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Section 4.5, Part A Solving Problems with Systems of Linear Equations 1 2

Steps in Solving Problems Involving Systems of Two Linear Equations in Two Variables: 1)Understand the problem. Read and reread the problem. Choose two variables to represent the two unknowns. 2)Translate the problem into two equations. 3)Solve the system of equations. 4)Interpret the results. Check proposed solution in the problem. State your conclusion. Check proposed solution in the problem !!!!!

Example continued One number is 4 more than twice the second number. Their total is 25. Find the numbers. Read and reread the problem. Since we are looking for two numbers, we let x = first number y = second number 1.UNDERSTAND

continued 2. TRANSLATE continued One number is 4 more than twice the second number. x = 2y + 4 Their total is 25. x + y = 25

3.SOLVE continued Using the substitution method, we substitute the solution for x from the first equation into the second equation. x + y = 25 (2y +4) + y = 25 Replace x with 4 + 2y. 3y + 4 = 25 Simplify. 3y = 21 Subtract 4 from both sides. y = 7 Divide both sides by 3. We are solving the system x = 2y +4 x + y = 25

4.INTERPRET continued Check: Substitute x = 18 and y = 7 into both of the equations. First equation: x = 4 + 2y 18 = 4 + 2(7) True Second equation: x + y = = 25 True State: The two numbers are 18 and 7. Now we substitute 7 for y into the first equation. x = 4 + 2y = 4 + 2(7) = = 18

Example continued Hilton University Drama club sold 311 tickets for a play. Student tickets cost 50 cents each; non- student tickets cost $1.50. If the total receipts were $385.50, find how many tickets of each type were sold. 1.UNDERSTAND Read and reread the problem. Since we are looking for two numbers, we let s = the number of student tickets n = the number of non-student tickets

continued 2.TRANSLATE Hilton University Drama club sold 311 tickets for a play. s + n = 311 total receipts were $ s Total receipts = Admission for students 1.50n Admission for non-students + continued

3.SOLVE continued We are solving the system s + n = s n = Since the equations are written in standard form (and we might like to get rid of the decimals anyway), we’ll solve by the addition/elimination method. (Substitution could be used instead, if you prefer to do it that way.) Question: If we wanted to eliminate s, what would we multiply the 0.50s in the second equation by to make it become -1s? Answer: Multiply the second equation by –2. s + n = 311 –2(0.50s n) = –2(385.50) s + n = 311 –s – 3n = –771 –2n = –460 n = 230 continued

4.INTERPRET Check: Substitute s = 81 and n = 230 into both of the equations. s + n = 311 First Equation = 311 True 0.50s n = Second Equation = True 0.50(81) (230) = State: There were 81 student tickets and 230 non student tickets sold. Now we substitute 230 for n into the first equation to solve for s. s + n = 311 s = 311 s = 81 continued

How could we set this problem up using two variables? x = ? Ounces of 16% solution y = ? Ounces of 8 % solution Equation 1? x + y = 32 Equation 2? 0.16 ● x ● y = 0.11●32 How would you check these answers?

The assignment on this material (HW 4.5A) Is due at the start of the next class session. You may now OPEN your LAPTOPS and begin working on the homework assignment.