Systems of Nonlinear Equations in Two Variables

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.

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.

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 + 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.

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 4 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 = -3 + 3 = 0, so (0, -3) is a solution. If y = -1: x = -1 + 3 = 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)}.

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 1 Write both equations in the form Ax2 + By2 = C. Both equations are already in this form, so we can skip this step. Step 2 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

Text Example cont. Solution Steps 3 and 4 Add equations and solve for the remaining variable. +1 = x 1 x2 3 3x2 -10 y2 – -x2 13 + 4x2 Add. Step 5 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.

Text Example cont. Solution Step 6 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)

Text Example Solve the system: y = x2 + 3 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.

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 = -4 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.

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 = x2 + 3 This is Equation 1. 3 = x2 + 3 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)}.

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=-12 -14y+2y2=-12 2y2-14y+12=0

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) =-14+12 =-2 y=1 x=-14+2y =-14+2(1) =-14+2 =-12 (-2, 6) and (-12,1)

Systems of Linear Equations in Three Variables

Systems of Linear Equations in Three Variables and Their Solutions An equation such as x + 2y - 3z = 9 is called a linear equation in three variables. In general, any equation of the form Ax + By + Cz = D where A, B, C, and D are real numbers such that A, B, and C are not all 0, is a linear equation in the variables x, y, and z. The graph of this linear equation in three variables is a plane in three-dimensional space.

Text Example Show that the ordered triple (-1, 2, -2) is a solution of the system: x + 2y – 3z = 9 2x – y + 2z = -8 -x + 3y – 4z = 15. Solution Because -1 is the x-coordinate, 2 is the y-coordinate, and -2 is the z-coordinate of (-1, 2, -2), we replace x by -1, y by 2, and z by -2 in each of the three equations. x + 2y – 3z = 9 -1 + 2(2) – 3(-2) = 9 -1 + 4 + 6 = 9 9 = 9 ? true 2x – y + 2z = -8 2(-1) – 2 + 2(-2) = -8 -2 – 2 – 4 = -8 -8 = -8 ? true -x + 3y – 4z = 15 -(-1) + 3(2) – 4(-2) = 15 1 + 6 + 8 = 15 15 = 15 ? true The ordered triple (-1, 2, -2) satisfies the three equations. Thus, the solution set is {(-1, 2, -2)}.

Solving Linear Systems in Three Variables by Eliminating Variables Reduce the system to two equations in two variables. This is usually accomplished by taking two different pairs of equations and using the addition method to eliminate the same variable from each pair. Solve the resulting system of two equations in two variables using addition or substitution. The result is an equation in one variable that gives the value of that variable. Back-substitute the value of the variable found in step 2 into either of the equations in two variables to find the value of the second variable. Use the values of the two variables from steps 2 and 3 to find the value of the third variable by back-substituting into one of the original equations. Check the proposed solution in each of the original equations.

Text Example Solve the system: 3 = 3z + 5y -2x -3 2z 3y 3x 4z – 2y 5x Equation 1 Equation 2 Equation 3 Solution There are many ways to proceed. Because our initial goal is to reduce the system to two equations in two variables, the central idea is to take two different pairs of equations and eliminate the same variable from each pair. Step 1 Reduce the system to two equations in two variables. We choose any two equations and use the addition method to eliminate a variable. Let's eliminate z from Equations 1 and 2. We do so by multiplying Equation 2 by 2. Then we add equations. Equation 1 No change. Multiply by 2. -3 = 2z + 3y 3x 3 4z – 2y 5x -3 = 4y + 11x -6 4z 6y 6x 3 – 2y 5x Equation 2 Add: Equation 4

Text Example cont. Solution Now we must eliminate the same variable from another pair of equations. We can eliminate z from Equations 2 and 3. First, we multiply Equation 2 by –3. Next, we multiply Equation 3 by 2. Finally, we add equations. Equation 2 Multiply by -3. Multiply by 2. 3 = 3z + 5y -2x -3 2z 3y 3x 15 = y + -13x 6 6z 10y -4x 9 – 9y -9x Equation 3 Add: Equation 5 Equations 4 and 5 give us a system of two equations in two variables. Step 2 Solve the resulting system of two equations in two variables. We choose any two equations and use the addition method to eliminate a variable. Let's eliminate z from Equations 1 and 2. We do so by multiplying Equation 2 by 2. Then we add equations. Equation 4 No change. Multiply by -4. 15 = y + -13x -3 4y 11x -1 = x -63 63x -60 4y – 52x -3 + 11x Equation 5 Add: Divide both sides by 63.

