Name: Period: Date: 1. Draw the graph of the following system: 3 x+ 5 y+ 13 = 0 29 x 11 y 7 = 0 3 13 y = x 3x+ 5y+ 13= 0 5 5 29x 11y 7 = 0 29 7 y = x 11 11 Practice Test Page 1
2. Determine the ordered triple below that is a solution of the given system of equations. x 4 y+ 5z = 39 5 x 5 y 7 z = 31 5 x+ y+ 7z = 93 ( xyz,, ) = ( 7, 2,8) 3. An augmented matrix that represents a system of linear equations (in variables x, y, and z) has been reduced using Gauss-Jordan elimination (i.e. Reduced Row-Echelon Form.) Write the solution represented by the augmented matrix. 1 0 0 8 0 1 0 7 001 9 ( xyz,, ) = ( 8, 7,9) 4. Perform the indicated row operations on the matrix. Show the final result. 1 1 4 7 6 8 7 5 12 Add 7 times R 1 to R 2. Add 7 times R 1 to R 3. 1 1 4 0 1 20 0 2 40 Version Practice Page 2
5. Find the equilibrium point of the demand and supply equations. (The equilibrium point is the price p and number of units x that satisfy both the demand and supply equations.) Demand Supply p = 71 0.07x p = 0.5x 100 We are looking for the combination of ( xp, ) that will be the same in both equations. Step 1: Set the two expressions for P equal to each other: 71 0.07x= 0.5x 100 Step 2: (Optional) If you don t like decimals, multiply both sides of the equation by 100. 7100 7x= 50x 10000 Step 3: Solve for x. 17100 = 57x 17100 = x 57 300 x Step 4: Substitute 300 for x in one of the equations and solve for P. ( ) P = 0.5 300 100 P = 150 100 P = 50 Step 5: Finalize the answer. ( xp, ) = ( 300,50) Version Practice Page 3
6. Determine the ordered pair that is a solution of the system. 6 x 4 y = 3 2 x 9 y = 9 Step 1: Multiply the bottom equation by 3. 6x 4y = 3 6x 27 y = 27 Step 2: Add the two equations. 31y = 30 y = Step 3: Substitute 30 31 Step 4: Finalize. 30 31 30 6x 4 = 3 31 120 6x = 3 31 120 93 6x = 31 31 27 6x = 31 27 x = 186 9 x = 62 9 30, =, 62 31 ( xy) for y and solve. Version Practice Page 4
7. Write the system of linear equations represented by the augmented matrix. Then use back-substitution to solve. (Use variables x, y, and z.) 1 6 8 26 0 1 3 5 0 0 1 2 x+ 6y+ 8z = 26 1y 3z = 5 z = 2 z = 2 1y 3z = 5 ( ) y 3 2 = 5 y + 6= 5 y = 1 x+ 6y+ 8z = 26 ( ) ( ) x + 6 1 + 8 2 = 26 x 6 16 = 26 x 22 = 26 x = 4 ( xyz,, ) = ( 4, 1, 2) Version Practice Page 5
8. Find the equation of the parabola 2 y = ax + bx + c that passes through the points. ( 3, 3 ), ( 1, 2 ), ( 1,1) Use the x- and y- coordinates of the given points to make a system of equations. = ( ) 2 + b( 3) + c = 3 ( ) 2 + b( 1) + c = 2 ( ) 2 + b( 1) + c = 1 a 3 a 1 a 1 = 9a 3b + c = 3 a b + c = 2 a + b + c = 1 rref 9 3 1 3 1 0 0 1 1 1 1 2 0 1 0 1.5 = 1 1 1 1 0 0 1 1.5 2 3 3 y = x + x 2 2 9. Fill in the blank using elementary row operations to form a row-equivalent matrix. 1 2 6 2 8 4 1 2 6 0 8 In order for the element in row 2 column 1 to turn from a 2 to a 0, row one must have been multiplied by - 2 and added to row two and row two replaced. Therefore, when the 2 in row one column two is multiplied by - 2 we get - 4. When - 4 is added to the - 8 in row two column two, the result is: - 12. Version Practice Page 6
10. Determine which ordered pair is a solution of the system. 2 x+ 2 y = 4 7 x+ 6 y = 48 From equation 1 we get: x= 4 2y 2 Substitute this expression for x in the second equation. 7( 4 2y ) 2 + 6y = 48 28 14y 2 + 6y = 48 28 14y 2 + 6y + 48 = 0 14y 2 + 6y + 20 = 0 ( ) = 0 2 7 y 2 3y 10 7 y 2 3y 10 = 0 ( 7 y 10) ( y +1) = 0 y = 10 7, 1 Substitute these values into either of the two equations to find the value of x. x = 4 2 10 2 7 x = 4 2 100 49 x = 4 200 49 x = 196 49 200 49 x = 396 49 ( x, y) = 396 49,10 7 x = 4 2( 1) 2 x = 4 2( 1) x = 4 2 x = 6 ( x, y) = ( 6, 1) Version Practice Page 7
11. Write the form of the partial fraction decomposition of the rational expression. Do not solve for the constants. 2 x 2 2x x 2 2 2 = 2x x x ( 2) A B + x x 2 12. Identify the elementary row operation being performed to obtain the new rowequivalent matrix. Original Matrix New Row-Equivalent Matrix 3 5 9 3 8 1 0 3 8 0 3 8 After careful observation one should note the second row has stayed the same but row 1 has changed. Noting the - 5 in row one column two was changed to - 8, we can assume - 3 was added to - 5. In order for the 3 from row two column two to become a - 3, the second row must have been multiplied by - 1. Thus, Add 1 times Row 2 to Row1. Version Practice Page 8
13. Solve the system of linear equations. x + y + z + w= 10 3 x 4 y+ 4 z w= 14 4 x 2 z+ 4 w= 4 3 x 4 y+ 4 z+ 4 w= 7 Once again, use the matrix features of your graphing calculator to solve. rref 1 1 1 1 10 1 0 0 0 1 3 4 4 1 14 0 1 0 0 4 = 4 0 2 4 4 0 0 1 0 2 3 4 4 4 7 0 0 0 1 3 Therefore: ( xyzw,,, ) = ( 1, 4, 2, 3) Version Practice Page 9
1. Answer Key Similar to Exercise: 7.2.18 2. (7, 2, 8) Similar to Exercise: 7.3.2b 3. x = 8, y = 7, z = 9 Similar to Exercise: 7.4.48 4. 1 1 4 0 1 20 0 2 40 Similar to Exercise: 7.4.29 300,50 5. ( ) Similar to Exercise: 7.2.70 6. 9 30, 62 31 Similar to Exercise: 7.1.1b 7. x = 4, y = 1, z = 2 Similar to Exercise: 7.4.44 8. 2 3 3 y = x + x 2 2 9. 1 2 6 0 12 8 Similar to Exercise: 7.4.16 10. 396 10 ( 6, 1) &, 49 7 Similar to Exercise: 7.1.2b 11. A B + x x 2 Similar to Exercise: 7.3.57 12. Add 1 times R 2 to R 1. Similar to Exercise: 7.4.22 13. Version Practice Page 10
( 1, 4, 2, 3) Similar to Exercise: 7.3.44 Version Practice Page 11