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What is the solution to the following system of equations?
x−y+z=−1
x+y+3z=-3
2x−y+2z=0
To solve this system of equations, we can use the method of substitution or elimination. Let's use the method of substitution.
From the first equation, we can isolate x:
x - y + z = -1
x = y - z - 1
Substitute this value of x into the second equation:
x + y + 3z = -3
(y - z - 1) + y + 3z = -3
2y + 2z - 1 = -3
2y + 2z = -2
Now, let's isolate y in the third equation:
2x - y + 2z = 0
2(y - z - 1) - y + 2z = 0
2y - 2z - 2 - y + 2z = 0
y - 2 = 0
y = 2
Substitute the value of y into the equation 2y + 2z = -2:
2(2) + 2z = -2
4 + 2z = -2
2z = -6
z = -3
Now that we have y = 2 and z = -3, we can substitute these values into the first equation to find x:
x - y + z = -1
x - 2 - 3 = -1
x - 5 = -1
x = 4
Therefore, the solution to the system of equations is x = 4, y = 2, and z = -3.
To find the solution to the system of equations, we can use the method of substitution or elimination. Let's solve it using the substitution method.
Step 1: Solve one equation for one variable in terms of the others.
Let's solve the first equation for x:
x = -1 + y - z
Step 2: Substitute the expression from Step 1 into the other equations.
Substituting x = -1 + y - z into the second equation:
(-1 + y - z) + y + 3z = -3
Simplifying the equation:
2y + 2z = -2
Substituting x = -1 + y - z into the third equation:
2(-1 + y - z) - y + 2z = 0
Simplifying the equation:
y - 2z = 2
Step 3: Solve the resulting system of equations.
Now we have the following system of equations:
2y + 2z = -2
y - 2z = 2
We can solve this system of equations by eliminating one variable, let's eliminate y.
Multiply both sides of the second equation by 2:
2(y - 2z) = 2(2)
2y - 4z = 4
Now, we can add the two equations:
(2y + 2z) + (2y - 4z) = -2 + 4
4y - 2z = 2
Simplifying the equation:
2(2y - z) = 2
2y - z = 1
So now we have a new system of equations:
2y + 2z = -2
2y - z = 1
We can solve this system using the substitution method or elimination method again.
Let's solve using the elimination method:
Multiply the second equation by 2:
2(2y - z) = 2(1)
4y - 2z = 2
Now subtract the resulting equation from the first equation:
(2y + 2z) - (4y - 2z) = -2 - 2
-2y + 4z = -4
Simplifying the equation:
-2y + 4z = -4
Now divide both sides of the equation by -2:
-2y/-2 + 4z/-2 = -4/-2
y - 2z = 2
We can see that the equation y - 2z = 2 is the same as the second equation we had in the beginning of these steps, which means they are dependent equations. This implies that the system of equations has infinitely many solutions.
Therefore, there is no unique solution to the given system of equations.
To find the solution to the system of equations, we will use the method of Gaussian elimination. Here are the steps to solve the system of equations:
Step 1: Write the augmented matrix for the system of equations:
[1, -1, 1 | -1]
[1, 1, 3 | -3]
[2, -1, 2 | 0]
Step 2: Perform row operations to convert the matrix into row echelon form.
R2 = R2 - R1
R3 = R3 - 2R1
The new matrix becomes:
[1, -1, 1 | -1]
[0, 2, 2 | -2]
[0, 1, 0 | 2]
Step 3: Perform further row operations to obtain the reduced row echelon form.
R2 = R2/2
R3 = R3 - R2
The new matrix becomes:
[1, -1, 1 | -1]
[0, 1, 1 | -1]
[0, 0, -1 | 3]
Step 4: Solve for the variables using back substitution.
From the third row, we have -z = 3, which implies z = -3.
Substituting the value of z into the second row, we have 1 + 1y + 1(-3) = -1, which simplifies to y - 2 = -1. Solving for y, we get y = 1.
Finally, substituting the values of y and z into the first row, we have 1 - 1 + 1(-3) = -1, which simplifies to x - 2 = -1. Solving for x, we get x = 1.
Step 5: The solution to the system of equations is x = 1, y = 1, and z = -3.
Therefore, the answer is x = 1, y = 1, and z = -3.