Supplemental Worksheet Problems To Accompany: The Algebra 2 Tutor Section 8 Solving Systems of Equations in Three Variables

Transcription

1 Supplemental Worksheet Problems To Accompany: The Algebra 2 Tutor Please watch Section 8 of this DVD before working these problems. The DVD is located at: Sample Videos For this DVD Are Located Here: Page 1

2 1) Solve the system of equations: x+ y+ z = 4 x y+ z = 2 x y z = 0 Page 2

3 2) Solve the system of equations: x y+ z = 4 x+ 2y z = 1 x+ y 3z = 2 Page 3

4 3) Solve the system of equations: x+ 2y+ 2z = 10 2x+ y+ 2z 2x+ 2y+ 1 Page 4

5 Question Answer 1) Solve the system of equations: x+ y+ z = 4 x y+ z = 2 x y z = 0 Begin. z = x y We begin by solving the third equation for z. Subtract z from both sides of the third equation to arrive at an equation for z. x+ y+ z = 4 plug in z = x-y x+ y+ ( x- y) = 4 Use the equation for z that we just solved for and plug that value of z into the first and second equation. AND x y+ z = 2 plug in z=x-y x y+ ( x y) =2 2x = 4 2x 2y = 2 By plugging in z we have arrived at two equations in x and y only. Combine like terms in each of these two equations. (continued on next page) Page 5

6 Working with the first of these two equations, solve for x. To do this, divide both sides by 2. This is the final solution for x. 2x 2y = 2 plug in x=2... ( 2) 2 2y =2 Working with the second of these two equations ( 2x 2y = 2), plug in the value of x that we just found. 4 2y = 2 Do the multiplication on the left. 2y = 2 Subtract 4 from both sides. y = 1 x+ y+ z = 4 plug in x=2 and y= z = 4 Divide both sides by -2. This is the final solution for y. Now we have found the solution for x and y. To find z, just plug in our values of x and y back into one of the original equations. We choose to plug into the first of the original three equations. (continued on next page) Page 6

7 3+ z = 4 y = 1 Do the addition on the left. Subtract 3 from both sides. We now have the solutions for x, y, and z. Ans: y = 1 Page 7

8 Question Answer 2) Solve the system of equations: x y+ z = 4 x+ 2y z = 1 x+ y 3z = 2 Begin. x = y z+ 4 We begin by solving the first equation for x. x+ 2y z = 1 plug in x = y - z+4 y z+ 4+ 2y z = 1 Use the equation for x that we just solved for and plug that value of x into the second and third equation. AND x+ y 3z = 2 plug in x = y - z + 4 y z+ 4+ y 3z = 2 3y 2z = 5 2y 4z = 6 By plugging in x we have arrived at two equations in y and z only. Combine like terms in each of these two equations. (continued on next page) Page 8

9 -2 + 2y ( 3y - 2z = -5) - 4z = -6 We need to solve these two equations for y and z. We can use any method we like. We choose to use the addition method on these two equations. We will first multiply the first of these two equations by -2 then add them together. This will get rid of the z variable for us. -6y + 4z = y - 4z = -6-6y+ 4z = y - 4z = -6-4y = 4 Perform the multiplication on the first equation. Add the equations together. y = 1 Working with the equation we just found, -4y = 4, divide both sides by -4. This is the final solution for y. 2y 4z = 6 plug in y = ( 1) 2 4z = 6 To find x, plug in y = -1 into the intermediate equation we found,. 2y - 4z = z = 6 Do the multiplication on the left. 4z = 4 Add 2 to both sides. Divide both sides by -4. This is the final solution for z. (continued on next page) Page 9

10 x y+ z = 4 plug in y = -1 and z=1 ( ) x = 4 x + 2= 4 y = 1 Plug in the values of y and z into one of the original equations. We choose to plug into the first of the three original equations. Do the addition on the left. Subtract 2 from both sides. We now have the solutions for x, y, and z. Ans: y = 1 Page 10

11 Question Answer 3) Solve the system of equations: x+ 2y+ 2z = 10 2x+ y+ 2z 2x+ 2y+ 1 Begin. x = 10 2y 2z We begin by solving the first equation for x. 2x + y+ 2z plug in x = 10-2y - 2z ( 10-2y - 2z) 2 + y+ 2z Use the equation for x that we just solved for and plug that value of x into the second and third equation. AND 2x+ 2y+ 1 plug in x = 10-2y - 2z ( 10-2y - 2z) 2 + 2y+ 1 2 ( 10-2y - 2z) 20 4y 4z+ y+ 2z 20 3y 2z 3y 2z = 11 AND ( 10-2y - 2z) + y+ 2z 2 + 2y y 4z+ 2y y 3z = 11 2y 3z = 9 For each new equation we have just found, do the distribution inside the parenthesis and combine like terms. We have simplified the resulting equations and now we need to solve the two purple equations at left for y and z. (continued on next page) Page 11

12 Question Answer 3y 2z = 11 2y 3z = 9-2 ( ( ) -3y - 2z = -11) + 3-2y - 3z = -9 We must solve the purple equations to find y and z. We can use any method we want to solve this system. We choose to solve it by addition. To eliminate y from the system: Multiply the first equation by -2. Multiply the second equation by 3 Add the equations. + 6y + 4z = 22-6y - 9z = -27 Perform the multiplications. + 6y + 4z = 22-6y - 9z = -27-5z = -5 Add these equations together. To solve for z, divide both sides of this equation by -5. We have found the solution for z. 3y 2z = 11 Substitute: z=1... () 3y 21 = 11 To solve for y, substitute what we found for z back into either of the previous equations involving y and z. We choose to substitute back into: 3y 2z = 11 (continued on next page). Page 12

13 3y 2= 11 3y = 9 y = 3 Do the multiplication. Add 2 to both sides. Divide both sides by -3. x+ 2y+ 2z = 10 To find x, substitute what we have found for Sub: y=3 and z=1... y and z back into one of the original equations. We choose to substitute back x + 2( 3) + 2( 1) = 10 into the equation: x+ 2y+ 2z = 10 x = 10 Do the multiplications. x + 8= 10 Do the addition on the left. Subtract 8 from both sides. We have now found the solution for x. y = 3 We have solved for x, y, and z. Ans: y = 3 Page 13