System of Equations worksheet

1. Solution: x = 12/19, y = 91/38, z = 35/38

1. Your system of equations

   $\text{3x + 3y + z = 10; x − y + 3z = 1; 4x + y − z = 4}$

2. Write the system as a grid (an augmented matrix): the numbers in front of each letter, a bar, then the right-hand sides

   $\text{[ 3, 3, 1 | 10 ] [ 1, −1, 3 | 1 ] [ 4, 1, −1 | 4 ]}$

3. Multiply row 1 so its pivot becomes 1, which makes the next steps easy: R1 → 1/3·R1

   $\text{[ 1, 1, 1/3 | 10/3 ] [ 1, −1, 3 | 1 ] [ 4, 1, −1 | 4 ]}$

4. Subtract a multiple of row 1 from row 2 to turn that column's entry into 0: R2 → R2 − R1

   $\text{[ 1, 1, 1/3 | 10/3 ] [ 0, −2, 8/3 | −7/3 ] [ 4, 1, −1 | 4 ]}$

5. Subtract a multiple of row 1 from row 3 to turn that column's entry into 0: R3 → R3 − 4·R1

   $\text{[ 1, 1, 1/3 | 10/3 ] [ 0, −2, 8/3 | −7/3 ] [ 0, −3, −7/3 | −28/3 ]}$

6. Multiply row 2 so its pivot becomes 1, which makes the next steps easy: R2 → −1/2·R2

   $\text{[ 1, 1, 1/3 | 10/3 ] [ 0, 1, −4/3 | 7/6 ] [ 0, −3, −7/3 | −28/3 ]}$

7. Subtract a multiple of row 2 from row 1 to turn that column's entry into 0: R1 → R1 − R2

   $\text{[ 1, 0, 5/3 | 13/6 ] [ 0, 1, −4/3 | 7/6 ] [ 0, −3, −7/3 | −28/3 ]}$

8. Subtract a multiple of row 2 from row 3 to turn that column's entry into 0: R3 → R3 + 3·R2

   $\text{[ 1, 0, 5/3 | 13/6 ] [ 0, 1, −4/3 | 7/6 ] [ 0, 0, −19/3 | −35/6 ]}$

9. Multiply row 3 so its pivot becomes 1, which makes the next steps easy: R3 → −3/19·R3

   $\text{[ 1, 0, 5/3 | 13/6 ] [ 0, 1, −4/3 | 7/6 ] [ 0, 0, 1 | 35/38 ]}$

10. Subtract a multiple of row 3 from row 1 to turn that column's entry into 0: R1 → R1 − 5/3·R3

    $\text{[ 1, 0, 0 | 12/19 ] [ 0, 1, −4/3 | 7/6 ] [ 0, 0, 1 | 35/38 ]}$

11. Subtract a multiple of row 3 from row 2 to turn that column's entry into 0: R2 → R2 + 4/3·R3

    $\text{[ 1, 0, 0 | 12/19 ] [ 0, 1, 0 | 91/38 ] [ 0, 0, 1 | 35/38 ]}$

12. So the one solution is

    $\text{x = 12/19, y = 91/38, z = 35/38}$

2. Solution: x = 11/5, y = 12/5

1. Your system of equations

   $\text{2x + 4y = 14; 2x − y = 2}$

2. Cramer's rule: each unknown is one determinant divided by another. D uses the coefficients. D_k swaps column k for the right-hand sides

   $\text{x = Dx/D, y = Dy/D}$

3. Find D, the determinant of the coefficients. If it isn't 0, there's exactly one answer

   $\text{D}$$=\mathrm{-10}$

4. Replace column 1 (the x coefficients) with the constants and take the determinant: Dx = −22. Then x = Dx ÷ D

   $\text{x}$$=\frac{\mathrm{-22}}{\mathrm{-10}}$$=\frac{11}{5}$

5. Replace column 2 (the y coefficients) with the constants and take the determinant: Dy = −24. Then y = Dy ÷ D

   $\text{y}$$=\frac{\mathrm{-24}}{\mathrm{-10}}$$=\frac{12}{5}$

6. So the one solution is

   $\text{x = 11/5, y = 12/5}$

3. Solution: x = 12/5, y = 7/5

1. Your system of equations

   $\text{2x + 3y = 9; x − y = 1}$

2. Write the system as a grid (an augmented matrix): the numbers in front of each letter, a bar, then the right-hand sides

