Solve the following system of equations
3y + 2z = 5
2x − y − 3z = 2
2x + 2y − z = 7
2x − y − 3z = 2
2x + 2y − z = 7
-------------------subtract
-3 y - 2 z = -5
+3 y + 2 z = +5
---------------------add
0 = 0
no solution, lines parallel never cross
To solve the given system of equations, we will use the method of elimination or substitution. Let's use the method of elimination to solve this system.
Step 1: Eliminate the variable x by multiplying the second equation by 2 and the third equation by 2.
2x + 2y - z = 7 (equation 3)
2(2x - y - 3z = 2)
=> 4x - 2y - 6z = 4 (equation 2')
2(2x + 2y - z = 7)
=> 4x + 4y - 2z = 14 (equation 3')
Step 2: Subtract equation 2' from equation 3'.
(4x + 4y - 2z) - (4x - 2y - 6z) = 14 - 4
4x + 4y - 2z - 4x + 2y + 6z = 10
6y + 4z = 10
3y + 2z = 5 (equation 1)
Step 3: Now we have a system with two equations:
3y + 2z = 5 (equation 1)
6y + 4z = 10 (from Step 2)
Step 4: Multiply equation 1 by -2 and add it to equation 2.
(-2)(3y + 2z) + (6y + 4z) = (-2)(5) + 10
-6y - 4z + 6y + 4z = -10 + 10
0 = 0
Step 5: The result of 0 = 0 means that the two equations are equivalent and dependent, which indicates that there are infinitely many solutions.
So, the solution for the given system of equations is infinite. Each solution will have the form (x, y, z), where x can be any real number, and y and z can be expressed in terms of x.