1. Linear and Nonlinear Systems of Equations 7.1 JMerrill, 2010

2. Use substitution to solve the system of equations. Example: Solving Linear Systems by Substitution y = x – 1 x + y = 7 Step 1 If necessary, solve one equation for one variable. The first equation is already solved for y. Step 2 Substitute the expression into the other equation. x + y = 7 x + (x – 1) = 7 2x – 1 = 7 2x = 8 x = 4 Substitute (x –1) for y in the other equation. Combine like terms.

3. Step 3 Substitute the x-value into one of the original equations to solve for y. Example Continued y = x – 1 y = (4) – 1 y = 3 Substitute x = 4. The solution is the ordered pair (4, 3).

4. Example Continued Check A graph or table supports your answer.

5. Use substitution to solve the system of equations. Example: Solving Linear Systems by Substitution 2y + x = 4 3x – 4y = 7 Method 1 Isolate y. 2y + x = 4 5x = 15 3x + 2x – 8 = 7 First equation. Method 2 Isolate x. Isolate one variable. Second equation. Substitute the expression into the second equation. Combine like terms. 2y + x = 4 x = 3 x = 4 – 2y 3x – 4y= 7 3(4 – 2y)– 4y = 7 12 – 6y – 4y = 7 12 – 10y = 7 –10y = –5 5x – 8 = 7 First part of the solution 3x –4 +2 +2 = 7 y = + + 2 3x – 4y = 7

6. Substitute the value into one of the original equations to solve for the other variable. 2y + (3) = 4 2y = 1 Example Continued Substitute the value to solve for the other variable. Second part of the solution 2 + x = 4 1 + x = 4 x = 3 By either method, the solution is. Method 1Method 2

7. Use substitution to solve the system of equations. 5x + 6y = –9 2x – 2 = –y You Try (3, –4)

8. You can also solve systems of equations by elimination. With elimination, you get rid of one of the variables by adding or subtracting equations. This is actually 7.2, but if it’s easier, do it!

9. Use elimination to solve the system of equations. Example: Solving Linear Systems by Elimination 3x + 2y = 4 4x – 2y = –18 Step 1 Add the equations together to solve for one variable. 3x + 2y = 4 + 4x – 2y = –18 The y-terms have opposite coefficients. First part of the solution 7x = –14 x = –2 Add the equations to eliminate y.

10. Example Continued Step 2 Substitute the x-value into one of the original equations to solve for y. 3(–2) + 2y = 4 2y = 10 y = 5 Second part of the solution The solution to the system is (–2, 5).

11. Systems may have one solution, no solutions, or infinitely many solutions. When you try to solve these systems algebraically, the result will be an identity or a contradiction. An identity, such as 0 = 0, is always true and indicates infinitely many solutions. A contradiction, such as 1 = 3, is never true and indicates no solution. Remember!

12. Determine the solution, if it exists. Example: Solving Systems with No Solution 3x + y = 1 2y + 6x = –18 Isolate y. Substitute (1–3x) for y in the second equation. Solve the first equation for y. 3x + y = 1 2(1 – 3x) + 6x = –18 y = 1 –3x 2 – 6x + 6x = –18 2 = –18 Distribute. Simplify. Because 2 is never equal to –18, the equation is a contradiction. Therefore, the system is inconsistent and has no solution.

13. Determine the solution, if it exists. 56x + 8y = –32 7x + y = –4 Isolate y. Substitute (–4 –7x) for y in the first equation. Solve the second equation for y. 7x + y = –4 56x + 8(–4 – 7x) = –32 y = –4 – 7x 56x – 32 – 56x = –32 Distribute. Simplify. Because –32 is equal to –32, the equation is an identity. The system is consistent, dependent and has infinite number of solutions. Example: Solving Systems with Infinitely Many Solutions –32 = –32

14. Determine the solution, if it exists. 6x + 3y = –12 2x + y = –6 Because –18 is never equal to –12, the equation is a contradiction. Therefore, the system is inconsistent and has no solutions. You Try

15. A veterinarian needs 60 pounds of dog food that is 15% protein. He will combine a beef mix that is 18% protein with a bacon mix that is 9% protein. How many pounds of each does he need to make the 15% protein mixture? Example: Zoology Application Let x present the amount of beef mix in the mixture. Let y present the amount of bacon mix in the mixture.

16. Example Continued Write one equation based on the amount of dog food: Amount of beef mix plus amount of bacon mix equals xy 60. 60+ = Write another equation based on the amount of protein: Protein of beef mix plus protein of bacon mix equals 0.18x0.09y protein in mixture. 0.15(60) + =

17. Solve the system. x + y = 60 0.18x +0.09y = 9 x + y = 60 y = 60 – x First equation 0.18x + 0.09(60 – x) = 9 0.18x + 5.4 – 0.09x = 9 0.09x = 3.6 x = 40 Solve the first equation for y. Substitute (60 – x) for y. Distribute. Simplify. Example Continued

18. Substitute the value of x into one equation. Substitute x into one of the original equations to solve for y. 40 + y = 60 y = 20 Solve for y. The mixture will contain 40 lb of the beef mix and 20 lb of the bacon mix. Example Continued

19. Nonlinear Systems When graphing nonlinear systems, you may get: 2 solutions: Or no real solutions:

20. Solving Nonlinear Systems The process is exactly the same: ◦S◦Solve for one variable ◦S◦Substitute that equation into the other equation ◦F◦Factor and solve