# The ordered pair which is the

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The ordered pair which is the
Two linear equations in two variables will be considered together to form a system of two linear equations in two variables. Examples: To solve a system (such as the above examples) means to find all of the ordered pairs that simultaneously satisfy both equations in the system. For example, graphing the two equations on the same set of axes will show the intersection of the two lines. x axis y axis The ordered pair which is the intersection of the two lines is the solution of the system. Therefore {(2, 3)} is the solution set of the system. Next Slide

Since the graph of a linear equation is a straight line, there are three possibilities for the solution of a system of two linear equations. Case 1: Case 2: Case 3: one solution no solution infinitely many solutions Case 1: The graphs of the two equations are two lines intersecting in one point. There is exactly one solution, and the system is called a consistent system. Case 2: The graphs of the two equations are parallel lines. There is no solution, and the system is called an inconsistent system. Case 3: The graphs of the two equations are the same line and there are infinitely many solutions of the system. Any pair of real numbers that satisfies one of the equations also satisfies the other equation, and we say that the equations are dependent. Next Slide

One method of solving a system of equations is by graphing
One method of solving a system of equations is by graphing. However this method becomes impractical. Fortunately we have other methods for solving a system of equations. The Substitution Method Procedure for solving a system of two equations with two unknowns. Step 1. Solve one of the equations for one variable in terms of the other. Step 2. Substitute that expression obtained in Step 1 into the other equation. The result should be an equation with just one variable. Step 3. Solve the equation from Step 2. Step 4. Use the solution obtained in Step 3, along with the expression obtained in Step 1 to determine the solution of the system. Next Slide Step 5. Check the solution in both of the given equations.

Solution: Example 1. Solve the system by substitution: Choose an equation to solve for one variable. Solving for x in the second equation would be easier. Now Substitute 3y +10 for ‘x’ in the first equation. This will give an equation with only one variable. Then solve for y. Next, substitute –2 for y in the equation x=3y+10 to obtain the value for the variable x. The solution set is Then perform the check in both equations. Solve the system by substitution: Your Turn Problem #1

Solution: Example 2. Solve the system by substitution: Let’s solve for x in the second equation. (Either variable can be chosen in either equation.) Now Substitute for ‘x’ in the first equation and solve. Next, substitute 3 for y in the equation solved for x. The solution set is Perform check in both equations to verify answer. (Not shown.) Solve the system by substitution: Your Turn Problem #2

Solution: Example 3. Solve the system by substitution: Solve for y in the first equation. Now Substitute 3x – 6 for ‘y’ in the second equation and solve. Since 12  8, there is no solution. The system is inconsistent with an empty solution set. {} Answer: Solve the system by substitution: Your Turn Problem #3 Answer:

Solution: Example 4. Solve the system by substitution: y is already solved for in the second equation. This true statement, 6=6, indicates that a solution of one of the equations is also a solution of the other, so the solution set is an infinite set of ordered pairs. The two equations are dependent. Answer: To express the solution set, let x = k. The solution set is then: Solve the system by substitution: Your Turn Problem #4 Answer: A dependent system

Another method of solving a system of equations is by elimination
Another method of solving a system of equations is by elimination. This method is necessary because later it will be expanded to solve a system of three equations with three unknowns. The Elimination Method Procedure for solving a system of two equations with two unknowns. Step 1. Write both equations in the form Ax + By = C. Step 2. Multiply one or both equations by appropriate numbers so that the sum of the coefficients of either x or y is zero. Step 3. Add the new equations. The sum should be an equation with just one variable. Step 4. Solve the equation from Step 3. Step 5. Substitute the result of Step 4 into either of the given equations and solve for the other variable. Next Slide Step 6. Check the solution in both of the given equations.

