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Linear and Nonlinear Systems of Equations 7.1

Linear and Nonlinear Systems of Equations 7.1. JMerrill , 2010. y = x – 1. x + y = 7. Example: Solving Linear Systems by Substitution. Use substitution to solve the system of equations.

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Linear and Nonlinear Systems of Equations 7.1

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  1. Linear and Nonlinear Systems of Equations7.1 JMerrill, 2010

  2. y= x – 1 x + y = 7 Example: Solving Linear Systems by Substitution Use substitution to solve the system of equations. 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 Substitute (x –1) for y in the other equation. 2x– 1 = 7 Combine like terms. 2x = 8 x = 4

  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 Substitute x = 4. y = 3 The solution is the ordered pair (4, 3).

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

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

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

  7. 5x + 6y= –9 2x– 2 = –y You Try Use substitution to solve the system of equations. (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. 3x + 2y = 4 4x– 2y = –18 Example: Solving Linear Systems by Elimination Use elimination to solve the system of equations. Step 1 Add the equations together to solve for one variable. 3x+ 2y = 4 The y-terms have opposite coefficients. + 4x– 2y = –18 7x = –14 Add the equations to eliminate y. First part of the solution x = –2

  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 Second part of the solution y = 5 The solution to the system is (–2, 5).

  11. Remember! 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. 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.

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

  13. –32= –32  Example: Solving Systems with Infinitely Many Solutions Determine the solution, if it exists. 56x + 8y = –32 7x + y = –4 Isolatey. 7x + y = –4 y = –4 – 7x Solve the second equation for y. Substitute (–4 –7x) for y in the first equation. 56x+ 8(–4 – 7x) = –32 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.

  14. You Try 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.

  15. Example: Zoology Application 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? Let x present the amount of beef mix in the mixture. Let y present the amount of bacon mix in the mixture.

  16. Amount of beef mix amount of bacon mix plus equals 60. = x + y 60 Protein of beef mix protein of bacon mix protein in mixture. plus equals = + 0.18x 0.09y 0.15(60) Example Continued Write one equation based on the amount of dog food: Write another equation based on the amount of protein:

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

  18. Example Continued Substitute x into one of the original equations to solve for y. Substitute the value of x into one equation. 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.

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

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

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