Text Example cont. Solution Step 3 Use back-substitution in one of the equations in two variables to find the value of the second variable. We back-substitute -1 for x in either Equation 4 or 5 to find the value of y. -13x + y = 15 Equation 5  -13(-1) + y = 15 Substitute -1 for x.   13 + y = 15 Multiply.   y = 2 Subtract 13 from both sides. Step 4 Back-substitute the values found for two variables into one of the original equations to find the value of the third variable. We can now use any one of the original equations and back-substitute the values of x and y to find the value for z. We will use Equation 2. 3x + 3y + 2z = -3 Equation 2 3(-l) + 3(2) + 2z = -3 Substitute -1 for x and 2 for y.

Text Example cont. Solution 3 + 2z = -3 Multiply and then add. 2z = -6 Subtract 3 from both sides. z = -3 Divide both sides by 2.   With x = -1, y = 2, and z = -3, the proposed solution is the ordered triple (-1,2,-3). Step 5 Check. Check the proposed solution, (-1, 2, -3), by substituting the values for x, y, and z into each of the three original equations. These substitutions yield three true statements. Thus, the solution set is {(-1, 2, -3)}.

Text Example Solve the system: Solution 16 = z + 2y x 17 2z y 8 Equation 1 Equation 2 Equation 3 Solution Step 1 Reduce the system to two equations in two variables. Because Equation 1 contains only x and z, we could eliminate y from Equations 2 and 3. This will give us two equations in x and z. To eliminate y from Equations 2 and 3, we multiply Equation 2 by -2 and add Equation 3. Equation 2 Multiply by -2. No change. -3 = z + 2y x 17 2z y -18 = 3z – -x 16 z + 2y x -34 4z -2x Equation 3 Add: Equation 4 Equation 4 and the given Equation 1 provide us with a system of two equations in two variables.

Text Example cont. Solution Step 2 Solve the resulting system of two equations in two variables. We will solve Equations 1 and 4 for x and z. 5 = z -10 -2z -18 3z – -x 8 + x Equation 1 Equation 4 Add: Divide both sides by -2. Step 3 Use back-substitution in one of the equations in two variables to find the value of the second variable. To find x, we back-substitute 5 for z in either Equation 1 or 4. We will use Equation 1. x + z = 8 x + 5 = 8 x = 3 Equation 1 Substitute 5 for z. Subtract 5 from both sides.

Text Example cont. Solution Step 4 Back-substitute the values found for two variables into one of the original equations to find the value of the third variable. To find y, we back-substitute 3 for x and 5 for z into Equation 2 or 3. We can't use Equation 1 because y is missing in this equation. We will use Equation 2. x + y + 2z = 17 3 + y + 2(5) = 17 y + 13 = 17 y = 4 We found that z = 5, x = 3, and y = 4. Thus, the proposed solution is the ordered triple (3, 4, 5). Step 5 Check. Substituting 3 for x, 4 for y, and 5 for z into each of the three original equations yields three true statements. Consequently, the solution set is {(3,4,5)}.

Example Reduce the system to 2X2 -2(x - 3y + z = 1) 2x -3z = -8 Solve the system: x - 3y + z = 1 2x -3z = -8 3x +8y +2z = 1 Solution: Reduce the system to 2X2 -2(x - 3y + z = 1) 2x -3z = -8 -2x+6y-2z= -2 2x -3z= -8 6y - 5z = -10 -3(x - 3y + z = 1) 3x +8y +2z = 1 -3x +9y -3z = -3 3x +8y +2z = 1 17y - z = -2

Example cont. Solution: (-1, 0, 2) 6y - 5z = -10 -5(17y - z = -2) 6y - 5z = -10 -85y +5z = 10 -79y = 0 y = 0 6y - 5z = -10 17y - z = -2 6(0) - 5z = -10 -5z = -10 z = 2 x - 3(0) + 2 = 1 x+2 = 1 x = -1 Solution: (-1, 0, 2)

Partial Fractions

Steps in Partial Fraction Decomposition Set up the partial fraction decomposition with the unknown constants A, B, C, etc., in the numerator of the composition. Multiply both sides of the resulting equation by the least common denominator. Simplify the right-hand side of the equations. Write both sides in descending powers, equate coefficients of like powers of x, and equate constant terms. Solve the resulting linear system for A, B, C, etc. Substitute the values for A, B, C, etc., into the equation in the first step.

Text Example Find the partial fraction decomposition of Solution Step 1 Set up the partial fraction decomposition with the unknown constants. Because the linear factor x – 3 is repeated twice, we must include one fraction with a constant numerator for each power of x – 3.

