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12-1: Solving Systems of Equations

Learn how to solve systems of equations using various methods, including substitution and elimination. Understand the different types of solutions and how to recognize them. Practice solving equations with examples and assignments.

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12-1: Solving Systems of Equations

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  1. 12-1: Solving Systems of Equations Essential Question: Why are we in Algebra 2?

  2. 12-1: Solving Systems of Equations • Solutions of a System of Equations • A set of values that satisfy all the equations in the system. • Example 1: Determine whether (0, 7, 12) and (1, 2, 3) are solutions to the system of equations: • 2x – 5y + 3z = 1 • x + 2y – z = 2 • 3x + y + 2z = 11 • Trying (0, 7, 12) • 2(0) – 5(7) + 3(12) = 1 • (0) + 2(7) – (12) = 2 • 3(0) + (7) + 2(12) = 31 doesn’t work • (0, 7, 12) is not a solution

  3. 12-1: Solving Systems of Equations • Solution of a System of Equations • A set of values that satisfy all the equations in the system. • Example 1: Determine whether (0, 7, 12) and (1, 2, 3) are solutions to the system of equations: • 2x – 5y + 3z = 1 • x + 2y – z = 2 • 3x + y + 2z = 11 • Trying (1, 2, 3) • 2(1) – 5(2) + 3(3) = 1 • (1) + 2(2) – (3) = 2 • 3(1) + (2) + 2(3) = 11 • (1, 2, 3) is a solution

  4. 12-1: Solving Systems of Equations • Solving Systems with Graphs • You can solve systems of two variables by finding the point of intersection. • Steps: • Solve each equation for y • Graph each equation • Use the intersection tool on your graphing calculator to find the point of intersection • Example 2: Find a solution to the system • 2x – y = 1 • 3x + 2y = 4 • Graph on board • Solution is (0.86, 0.71) → • y = 2x – 1 • y = -3/2 x + 2

  5. 12-1: Solving Systems of Equations • Type of System and Number of Solutions • Because graphs in a linear system with two variables are lines, there are exactly three possibilities • Inconsistent System: The lines are parallel and have no point of intersection • Independent System: The lines intersect at a single point • Dependent System: The lines coincide (they’re the same line)

  6. 12-1: Solving Systems of Equations • Solving Systems Algebraically • Substitution Method • Solve one equation for x (or y) • Substitute the expression for x (or y) into the other equation • Solve for the remaining variable • Substitute the value found in Step 3 into one of the original equations, and solve for the other variable. • Verify the solution in each equation. • Example 3 • Solve the system of equations below by substitution • 3x – y = 12 • 2x + 3y = 2

  7. 12-1: Solving Systems of Equations • 3x – y = 12 • 2x + 3y = 2 • Solve the first equation for y • y = 3x – 12 • Substitute 3x – 12 for y in the second equation • 2x + 3(3x – 12) = 2 • Solve for x • 2x + 9x – 36 = 2 • 11x = 38 • x = 38/11 • Substitute 38/11 into the equation from step 1 • y = 3(38/11) – 12 • y = -18/11 • The solution is (38/11, -18/11)

  8. 12-1: Solving Systems of Equations • Solving Systems Algebraically • Elimination Method • Multiply one or both of the equations by a nonzero constant so that the coefficients of x (or y) are opposite of each other • Eliminate x (or y) by adding the equations, and solve for the remaining variable • Substitute the value found in Step 2 into one of the original equations, and solve for the other variable • Verify the solution in each equation • Example 4 • Solve the system of equations below by substitution • x – 3y = 4 • 2x + y = 1

  9. 12-1: Solving Systems of Equations • x – 3y = 4 • 2x + y = 1 • (Eliminating x) Multiply the top equation by -2 • -2x + 6y = -8 • 2x + y = 1 • Add the equations together • 7y = -7 • Solve for y • y = -1 • Substitute -1 for y into either equation to solve for x • x – 3(-1) = 4 • x + 3 = 4 • x = 1 • The solution is (1, -1)

