Solve the following systems:
2x + 2z =2
4x + 2y =3
2× + 2y + 2z =1
x+z = 1 so z = (1-x)
2 x + 1 y = 1.5 so y = (1.5 - 2x)
x + y + z = 0.5
so
x + (1.5-2x) + (1-x) = 0.5
x + 1.5 - 2x +1 - x = 0.5
-2 x + 2.5 = 0.5
2 x = 2
x = 1
etc
This was actually just like the last one. It is your turn to do them now. I already passed the course.
To solve the given system of equations, we can use the method of elimination or substitution. Let's use the elimination method to solve it.
Step 1: Multiply the first equation by 2 to make the coefficients of x in the first and third equations the same.
Equation 1: 2x + 2z = 2 → 4x + 4z = 4
Step 2: Multiply the second equation by -2 to make the coefficients of x in the second and third equations the same.
Equation 2: 4x + 2y = 3 → -8x - 4y = -6
Step 3: Add the modified Equation 1 and Equation 2 to eliminate the variable x.
(4x + 4z) + (-8x - 4y) = 4 + (-6)
-4x + 4z - 4y = -2 → -x + z - y = -1 (Equation 3)
Step 4: Subtract Equation 1 from Equation 3 to eliminate x.
(-x + z - y) - (2x + 2z) = -1 - 2
-3x - y = -3 → 3x + y = 3 (Equation 4)
Now, we have two equations with two variables:
Equation 4: 3x + y = 3
Equation 3: -x + z - y = -1
Step 5: Solve for y in Equation 4.
3x + y = 3 → y = 3 - 3x
Step 6: Substitute the value of y in Equations 3 and 1.
Equation 3: -x + z - (3 - 3x) = -1 → -x + z - 3 + 3x = -1
2x + z = 2 (Equation 5)
Equation 1: 2x + 2z = 2 (Equation 6)
Step 7: Multiply Equation 5 by -1/2 to get the variable z alone.
(-1/2)(2x + z) = (-1/2)(2) → -x - z/2 = -1
Step 8: Subtract Equation 6 from the modified Equation 5 to eliminate the variable x.
(-x - z/2) - (2x + 2z) = -1 - 2 → -x - z/2 - 2x - 2z = -3
Simplifying the equation:
-3x - (z + 2z) = -3 → -3x - 3z = -3 → x + z = 1 (Equation 7)
Now, we have two equations with two variables:
Equation 7: x + z = 1
Equation 5: -x - z/2 = -1
Step 9: Multiply Equation 7 by 2 to get rid of the fraction in Equation 5.
2(x + z) = 2(1) → 2x + 2z = 2 (Equation 8)
Step 10: Add the modified Equation 7 and Equation 5 to eliminate the variable x.
(-x - z/2) + (2x + 2z) = -1 + 2 → x + (2x) + z + (2z) = 1 + 2
Simplifying the equation:
3x + 3z = 3 → x + z = 1 (Equation 9)
Equation 9 is the same as Equation 7, which means they are equivalent equations. So, we can conclude that the system has infinitely many solutions.
To summarize:
The given system of equations has infinitely many solutions. The solution can be represented as (x, y, z) = (t, 3 - 3t, 1 - t), where t is any real number.