Text Example cont. Solution Step 2 Multiply both sides of the resulting equation by the least common denominator. We clear fractions, multiplying both sides by x(x – 3)2, the least common denominator. We use the distributive property on the right side. Dividing out common factors in numerators and denominators, we obtain x – 18 = A(x – 3)2 + Bx(x – 3) + Cx.

0x2 + 1x – 18 = (A + B)x2 + (-6A – 3B + C)x + 9A. Text Example cont. Solution Step 3 Simplify the right side of the equation. Square x – 3. Then apply the distributive property. x – 18 = A(x2 – 6x + 9) + Bx(x – 3) + Cx Square x – 3. x – 18 = Ax2 – 6Ax + 9A + Bx2 – 3Bx + Cx Apply the distributive property. Step 4 Write both sides in descending powers, equate coefficients of like powers of x, and equate constant terms. Square x – 3. Then apply the distributive property. x – 18 = Ax2 + Bx2 – 6Ax – 3Bx + Cx + 9A Express both sides in the same form. 0x2 + 1x – 18 = (A + B)x2 + (-6A – 3B + C)x + 9A.

Text Example cont. Solution Step 5 Solve the resulting system for A, B, and C. Dividing both sides of the last equation by 9, we obtain A = -2. Substituting –2 for A in the first equation, A + B = 0, gives –2 + B = 0 or B = 2. We find C by substituting –2 for A and 2 for B in the middle equation, -6A – 3B + C = 1. We obtain C = -5. Step 6 Substitute values of A, B, and C and write the partial fraction decomposition. With A = -2, B = 2, and C = -5, the required partial fraction decomposition is

Text Example Find the partial fraction decomposition of Solution Step 1 Set up the partial fraction decomposition with the unknown constants. We put a constant (A) over the linear factor and a linear expression (Bx + C) over the prime quadratic factor.

3x2 + 17x + 14 = A(x2 + 2x + 4) + (Bx + C)(x - 2). Text Example cont. Solution Step 2 Multiply both sides of the resulting equation by the least common denominator. We clear fractions, multiplying both sides by (x - 2) (x2 + 2x + 4), the least common denominator. We use the distributive property on the right side. Dividing out common factors in numerators and denominators, we obtain 3x2 + 17x + 14 = A(x2 + 2x + 4) + (Bx + C)(x - 2).

Text Example cont. Solution Step 3 Simplify the right side side of the equation. We multiply on the right side by distributing A over each term in parentheses and multiplying (Bx + C)(x – 2) using the FOIL method. 3x2 + 17x + 14 = Ax2 + 2Ax + 4A + Bx2 – 2Bx + Cx – 2C. Step 4 Write both sides in descending powers, equate coefficients of like powers of x, and equate constant terms. The left side, 2x2 + 17x + 14, is in descending powers of x. We write the right side in descending powers of x 3x2 + 17x + 14 = Ax2 + Bx2 + 2Ax – 2Bx + Cx + 4A – 2C. And express both sides in the same form. 3x2 + 17x + 14 = (A + B)x2 + (2A – 2B + C) x + (4A – 2C).

Text Example cont. Solution Equating coefficients of like powers of x and constant terms results in the following system of linear equations. A + B = 3 2A – 2B + C = 17 4A – 2C = 14 Step 5 Solve the resulting system for A, B, and C. Because the first equation involves A and B, we can obtain another equation in A and B by eliminating C from the second and third equations. Multiply the second equation by 2 and add equations. Solving in this manner, we obtain A = 5, B = -2, and C = 3.

Text Example cont. Solution Step 6 Substitute the values of A, B, and C and write the partial fraction decomposition. With A = 5, B = -2, and C = 3, the required partial fraction decomposition is

Example Find the partial fraction decomposition of Solution

Example Cont. Find the partial fraction decomposition of Solution

Systems of Nonlinear Equations in Two Variables

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.

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.

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 + 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.

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 4 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 = -3 + 3 = 0, so (0, -3) is a solution. If y = -1: x = -1 + 3 = 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)}.

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 1 Write both equations in the form Ax2 + By2 = C. Both equations are already in this form, so we can skip this step. Step 2 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

Text Example cont. Solution Steps 3 and 4 Add equations and solve for the remaining variable. +1 = x 1 x2 3 3x2 -10 y2 – -x2 13 + 4x2 Add. Step 5 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.

Text Example cont. Solution Step 6 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)

Text Example Solve the system: y = x2 + 3 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.

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 = -4 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.

