1. Systems of Nonlinear Equations in Two Variables

2. Systems of Nonlinear Equations and Their Solutions
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:
x2 = 2y + 10
3x – y = 9
y = x2 + 3
x2 + y2 = 9
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.

3. Text Example Solve by the substitution method: x – y = 3
The graph is a line.
The graph is a circle.
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.

4. Text Example cont. 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 ( 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 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.

5. Text Example cont. Solution
y2 + 4y + 3 = 0 Simplify by dividing both sides by 2.
(y + 3)(y + 1) = 0 Factor.
y + 3 = 0 or y + 1 = 0 Set each factor equal to 0.
y = -3 or y = -1 Solve for y.
Step Back-substitute the obtained values into the equation from step 1. Now that we have the y-coordin-ates of the solutions, we back-substitute -3 for y and -1 for y in the equation x = y + 3.
If y = -3: x = = 0, so (0, -3) is a solution.
If y = -1: x = = 2, so (2, -1) is a solution. -1 -5 -4 -3 -2 1 2 3 4 5 6 7 -6 -7
x – y = 3
(x – 2)2 + (y + 3)2 = 4
(2, -1)
(0, -3)
Step 5 Check the proposed solution in both of the system's given equations. Take a moment to show that each ordered pair satisfies both equations. The solution set of the given system is {(0, -3), (2, -1)}.

6. Text Example Solve the system: 4x2 + y2 = 13 x2 + y2 = 10
Equation 1.
Equation 2.
Solution We can use the same steps that we did when we solved linear systems by the addition method.
Step Write both equations in the form Ax2 + By2 = C. Both equations are already in this form, so we can skip this step.
Step If necessary, multiply either equation or both equations by appropriate numbers so that the sum of the x2-coefficients or the sum of the y2-coefficients is 0. We can eliminate y by multiplying Equation 2 by -1.
No change.
Multiply by -1. 10 = y2 + x2 13
4x2
-10 = y2 –
-x2 13 +
4x2

7. Text Example cont. Solution
Steps 3 and Add equations and solve for the remaining variable. +1 = x 1 x2 3
3x2
-10 y2 –
-x2 13 +
4x2
Add.
Step Back-substitute and find the values for the other variables. We must back-substitute each value of x into either one of the original equations. Let's use x2 + y2 = 10, Equation 2. If x = 1,
12 + y2 = 10 Replace x with 1 in Equation 2.
y2 = 9 Subtract 1 from both sides.
y = ±3 Apply the square root method.
(1, 3) and (1, -3) are solutions. If x = -1,
(-1)2 + y2 = 10 Replace x with -1 in Equation 2.
y2 = 9 The steps are the same as before.
y = ±3
(-1, 3) and (-1, -3) are solutions.

8. Text Example cont. Solution
Step Check. Take a moment to show that each of the four ordered pairs satisfies Equation 1 and Equation 2. The solution set of the given system is
{(1, 3), (1, -3), (-1, 3), (-1, -3)}. -1 -5 -4 -3 -2 1 2 3 4 5 6 7 -6 -7
4x2 + y2 = 13
x2 + y2 = 10
(-1, -3)
(-1, 3)
(1, 3)
(1, -3)

9. Text Example Solve the system:
y = x Equation 1 (The graph is a parabola.)
x2 + y2 = 9 Equation 2 (The graph is a circle.)
Solution We could use substitution because Equation 1 has y expressed in terms of x, but this would result in a fourth-degree equation. However, we can rewrite Equation 1 by subtracting x2 from both sides and adding the equations to eliminate the x2-terms. y + 12 = y2 9 x2 3
-x2
Subtract x2 from both sides of Equation 1.
This is Equation 2.
Add.
Add the equations.

10. Text Example cont. Solution We now solve this quadratic equation.
y + y2 = 12
y2 + y – 12 = 0 Subtract 12 from both circles and get the quadratic equation equal to 0.
(y + 4)(y – 3) = 0 Factor.
y + 4 = 0 or y – 3 = 0 Set each factor equal to 0.
y = or y = 3 Solve for y.
To complete the solution, we must back-substitute each value of y into either
one of the original equations. We will use y = x2 + 3, Equation 1. First, we
substitute -4 for y.
-4 = x2 + 3
-7 = x2 Subtract 3 from both sides.

11. Text Example cont. Solution
Because the square of a real number cannot be negative, the equation x2 = -7 does not have real-number solutions. Thus, we move on to our other value for y, 3, and substitute this value into Equation 1. -1 -5 -4 -3 -2 1 2 3 4 5 6 7 -6 -7
y = x2 + 3
x2 + y2 = 9
(0, 3)
y = x This is Equation 1.
3 = x Back-substitute 3 for y.
0 = x2 Subtract 3 from both sides.
0 = x Solve for x.
We showed that if y = 3, then x = 0. Thus, (0, 3) is the solution. Take a moment to show that (0, 3) satisfies Equation 1 and Equation 2. The solution set of the given system is {(0, 3)}.

12. Example Solve the following system of equations x y=-12 x-2y+14 = 0
Solution:
so x=-14+2y and (-14+2y)y=-12
(-14+2y)y= y+2y2=-12 2y2-14y+12=0

13. Example Solve the following system of equations x y=-12 x-2y+14 = 0
Solution:
2y2-14y+12=0 2(y2-7y+6)=0 2(y-6)(y-1)=0 (y-6)=0, (y-1)=0 y=6, y=1
y=6 x=-14+2y =-14+2(6) = =-2
y=1 x=-14+2y =-14+2(1) =-14+2 =-12
(-2, 6) and (-12,1)

14. Systems of Nonlinear Equations in Two Variables