   $\text{[ 2, 3 | 9 ] [ 1, −1 | 1 ]}$

3. Multiply row 1 so its pivot becomes 1, which makes the next steps easy: R1 → 1/2·R1

   $\text{[ 1, 3/2 | 9/2 ] [ 1, −1 | 1 ]}$

4. Subtract a multiple of row 1 from row 2 to turn that column's entry into 0: R2 → R2 − R1

   $\text{[ 1, 3/2 | 9/2 ] [ 0, −5/2 | −7/2 ]}$

5. Multiply row 2 so its pivot becomes 1, which makes the next steps easy: R2 → −2/5·R2

   $\text{[ 1, 3/2 | 9/2 ] [ 0, 1 | 7/5 ]}$

6. Subtract a multiple of row 2 from row 1 to turn that column's entry into 0: R1 → R1 − 3/2·R2

   $\text{[ 1, 0 | 12/5 ] [ 0, 1 | 7/5 ]}$

7. So the one solution is

   $\text{x = 12/5, y = 7/5}$

4. Solution: x = 3/2, y = 5/2, z = 1/2

1. Your system of equations

   $\text{2x + 3y + z = 11; x − y + 4z = 1; 2x + y − z = 5}$

2. Cramer's rule: each unknown is one determinant divided by another. D uses the coefficients. D_k swaps column k for the right-hand sides

   $\text{x = Dx/D, y = Dy/D, z = Dz/D}$

3. Find D, the determinant of the coefficients. If it isn't 0, there's exactly one answer

   $\text{D}$$=24$

4. Replace column 1 (the x coefficients) with the constants and take the determinant: Dx = 36. Then x = Dx ÷ D

   $\text{x}$$=\frac{36}{24}$$=\frac{3}{2}$

5. Replace column 2 (the y coefficients) with the constants and take the determinant: Dy = 60. Then y = Dy ÷ D

   $\text{y}$$=\frac{60}{24}$$=\frac{5}{2}$

6. Replace column 3 (the z coefficients) with the constants and take the determinant: Dz = 12. Then z = Dz ÷ D

   $\text{z}$$=\frac{12}{24}$$=\frac{1}{2}$

7. So the one solution is

   $\text{x = 3/2, y = 5/2, z = 1/2}$

5. Solution: x = 6/5, y = 7/2, z = 11/10

1. Your system of equations

   $\text{2x + 3y + z = 14; x − y + 3z = 1; 3x + y − z = 6}$

2. Write the system as a grid (an augmented matrix): the numbers in front of each letter, a bar, then the right-hand sides