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 = x2 + 3 This is Equation 1. 3 = x2 + 3 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)}.

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=-12 -14y+2y2=-12 2y2-14y+12=0

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) =-14+12 =-2 y=1 x=-14+2y =-14+2(1) =-14+2 =-12 (-2, 6) and (-12,1)

Systems of Inequalities

Graphing a Linear Inequality in Two Variables Replace the inequality symbol with an equal sign and graph the corresponding linear equation. Draw a solid line if the original inequality contains a < or > symbol. Draw a dashed line if the original inequality contains a < or > symbol. Choose a test point in one of the half-planes that is not on the line. Substitute the coordinates of the test point into the inequality If a true statement results, shade the half-plane containing this test point. If a false statement results, shade the half-plane not containing this test point.

Text Example Graph: 3x – 5y < 15. Solution Step 1 Replace the inequality symbol by = and graph the linear equation. We need to graph 3x – 5y = 15. We can use intercepts to graph this line.   We set y = 0 to find We set x = 0 to find the x-intercept: the y-intercept: 3x – 5y = 15 3x – 5y = 15 3x – 5 • 0 = 15 3 – 0 • 5y = 15 3x = 15 -5y = 15 x = 5 y = -3

Text Example cont. Solution The x-intercept is 5, so the line passes through (5, 0). The y-intercept is -3, so the line passes through (0, -3). The graph is indicated by a dashed line because the inequality 3x – 5y < 15 contains a < symbol, rather than <. The graph of the line is shown below. -1 -3 -2 1 2 3 4 5 6 7 -4 -5 3x – 5y = 15

Text Example cont. Solution Step 2 Choose a test point in one of the half-planes that is not on the line. Substitute its coordinates into the inequality. The line 3x – 5y = 15 divides the plane into three parts – the line itself and two half-planes. The points in one half-plane satisfy 3x – 5y > 15. The points in the other half-plane satisfy 3x – 5y < 15. We need to find which half-plane is the solution. To do so, we test a point from either half-plane. The origin, (0, 0), is the easiest point to test. 3x – 5y < 15 This is the given inequality. Is 3 • 0 – 5 • 0 < 15? Test (0, 0) by substituting 0 for x and y. 0 – 0 < 15 0 < 15, true

Text Example cont. Solution Step 3 If a true statement results, shade the half-plane containing the test point. Because 0 is less than 15, the test point (0, 0) is part of the solution set. All the points on the same side of the line 3x - 5y = 15 as the point (0, 0) are members of the solution set. The solution set is the half-plane that contains the point (0, 0), indicated by shading this half-plane. The graph is shown using green shading and a dashed blue line. -1 -3 -2 1 2 3 4 5 6 7 -4 -5

Text Example Graph: x2 + y2 < 9. Solution -5 -4 -3 -2 -1 1 2 3 4 5 Solution Step 1 Replace the inequality symbol by = and graph the nonlinear equation. We need to graph x2 + y2 = 9. The graph is a circle of radius 3 with its center at the origin. The graph is shown below as a solid circle because equality is included in the < symbol.

Text Example cont. Solution Step 2 Choose a test point in one of the regions that is not on the circle. Substitute its coordinates into the inequality. The circle divides the plane into three parts – the circle itself, the region inside the circle, and the region outside the circle. We need to determine whether the region inside or outside the circle is the solution. To do so, we will use the test point (0, 0) from inside the circle. x2 + y2 < 9 This is the given inequality. Is 02 + 02 < 9? Test (0, 0) by substituting 0 for x and 0 for y. 0 + 0 < 9 0 < 9, true

Text Example cont. Solution Step 3 If a true statement results, shade the region containing the test point. The true statement tells us that all the points inside the circle satisfy x2 + y2 < 9. The graph is shown using green shading and a solid blue circle. -5 -4 -3 -2 -1 1 2 3 4 5

Example Graph: 2x+y<8 x+y>4 Solution: First graph 2x+y=8 x+y=4

Example cont. Test point (0,0) 2x+y<8 2(0) + 0 = 0 <8 true Graph: 2x+y<8 x+y>4 Solution: Test point (0,0) 2x+y<8 2(0) + 0 = 0 <8 true

Example cont. Test point (0,0) Graph: 2x+y<8 x+y>4 Solution: x+y>4 0+0=0 >4 false

Example cont. Graph: 2x+y<8 x+y>4 Solution:

Text Example Graph the solution set: x – y < 2 -2 < x < 4 Solution We begin by graphing x - y < 2, the first given inequality. The line x – y = 2 has an x-intercept of 2 and a y-intercept of -2. The test point (0, 0) makes the inequality x – y < 2 true, and its graph is shown below. -5 -4 -3 -2 -1 1 2 3 4 5

