1. Review 3-1 and 3-2 SLIDE SHOW

2. Solve the linear system using the substitution method. 3 x + 4y  – 4 Equation 1 x + 2y  2 Equation 2 x + 2y  2 x  – 2y + 2 3(– 2y + 2) + 4y  – 4 y  5y  5 S OLUTION -6y+6+4y  – 4 -2y = -10 x  – 2y + 2 x  – 2(5) + 2 x  – 8x  – 8 The solution is (– 8, 5). 1.

3. x – 2 y  3 2 x – 4 y  7 2(2 y + 3) – 4 y  7 6  7 Because the statement 6 = 7 is never true, there is no solution. 2. Solve the linear system using the substitution method. x  2 y + 3

4. Solve the linear system using the linear combination/elimination method. 2 x – 4y  13 Equation 1 4 x – 5y  8 Equation 2 – 4x + 8y  – 26 4 x – 5y  8 2 x – 4y  13 4 x – 5y  8 3y  –18 y  – 6y  – 6 Multiply the first equation by – 2 so that x -coefficients differ only in sign. S OLUTION – 2 3.

5. The Linear Combination Method: Multiplying One Equation 2 x – 4y  13 2 x – 4(– 6)  13 2 x + 24  13 x  – 11 2 The solution is –, – 6. (- 5 1 2 ) y  – 6y  – 6 Add the revised equations and solve for y. Write Equation 1. Substitute – 6 for y. Simplify. Solve for x. Substitute the value of y into one of the original equations. Solve the linear system using the linear combination method. 2 x – 4y  13 Equation 1 4 x – 5y  8 Equation 2

6. 6 x – 10 y  12 – 15 x + 25 y  – 30 Solve the linear system 6 x – 10 y  12 – 15 x + 25 y  – 30 30 x – 50 y  60 – 30 x + 50 y  – 60 0  0 Add the revised equations. Since no coefficient is 1 or –1, use the linear combination method. Because the equation 0 = 0 is always true, there are infinitely many solutions. S OLUTION 5 2 4.

7. 7 x – 12 y  – 22 Equation 1 – 5 x + 8 y  14 Equation 2 Solve the linear system using the linear combination method. 7 x – 12 y  – 22 – 5 x + 8 y  14 14 x – 24y  – 44 – 15 x + 24y  42 Add the revised equations and solve for x. – x  – 2– x  – 2 x  2 Multiply the first equation by 2 and the second equation by 3 so that the coefficients of y differ only in sign. S OLUTION 2 3 5.

8. – 5 x + 8 y  14 y = 3 – 5 (2) + 8 y  14 The solution is (2, 3). x  2 Add the revised equations and solve for x. Write Equation 2. Substitute 2 for x. Solve for y. Substitute the value of x into one of the original equations. Solve for y. 7 x – 12 y  – 22 Equation 1 – 5 x + 8 y  14 Equation 2 Solve the linear system using the linear combination method. Check the solution algebraically or graphically.

9. 6. If Brian bought 6 markers and 12 pens for $21.60 and then had to go back and buy 18 more markers and 20 pens for $50.40. How much was each item? 7. If Maria bought 33 books notebooks for $393. Each book costs $23.50 and each notebook costs $2.25. How many of each did she purchase? 6x + 12y = 21.60 18x + 20y = 50.40 x = 1.8 y =.9 Markers costs $1.80 Pens costs $0.90 x + y = 33 23.5x + 2.25y = 393 x = 15 y = 18 15 Books 18 Notebooks

10. Homework Textbook Page 155 Quiz 1 1-12 all