Solution: Example 5. Solve the system by elimination: Choose a variable to eliminate. Eliminating x would be easier. Multiply first equation by –1. Add the equations together then solve for y. Next, substitute 1 for y in either equation to obtain the value for the variable x. Answer: Then perform the check in both equations. Solve the system by elimination: Your Turn Problem #5

Solution: Example 6. Solve the system by elimination: Eliminate y by multiplying second row by 5. Add the equations together then solve for x. Next, substitute 5 for x in either equation to obtain the value for the variable y. Then perform the check in both equations. Answer: Solve the system by elimination: Your Turn Problem #6

Solution: Example 7. Solve the system by elimination: Eliminate x by multiplying first row by –2 and second row by 3. Add the equations together then solve for y. Next, substitute 7 for y in either equation to obtain the value for the variable x. Then perform the check in both equations. Answer: Solve the system by elimination: Your Turn Problem #7

Solution: Example 8. Solve the system by elimination: Eliminate x by multiplying first row by –4. Add the equations together then solve for y. Actually both variables cancel and we have 0 = -9 which is not true. Therefore there is no solution. The system is inconsistent with an empty solution set. Answer: Solve the system by elimination: Your Turn Problem #8 Answer:

Solution: Example 9. Solve the system by elimination: Eliminate x by multiplying first row by 2. Add the equations together then solve for y. Actually both variables cancel and we have 0 = 0 which is true. Therefore there are an infinite number of solutions. The system is dependent. Answer: To express the solution set, let x = k. The solution set is then: Solve the system by elimination: Your Turn Problem #9 Answer:

Applications of Linear Systems of Equations
Many problems involve more than one unknown quantity. Although some problems with two unknowns can be solved using just one variable, many times it is easier to use two variables. To solve a problem with two unknowns, we write two equations that relate the unknown quantities. The system formed by the pair of equations then can be solved using the previous methods. Procedure: Solving an Applied Problem by Writing a System of Equations Step 1. Determine what you are to find. Assign a variable for each unknown and write down what it represents. Step 2. Write down other information. If appropriate, draw a figure or a diagram and label it using the variables from Step 1. Use a chart or a box to summarize the information. Step 3. Write a system of equations. Write as many equations as there are unknowns. Step 4. Solve the system. Step 5. Answer the question(s). Be sure you have answered all questions posed. Step 6. Check. Check your solutions in the original problem. Be sure your answer makes sense.

Solution: Example 10. Solve by using a system of equations: The sum of two numbers is 37 and their difference is –3. Let x = first number and let y = second number We need two equations with two unknowns. Then use either method to solve the system Answer: The two numbers are 17 and 20. Their sum is 37 and their difference is –3. Your Turn Problem #10 Solve by using a system of equations: The sum of two numbers is 51 and their difference is 15. Answer: The two numbers are 33 and 18.

Solution: Example 11. Solve by using a system of equations: Laura and Tony are planning to move and need to buy some cardboard boxes. At Smart and Final, they can buy either 10 small and 12 medium boxes fro \$42, or 5 small and 10 medium boxes for \$29. Find the cost of each size box. Let x = cost of a small box and let y = cost of a medium box We need two equations with two unknowns. Then use either method to solve the system Answer: The cost of a small box is \$1.80 and the cost of a medium box is \$2.00. Your Turn Problem #11 Solve by using a system of equations: At the Rancho Nut Ranch, 5 pounds of peanuts and 6 pounds of cashews cost \$70, and 3 pounds of peanuts and 7 pounds of cashews cost \$76. Find the prices charged by the ranch for a pound of peanuts and a pound of cashews. Answer: Peanuts are \$2.00 per pound Cashews are \$10.00 per pound

Example 12. Solve by using a system of equations:
Tickets to a play at Cucamonga Jr. High cost \$2.50 for general admission and \$2.00 with a student identification. If 92 people paid to see a performance and \$203 was collected, how many of each type of admission were sold? Solution: Let’s use substitution. Answer: # of generals admission = 38 # of students = 54 Let x = number of general admission tickets sold let y = number of student tickets sold We need two equations with two unknowns. Your Turn Problem #12 Solve by using a system of equations: On a 10-day trip Dan rented a car for \$31 a day at weekend rates and \$40 a day at weekday rates. If his rental bill was \$364, how many days did he rent a each rate? The End. B.R. Answer: Dan rented the car for 4 weekend days and 6 weekdays.