   $\text{[ 2, 3, 1 | 14 ] [ 1, −1, 3 | 1 ] [ 3, 1, −1 | 6 ]}$

3. Multiply row 1 so its pivot becomes 1, which makes the next steps easy: R1 → 1/2·R1

   $\text{[ 1, 3/2, 1/2 | 7 ] [ 1, −1, 3 | 1 ] [ 3, 1, −1 | 6 ]}$

4. Subtract a multiple of row 1 from row 2 to turn that column's entry into 0: R2 → R2 − R1

   $\text{[ 1, 3/2, 1/2 | 7 ] [ 0, −5/2, 5/2 | −6 ] [ 3, 1, −1 | 6 ]}$

5. Subtract a multiple of row 1 from row 3 to turn that column's entry into 0: R3 → R3 − 3·R1

   $\text{[ 1, 3/2, 1/2 | 7 ] [ 0, −5/2, 5/2 | −6 ] [ 0, −7/2, −5/2 | −15 ]}$

6. Multiply row 2 so its pivot becomes 1, which makes the next steps easy: R2 → −2/5·R2

   $\text{[ 1, 3/2, 1/2 | 7 ] [ 0, 1, −1 | 12/5 ] [ 0, −7/2, −5/2 | −15 ]}$

7. Subtract a multiple of row 2 from row 1 to turn that column's entry into 0: R1 → R1 − 3/2·R2

   $\text{[ 1, 0, 2 | 17/5 ] [ 0, 1, −1 | 12/5 ] [ 0, −7/2, −5/2 | −15 ]}$

8. Subtract a multiple of row 2 from row 3 to turn that column's entry into 0: R3 → R3 + 7/2·R2

   $\text{[ 1, 0, 2 | 17/5 ] [ 0, 1, −1 | 12/5 ] [ 0, 0, −6 | −33/5 ]}$

9. Multiply row 3 so its pivot becomes 1, which makes the next steps easy: R3 → −1/6·R3

   $\text{[ 1, 0, 2 | 17/5 ] [ 0, 1, −1 | 12/5 ] [ 0, 0, 1 | 11/10 ]}$

10. Subtract a multiple of row 3 from row 1 to turn that column's entry into 0: R1 → R1 − 2·R3

    $\text{[ 1, 0, 0 | 6/5 ] [ 0, 1, −1 | 12/5 ] [ 0, 0, 1 | 11/10 ]}$

11. Subtract a multiple of row 3 from row 2 to turn that column's entry into 0: R2 → R2 + R3

    $\text{[ 1, 0, 0 | 6/5 ] [ 0, 1, 0 | 7/2 ] [ 0, 0, 1 | 11/10 ]}$

12. So the one solution is

    $\text{x = 6/5, y = 7/2, z = 11/10}$

6. Solution: x = 3, y = 1

1. Your system of equations

   $\text{2x + 3y = 9; x − y = 2}$

2. Cramer's rule: each unknown is one determinant divided by another. D uses the coefficients. D_k swaps column k for the right-hand sides

   $\text{x = Dx/D, y = Dy/D}$

3. Find D, the determinant of the coefficients. If it isn't 0, there's exactly one answer

   $\text{D}$$=\mathrm{-5}$

4. Replace column 1 (the x coefficients) with the constants and take the determinant: Dx = −15. Then x = Dx ÷ D

   $\text{x}$$=\frac{\mathrm{-15}}{\mathrm{-5}}$$=3$

5. Replace column 2 (the y coefficients) with the constants and take the determinant: Dy = −5. Then y = Dy ÷ D

   $\text{y}$$=\frac{\mathrm{-5}}{\mathrm{-5}}$$=1$

6. So the one solution is

   $\text{x = 3, y = 1}$

7. Solution: x = 1, y = 3, z = 2

1. Your system of equations

   $\text{2x + 3y + z = 13; x − y + 2z = 2; 4x + y − z = 5}$

2. Write the system as a grid (an augmented matrix): the numbers in front of each letter, a bar, then the right-hand sides

