To solve the given system of equations, we can use methods like substitution or elimination. Here, I will demonstrate the elimination method.
The equations are:
- 1. 2x + 4y + 3z = 7
- 2. 3x + y + 4z = 12
- 3. x + 3y + 4z = 4
Let’s first manipulate the equations to eliminate one variable. We’ll start by eliminating one variable (let’s say z) from the first two equations.
From Equation 1:
2x + 4y + 3z = 7 ==> 3z = 7 – 2x – 4y ==> z = (7 – 2x – 4y)/3
Substituting this expression for z into Equation 2:
3x + y + 4((7 – 2x – 4y)/3) = 12
Multiply through by 3 to eliminate the fraction:
9x + 3y + 4(7 – 2x – 4y) = 36
9x + 3y + 28 – 8x – 16y = 36
Combine like terms:
x – 13y + 28 = 36
Simplifying gives:
x – 13y = 8 (Equation 4)
Now, substitute z back into Equation 3:
x + 3y + 4((7 – 2x – 4y)/3) = 4
Again, multiply through by 3:
3x + 9y + 4(7 – 2x – 4y) = 12
3x + 9y + 28 – 8x – 16y = 12
Combining like terms yields:
-5x – 7y + 28 = 12
Simplifying gives:
-5x – 7y = -16
Or multiplying by -1:
5x + 7y = 16 (Equation 5)
Now we have a simpler system of two equations:
- 4. x – 13y = 8
- 5. 5x + 7y = 16
Next, we can solve this new system. From Equation 4, we can express x in terms of y:
x = 8 + 13y
Substituting this into Equation 5:
5(8 + 13y) + 7y = 16
40 + 65y + 7y = 16
72y = 16 – 40
72y = -24
y = -1/3
Now, we can find x:
x = 8 + 13(-1/3)
x = 8 – 13/3 = 24/3 – 13/3 = 11/3
Lastly, we find z using the expression we derived for z:
z = (7 – 2(11/3) – 4(-1/3))/3
z = (7 – 22/3 + 4/3)/3
z = (21/3 – 22/3 + 4/3)/3 = (3/3)/3 = 1/3
Thus, the solution to the system of equations is:
- x = 11/3
- y = -1/3
- z = 1/3