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What is the solution to the following system of equations? (If using a matrix calculator, use the image icon below to upload your work)
x−3y+2z=11
-x+4y+3z=5
2x−2y−4z=2
To find the solution to the given system of equations, we can use the method of Gaussian elimination.
Step 1: Write the augmented matrix for the system of equations:
| 1 -3 2 | 11 |
| -1 4 3 | 5 |
| 2 -2 -4 | 2 |
Step 2: Apply row operations to the matrix to transform it into row-echelon form. Let's start with the first column:
R2 = R2 + R1
R3 = R3 - 2R1
This yields the following matrix:
| 1 -3 2 | 11 |
| 0 1 5 | 16 |
| 0 4 -8 | -20 |
Step 3: Proceed to the second column by using row operations:
R3 = R3 - 4R2
This gives us the following matrix:
| 1 -3 2 | 11 |
| 0 1 5 | 16 |
| 0 0 -28 | -84 |
Step 4: To simplify calculation, let's divide the third row by -28:
R3 = (-1/28) * R3
The matrix becomes:
| 1 -3 2 | 11 |
| 0 1 5 | 16 |
| 0 0 1 | 3 |
Step 5: Now, we can backward substitute to find the values of x, y, and z. Starting from the bottom row:
z = 3
Substitute the value of z back into the second equation:
y + 5z = 16
y + 5(3) = 16
y + 15 = 16
y = 1
Finally, substitute the values of y and z into the first equation:
x - 3y + 2z = 11
x - 3(1) + 2(3) = 11
x - 3 + 6 = 11
x + 3 = 11
x = 8
Therefore, the solution to the given system of equations is:
x = 8,
y = 1,
z = 3
To solve the system of equations, we can use the method of elimination.
First, let's eliminate the x variable by adding the first equation to the second equation:
(x - 3y + 2z) + (-x + 4y + 3z) = 11 + 5
Simplifying this equation, we get:
y + 5z = 16 ...........(1)
Next, let's eliminate the x variable by adding twice the first equation to the third equation:
2(x - 3y + 2z) + (2x - 2y - 4z) = 2 + 2
Simplifying this equation, we get:
-4y + 2z = 6 .............(2)
Now we have two equations in terms of y and z:
y + 5z = 16 ...........(1)
-4y + 2z = 6 .............(2)
To eliminate y, we can multiply the first equation by 4 and the second equation by 5:
4(y + 5z) = 4(16)
-5(-4y + 2z) = 5(6)
This gives us:
4y + 20z = 64 ............(3)
20y - 10z = 30 ...........(4)
Now we can add equations (3) and (4) to eliminate y:
(4y + 20z) + (20y - 10z) = 64 + 30
Simplifying this equation, we get:
24y + 10z = 94 ...........(5)
Next, let's multiply equation (1) by 24:
24(y + 5z) = 24(16)
This gives us:
24y + 120z = 384 ............(6)
Now we can subtract equation (6) from equation (5) to eliminate y:
(24y + 10z) - (24y + 120z) = 94 - 384
Simplifying this equation, we get:
-110z = -290
Dividing both sides by -110, we find:
z = 2.63636363636... ≈ 2.64
Substitute this value of z back into equation (1):
y + 5(2.64) = 16
Simplifying this equation, we get:
y + 13.2 = 16
Subtracting 13.2 from both sides, we have:
y = 2.8
Finally, substitute the values of y and z into one of the original equations (let's use the first equation):
x - 3(2.8) + 2(2.64) = 11
Simplifying this equation, we get:
x - 8.4 + 5.28 = 11
Combining like terms, we have:
x - 3.12 = 11
Adding 3.12 to both sides, we find:
x = 14.12
Therefore, the solution to the system of equations is:
x = 14.12
y = 2.8
z = 2.64
To solve the system of equations, we can use the method of elimination or substitution. Let's use the method of elimination to solve the system of equations step-by-step.
Step 1: Write the system of equations:
x - 3y + 2z = 11 ...(1)
-x + 4y + 3z = 5 ...(2)
2x - 2y - 4z = 2 ...(3)
Step 2: Multiply equation (1) by 2:
2(x - 3y + 2z) = 2 * 11
2x - 6y + 4z = 22 ...(4)
Step 3: Add equation (2) to equation (4):
(-x + 4y + 3z) + (2x - 6y + 4z) = 5 + 22
x - 2y + 7z = 27 ...(5)
Step 4: Multiply equation (2) by 2:
2(-x + 4y + 3z) = 2 * 5
-2x + 8y + 6z = 10 ...(6)
Step 5: Multiply equation (3) by 2:
2(2x - 2y - 4z) = 2 * 2
4x - 4y - 8z = 4 ...(7)
Step 6: Add equation (5) to equation (6):
(x - 2y + 7z) + (-2x + 8y + 6z) = 27 + 10
- x + 6y + 13z = 37 ...(8)
Step 7: Add equation (7) to equation (8):
(4x - 4y - 8z) + (-x + 6y + 13z) = 4 + 37
3x + 2y + 5z = 41 ...(9)
Step 8: Multiply equation (8) by 3:
3(-x + 6y + 13z) = 3 * 37
-3x + 18y + 39z = 111 ...(10)
Step 9: Add equation (9) to equation (10):
(3x + 2y + 5z) + (-3x + 18y + 39z) = 41 + 111
20y + 44z = 152 ...(11)
Step 10: Multiply equation (11) by 5:
5(20y + 44z) = 5 * 152
100y + 220z = 760 ...(12)
Step 11: Multiply equation (11) by 2:
2(20y + 44z) = 2 * 152
40y + 88z = 304 ...(13)
Step 12: Subtract equation (13) from equation (12):
(100y + 220z) - (40y + 88z) = 760 - 304
60y + 132z = 456 ...(14)
Step 13: Divide equation (14) by 4:
(60y + 132z) / 4 = 456 / 4
15y + 33z = 114 ...(15)
Step 14: Multiply equation (15) by 5:
5(15y + 33z) = 5 * 114
75y + 165z = 570 ...(16)
Step 15: Subtract equation (15) from equation (16):
(75y + 165z) - (15y + 33z) = 570 - 114
60y + 132z = 456 ...(17)
Step 18: Divide equation (17) by 4:
(60y + 132z) / 4 = 456 / 4
15y + 33z = 114 ...(18)
Step 19: Subtract equation (15) from equation (18):
(15y + 33z) - (15y + 33z) = 114 - 114
0 = 0
Step 20: The system of equations has infinitely many solutions since equation (0 = 0) holds true. This means that any value of y and z can be chosen, and x can be expressed in terms of y and z using any of the original equations.
Therefore, the solution to the system of equations is x = 0, y = any value, and z = any value.