   $\text{[ 2, 3, 1 | 13 ] [ 1, −1, 2 | 2 ] [ 4, 1, −1 | 5 ]}$

3. Multiply row 1 so its pivot becomes 1, which makes the next steps easy: R1 → 1/2·R1

   $\text{[ 1, 3/2, 1/2 | 13/2 ] [ 1, −1, 2 | 2 ] [ 4, 1, −1 | 5 ]}$

4. Subtract a multiple of row 1 from row 2 to turn that column's entry into 0: R2 → R2 − R1

   $\text{[ 1, 3/2, 1/2 | 13/2 ] [ 0, −5/2, 3/2 | −9/2 ] [ 4, 1, −1 | 5 ]}$

5. Subtract a multiple of row 1 from row 3 to turn that column's entry into 0: R3 → R3 − 4·R1

   $\text{[ 1, 3/2, 1/2 | 13/2 ] [ 0, −5/2, 3/2 | −9/2 ] [ 0, −5, −3 | −21 ]}$

6. Multiply row 2 so its pivot becomes 1, which makes the next steps easy: R2 → −2/5·R2

   $\text{[ 1, 3/2, 1/2 | 13/2 ] [ 0, 1, −3/5 | 9/5 ] [ 0, −5, −3 | −21 ]}$

7. Subtract a multiple of row 2 from row 1 to turn that column's entry into 0: R1 → R1 − 3/2·R2

   $\text{[ 1, 0, 7/5 | 19/5 ] [ 0, 1, −3/5 | 9/5 ] [ 0, −5, −3 | −21 ]}$

8. Subtract a multiple of row 2 from row 3 to turn that column's entry into 0: R3 → R3 + 5·R2

   $\text{[ 1, 0, 7/5 | 19/5 ] [ 0, 1, −3/5 | 9/5 ] [ 0, 0, −6 | −12 ]}$

9. Multiply row 3 so its pivot becomes 1, which makes the next steps easy: R3 → −1/6·R3

   $\text{[ 1, 0, 7/5 | 19/5 ] [ 0, 1, −3/5 | 9/5 ] [ 0, 0, 1 | 2 ]}$

10. Subtract a multiple of row 3 from row 1 to turn that column's entry into 0: R1 → R1 − 7/5·R3

    $\text{[ 1, 0, 0 | 1 ] [ 0, 1, −3/5 | 9/5 ] [ 0, 0, 1 | 2 ]}$

11. Subtract a multiple of row 3 from row 2 to turn that column's entry into 0: R2 → R2 + 3/5·R3

    $\text{[ 1, 0, 0 | 1 ] [ 0, 1, 0 | 3 ] [ 0, 0, 1 | 2 ]}$

12. So the one solution is

    $\text{x = 1, y = 3, z = 2}$

8. Solution: x = 7/2, y = 5/4

1. Your system of equations

   $\text{3x + 2y = 13; x − 2y = 1}$

2. Write the system as a grid (an augmented matrix): the numbers in front of each letter, a bar, then the right-hand sides

   $\text{[ 3, 2 | 13 ] [ 1, −2 | 1 ]}$

3. Multiply row 1 so its pivot becomes 1, which makes the next steps easy: R1 → 1/3·R1

   $\text{[ 1, 2/3 | 13/3 ] [ 1, −2 | 1 ]}$

4. Subtract a multiple of row 1 from row 2 to turn that column's entry into 0: R2 → R2 − R1

   $\text{[ 1, 2/3 | 13/3 ] [ 0, −8/3 | −10/3 ]}$

5. Multiply row 2 so its pivot becomes 1, which makes the next steps easy: R2 → −3/8·R2

   $\text{[ 1, 2/3 | 13/3 ] [ 0, 1 | 5/4 ]}$

6. Subtract a multiple of row 2 from row 1 to turn that column's entry into 0: R1 → R1 − 2/3·R2

   $\text{[ 1, 0 | 7/2 ] [ 0, 1 | 5/4 ]}$

7. So the one solution is

   $\text{x = 7/2, y = 5/4}$

9. Solution: x = 23/47, y = 136/47, z = 40/47

1. Your system of equations

   $\text{3x + 3y + z = 11; x − y + 4z = 1; 4x + y − z = 4}$

2. Write the system as a grid (an augmented matrix): the numbers in front of each letter, a bar, then the right-hand sides

