To solve this system of equations, we can use the method of elimination or substitution. Here, we will use elimination.
First, let's cancel out the x-coefficients in the first and second equations.
Multiply the first equation by -1:
-1(3x+y-4z) = -1(-30)
-3x - y + 4z = 30
Now, add this new equation to the second equation:
-3x - y + 4z + (3x + 2y + 2z) = 30 + (-8)
Simplifying, we get:
y + 6z = 22 ------> Equation 3
Next, let's cancel out the x-coefficients in the first and third equations.
Multiply the first equation by -5:
-5(3x+y-4z) = -5(-30)
-15x - 5y + 20z = 150
Now, add this new equation to the third equation:
-15x - 5y + 20z + (5x + 5y + z) = 150 + (-26)
Simplifying, we get:
-10x + 21z = 124 ------> Equation 4
Now, we have a system of two equations with two unknowns:
Equation 3: y + 6z = 22
Equation 4: -10x + 21z = 124
We can solve this system using substitution or elimination. Let's use substitution method here.
Let's solve Equation 3 for y:
y = 22 - 6z
Now substitute this value of y in Equation 4:
-10x + 21z = 124
Substitute y = 22 - 6z:
-10x + 21z = 124
-10x + 21z = 124
Now, we can solve this equation for x. However, since the equation is the same, we cannot determine a unique value for x. This means that the system is either inconsistent (no solution) or dependent (infinitely many solutions). We need one more equation to find a unique solution.
{3x+y-4z=-30
{3x+2y+2z=-8
{5x+5y+z=-26
1 answer