To solve the given system of equations, we’ll first rewrite them in a clearer format:
- Equation 1: 2y + 6z = 3y + 2z + 7
- Equation 2: 4 + 3x + 2y + 5z = 0
Step 1: Simplify Equation 1
Start with the first equation:
2y + 6z = 3y + 2z + 7
Rearranging gives us:
2y – 3y + 6z – 2z = 7
This simplifies to:
-y + 4z = 7
So we can represent Equation 1 as:
(1) -y + 4z = 7
Let’s rearrange this to express y:
(2) y = 4z – 7
Step 2: Substitute in Equation 2
Now substitute Equation (2) into Equation (2):
4 + 3x + 2(4z – 7) + 5z = 0
Expanding this gives:
4 + 3x + 8z – 14 + 5z = 0
Combine like terms:
3x + 13z – 10 = 0
This simplifies to:
(3) 3x + 13z = 10
Step 3: Solve for x
Rearranging Equation (3) gives:
3x = 10 – 13z
Divide by 3:
x = (10 – 13z) / 3
Step 4: Summary of Results
At this point, we have:
- y = 4z – 7
- x = (10 – 13z) / 3
To find specific values of x, y, and z, we can choose a value for z. For example, if we let:
- z = 1:
y = 4(1) – 7 = -3
x = (10 – 13(1)) / 3 = -1 - z = 2:
y = 4(2) – 7 = 1
x = (10 – 13(2)) / 3 = -8/3
Conclusion: The solutions create a family of values based on the choice of z, demonstrating that this system has infinitely many solutions expressed in terms of z.