   $\text{[ 3, 3, 1 | 11 ] [ 1, −1, 4 | 1 ] [ 4, 1, −1 | 4 ]}$

3. Multiply row 1 so its pivot becomes 1, which makes the next steps easy: R1 → 1/3·R1

   $\text{[ 1, 1, 1/3 | 11/3 ] [ 1, −1, 4 | 1 ] [ 4, 1, −1 | 4 ]}$

4. Subtract a multiple of row 1 from row 2 to turn that column's entry into 0: R2 → R2 − R1

   $\text{[ 1, 1, 1/3 | 11/3 ] [ 0, −2, 11/3 | −8/3 ] [ 4, 1, −1 | 4 ]}$

5. Subtract a multiple of row 1 from row 3 to turn that column's entry into 0: R3 → R3 − 4·R1

   $\text{[ 1, 1, 1/3 | 11/3 ] [ 0, −2, 11/3 | −8/3 ] [ 0, −3, −7/3 | −32/3 ]}$

6. Multiply row 2 so its pivot becomes 1, which makes the next steps easy: R2 → −1/2·R2

   $\text{[ 1, 1, 1/3 | 11/3 ] [ 0, 1, −11/6 | 4/3 ] [ 0, −3, −7/3 | −32/3 ]}$

7. Subtract a multiple of row 2 from row 1 to turn that column's entry into 0: R1 → R1 − R2

   $\text{[ 1, 0, 13/6 | 7/3 ] [ 0, 1, −11/6 | 4/3 ] [ 0, −3, −7/3 | −32/3 ]}$

8. Subtract a multiple of row 2 from row 3 to turn that column's entry into 0: R3 → R3 + 3·R2

   $\text{[ 1, 0, 13/6 | 7/3 ] [ 0, 1, −11/6 | 4/3 ] [ 0, 0, −47/6 | −20/3 ]}$

9. Multiply row 3 so its pivot becomes 1, which makes the next steps easy: R3 → −6/47·R3

   $\text{[ 1, 0, 13/6 | 7/3 ] [ 0, 1, −11/6 | 4/3 ] [ 0, 0, 1 | 40/47 ]}$

10. Subtract a multiple of row 3 from row 1 to turn that column's entry into 0: R1 → R1 − 13/6·R3

    $\text{[ 1, 0, 0 | 23/47 ] [ 0, 1, −11/6 | 4/3 ] [ 0, 0, 1 | 40/47 ]}$

11. Subtract a multiple of row 3 from row 2 to turn that column's entry into 0: R2 → R2 + 11/6·R3

    $\text{[ 1, 0, 0 | 23/47 ] [ 0, 1, 0 | 136/47 ] [ 0, 0, 1 | 40/47 ]}$

12. So the one solution is

    $\text{x = 23/47, y = 136/47, z = 40/47}$

10. Solution: x = 1, y = 12/5, z = 4/5

1. Your system of equations

   $\text{3x + 3y + z = 11; x − y + 3z = 1; 2x + 2y − z = 6}$

2. Write the system as a grid (an augmented matrix): the numbers in front of each letter, a bar, then the right-hand sides

   $\text{[ 3, 3, 1 | 11 ] [ 1, −1, 3 | 1 ] [ 2, 2, −1 | 6 ]}$

3. Multiply row 1 so its pivot becomes 1, which makes the next steps easy: R1 → 1/3·R1

   $\text{[ 1, 1, 1/3 | 11/3 ] [ 1, −1, 3 | 1 ] [ 2, 2, −1 | 6 ]}$

4. Subtract a multiple of row 1 from row 2 to turn that column's entry into 0: R2 → R2 − R1

   $\text{[ 1, 1, 1/3 | 11/3 ] [ 0, −2, 8/3 | −8/3 ] [ 2, 2, −1 | 6 ]}$

5. Subtract a multiple of row 1 from row 3 to turn that column's entry into 0: R3 → R3 − 2·R1

   $\text{[ 1, 1, 1/3 | 11/3 ] [ 0, −2, 8/3 | −8/3 ] [ 0, 0, −5/3 | −4/3 ]}$

6. Multiply row 2 so its pivot becomes 1, which makes the next steps easy: R2 → −1/2·R2

   $\text{[ 1, 1, 1/3 | 11/3 ] [ 0, 1, −4/3 | 4/3 ] [ 0, 0, −5/3 | −4/3 ]}$

7. Subtract a multiple of row 2 from row 1 to turn that column's entry into 0: R1 → R1 − R2

   $\text{[ 1, 0, 5/3 | 7/3 ] [ 0, 1, −4/3 | 4/3 ] [ 0, 0, −5/3 | −4/3 ]}$

8. Multiply row 3 so its pivot becomes 1, which makes the next steps easy: R3 → −3/5·R3

   $\text{[ 1, 0, 5/3 | 7/3 ] [ 0, 1, −4/3 | 4/3 ] [ 0, 0, 1 | 4/5 ]}$

9. Subtract a multiple of row 3 from row 1 to turn that column's entry into 0: R1 → R1 − 5/3·R3

   $\text{[ 1, 0, 0 | 1 ] [ 0, 1, −4/3 | 4/3 ] [ 0, 0, 1 | 4/5 ]}$

10. Subtract a multiple of row 3 from row 2 to turn that column's entry into 0: R2 → R2 + 4/3·R3

    $\text{[ 1, 0, 0 | 1 ] [ 0, 1, 0 | 12/5 ] [ 0, 0, 1 | 4/5 ]}$

11. So the one solution is

    $\text{x = 1, y = 12/5, z = 4/5}$