  10. 12-1: Solving Systems of Equations • Recognizing a System with Infinitely Many or No Solutions • Solve the system • 2x – 4y = 6 • -3x + 6y = -9 • Multiply the top equation by 3 and the bottom equation by 2 • 6x – 12y = 18 • -6x + 12y = -18 • Add the two equations • 0 = 0 • This is a true statement, which means the equations are the same line, which means there are an infinite number of solutions • If your ending statement would come out false, that would mean there is no solution

  11. 12-1: Solving Systems of Equations • Assignment • Page 788 – 789 • Problems 1 – 25 (odds)

  12. 12-1: Solving Systems of Equations • Ex 7: Solving Larger Systems by Elimination • Solve the system by elimination • [1] 2x + y – z = -1 • [2] -x – 3y + z = 5 • [3] x + 4y – 2z = -10 • Eliminate z by adding [1] and [2] • [1] 2x + y – z = -1 • [2] -x – 3y + z = 5 • [4] x – 2y = 4 • Eliminate z by using any two other equations, like [2] & [3] • [2] -x – 3y + z = 5 → multiply by 2 → -2x – 6y + 2z = 10 • [3] x + 4y – 2z = -10x + 4y – 2z = -10 • [5] -x – 2y = 0 (Continued)

  13. 12-1: Solving Systems of Equations • Bringing everything over so far… • [1] 2x + y – z = -1 [4] x – 2y = 4 • [2] -x – 3y + z = 5 [5] -x – 2y = 0 • [3] x + 4y – 2z = -10 • Eliminate x by adding [4] and [5] • [4] x – 2y = 4 • [5] -x – 2y = 0 • -4y = 4 • y = -1 • Substitute y = -1 in [4] to solve for x • x – 2(-1) = 4 • x + 2 = 4 • x = 2 • Substitute y = -1 and x = 2 in [2] to solve for z • –(2) – 3(-1) + z = 5 • -2 + 3 + z = 5 • 1 + z = 5 • z = 4 • Solution is (2, -1, 4)

  14. 12-1: Solving Systems of Equations • Ex 8: Applications of Systems • A ball game is attended by 575 people, and total ticket sales are $2575. If tickets cost $5 for adults and $3 for children, how many adults and how many children attended the game? • Let x = number of adultsLet y = number of children • Two equations • x + y = 575 • 5x + 3y = 2575

  15. 12-1: Solving Systems of Equations • x + y = 575 • 5x + 3y = 2575 • (Eliminating y) Multiply the top equation by -3 • -3x – 3y = -1725 • 5x + 3y = 2575 • Add the equations together • 2x = 850 • Solve for x • x = 425 • Substitute 425 for x into either equation to solve for y • (425) + y = 575 • y = 150 • There were 425 adults (x) and 150 children (y) in attendance

  16. 12-1: Solving Systems of Equations • Ex 9: Mixture Application • A café sells two kinds of coffee in bulk. The Costa Rican sells for $4.50 per pound, and the Kenyan sells for $7.00 per pound. The owner wishes to mix a blend that would sell for $5.00 per pound. How much of each type of coffee should be used in the blend? • Let x = Costa Rican blendLet y = Kenyan blend • Two equations • x + y = 1 (making one pound) • 4.50x + 7.00y = 5

  17. 12-1: Solving Systems of Equations • x + y = 1 • 4.5x + 7y = 5 • (Eliminating y) Multiply the top equation by -7 • -7x – 7y = -7 • 4.5x + 7y = 5 • Add the equations together • -2.5x = -2 • Solve for x • x = 0.8 • Substitute .8 for x into either equation to solve for y • (.8) + y = 1 • y = .2 • The owner should use 80% Costa Rican and 20% Kenyan in the new blend

  18. 12-1: Solving Systems of Equations • Assignment • Page 789 – 790 • Problems 33 & 34, 41 – 49 (odds)

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