Text Example cont. Solution Now let's consider the second given inequality -2 < x < 4. Replacing the inequality symbols by =, we obtain x = -2 and x = 4, graphed as vertical lines. The line of x = 4 is not included. Using (0, 0) as a test point and substituting the x-coordinate, 0, into -2 < x < 4, we obtain the true statement -2 < 0 < 4. We therefore shade the region between the vertical lines. -5 -4 -3 -1 1 2 3 5 4 -2

Text Example cont. Solution Finally, let's consider the third given inequality, y < 3. Replacing the inequality symbol by =, we obtain y = 3, which graphs as a horizontal line. Because (0, 0) satisfies y < 3 (0 < 3 is true), the graph consists of the half-plane below the line y = 3. -5 -4 -3 -1 1 2 3 5 4 -2

Linear Programming

Objective Functions in Linear Programming Many problems involve quantities that must be maximized or minimized. Businesses are interested in maximizing profit. An operation in which bottled water and medical kits are shipped to earthquake victims needs to maximize the number of victims helped by this shipment. An objective function is an algebraic expression in two or more variables describing a quantity that must be maximized or minimized.

Text Example Bottled water and medical supplies are to be shipped to victims of an earth-quake by plane. Each container of bottled water will serve 10 people and each medical kit will aid 6 people. If x represents the number of bottles of water to be shipped and y represents the number of medical kits, write the objective function that describes the number of people that can be helped. Solution Because each bottle of water serves 10 people and each medical kit aids 6 people, we have = 10x + 6y. Using z to represent the objective function, we have z = 10x + 6y. Unlike the functions that we have seen so far, the objective function is an equation in three variables. For a value of x and a value of y, there is one and only one value of z. Thus, z is a function of x and y. 6 times the number of medical kits. 10 times the number of bottles of water plus is The number of People helped

must be less than or equal to Text Example Planes can carry a total volume for supplies that does not exceed 6000 cubic feet. Each water bottle is 1 cubic foot and each medical kit also has a volume of 1 cubic foot. With x still representing the number of water bottles and y the number of medical kits, write an inequality that describes this second constraint. Solution Because each plane can carry a volume of supplies that does not exceed 6000 cubic feet, we have The total volume of the water bottles The total volume of the medical kits must be less than or equal to 6000 cubic feet. plus lx + ly < 6000. Each bottle is 1 cubic foot. Each kit is 1 cubic foot. The plane's volume constraint is described by the inequality x + y < 6000.

Solving a Linear Programming Problem Let z=ax + by be an objective function that depends on x and y. Furthermore, z is subject to a number of constraints on x and y. If a maximum or minimum value of z exists, it can be determined as follows: Graph the system of inequalities representing the constraints Find the value of the objective function at each corner, or vertex, of the graphed region. The maximum and minimum of the objective function occur at one or more of the corner points.

Example Given the objective function and a system of linear inequalities. Objective function z = 3x+2y Constraints x>0, y>0, 2x+y<8, x+y>4 Graph the system of inequalities representing the constraints. Find the value of the objective function at each corner of the graphed region. Use the values that you found in the prior step to determine the maximum value of the objective function and the values of x and y for which the maximum occurs.

Example cont. Graph x=0, y=0 2x+y=8 x+y=4

Example cont. x>0

Example cont. y>0

Example cont. x>0 y>0

Example cont. 2x+y<8

Example cont. x+y>4

Example cont. Corners (4,0), (0,4), (0,8)

Example cont. Objective function z = 3x+2y The maximum value of the objective function is 16 and it occurs when x = 0 and y = 8

Text Example Find the maximum value of the objective function z = 2x + y subject to the constraints: x > 0, y > 0, x + 2y < 5, x – y < 2. -1 -3 -2 1 3 4 5 6 7 2 -4 -5 x – y = 2 x + 2y = 5 (0, 2.5) (0, 0) (3, 1) (2, 0) Solution We begin by graphing the region in quadrant I (x > 0, y > 0) formed by the constraints. Now we evaluate the objective function at the four vertices of this region. Objective Function: z = 2x + y At (0, 0): z = 2 • 0 + 0 = 0 At (2, 0): z = 2 • 2 + 0 = 4 At (3, 1): z = 2 • 3 + 1 = 7 At (0, 2.5): z = 2 • 0 + 2.5 = 2.5 The maximum value of z. Thus, the maximum value of z is 7, and this occurs when x = 3 and y = 1.