2, 6) is a solution. If y is 1, so (-12, 1) is a solution. 5) Check the proposed solutions in both of the system’s given equations. We begin by checking (-2, 6). Replace x with -2 and y with 6. ? ? ? true true The ordered pair (-2, 6) satisfies both equations. Thus, (-2, 6) is a solution of the system. Blitzer, Intermediate Algebra, 5e – Slide #6 Section 10.5

Solving Systems of Nonlinear Equations CONTINUED Now let’s check (-12, 1). Replace x with -12 and y with 1. ? ? ? true true The ordered pair (-12, 1) satisfies both equations. Thus, (-12, 1) is a solution of the system. Thus, the solution set is {(-2, 6), (-12, 1)}. The graph of the equations in the system and the solutions as intersection points follows. Blitzer, Intermediate Algebra, 5e – Slide #7 Section 10.5

Solving Systems of Nonlinear Equations CONTINUED (-2, 6) (-12, 1) Blitzer, Intermediate Algebra, 5e – Slide #8 Section 10.5

Solving Systems of Nonlinear Equations EXAMPLE Solve by the substitution method: Equation 1 Equation 2 SOLUTION We can use the same steps that we did when we solved linear systems by the addition method. 1) Write both equations in the form . Both equations are already in this form, so we can skip this step. 2) If necessary, multiply either equation or both equations by appropriate numbers so that the sum of the -coefficients or the sum of the -coefficients is 0. We can eliminate by multiplying Equation 2 by -2. Blitzer, Intermediate Algebra, 5e – Slide #9 Section 10.5

Solving Systems of Nonlinear Equations CONTINUED No Change Multiply by -2. 3) & 4) Add equations and solve for the remaining variable. Add equations. Multiply both sides by -1. Use the square root property. Blitzer, Intermediate Algebra, 5e – Slide #10 Section 10.5

Solving Systems of Nonlinear Equations CONTINUED 5) Back-substitute and find the values for the other variable. We must back-substitute each value of x into either one of the original equations. Let’s use Equation 2. If x = 1, Replace x with 1 in Equation 2. Simplify. Subtract 2 from both sides. Multiply both sides by -1. Apply the square root property. (1, 2) and (1, -2) are solutions. If x = -1, Replace x with -1 in Equation 2. Simplify. Subtract 2 from both sides. Blitzer, Intermediate Algebra, 5e – Slide #11 Section 10.5

Solving Systems of Nonlinear Equations CONTINUED Multiply both sides by -1. Apply the square root property. (-1, 2) and (-1, -2) are solutions. 6) Check. Take a moment to show that each of the four ordered pairs satisfies the given equations, The solutions are (1, 2), (1, -2), (-1, 2), and (-1, -2), and the solution set of the given system is {(1, 2), (1, -2), (-1, 2), (-1, -2)}. Blitzer, Intermediate Algebra, 5e – Slide #12 Section 10.5

Solving Systems of Nonlinear Equations EXAMPLE A system for tracking ships indicates that a ship lies on a hyperbolic path described by . The process is repeated and the ship is found to lie on a hyperbolic path described by . If it is known that the ship is located in the first quadrant of the coordinate system, determine its exact location. SOLUTION Since both equations represent the path of the ship, we will determine the points they have in common by solving the system of equations 1) Use variables to represent unknown quantities. This has already been done so we may skip this step. Blitzer, Intermediate Algebra, 5e – Slide #13 Section 10.5

Solving Systems of Nonlinear Equations CONTINUED 2) Write a system of equations describing the problem’s conditions. This has already been done so we may skip this step. 3) Solve the system and answer the problem’s question. We must solve the system Equation 1 Equation 2 We will use the addition method. We will multiply Equation 1 by 2 and then add the equations. Multiply by 2. No Change Blitzer, Intermediate Algebra, 5e – Slide #14 Section 10.5

Solving Systems of Nonlinear Equations CONTINUED This is the resultant equation. Divide both sides by 3. Apply the square root property. Since we are only interested in solutions that lie in the first quadrant, we will ignore solutions having y = -1. We now solve for x given y = 1. We will use Equation 2. Replace y with 1 in Equation 2. Simplify. Add 1 to both sides. Divide both sides by 2. Apply the square root property. Blitzer, Intermediate Algebra, 5e – Slide #15 Section 10.5

Solving Systems of Nonlinear Equations CONTINUED Since we are only interested in solutions that lie in the first quadrant, we will ignore solutions having x = -1. Therefore, the only viable solution occurs when x = 1 and when y = 1. Therefore the solution is (1, 1) and the solution set is written {(1, 1)}. A Study Tip: When solving nonlinear systems, extra solutions may be introduced that do not satisfy both equations in the system. Therefore, you should get into the habit of checking all proposed pairs in each of the system’s two equations. Blitzer, Intermediate Algebra, 5e – Slide #16 Section 10.5