Question

{3x+y-4z=-30
{3x+2y+2z=-8
{5x+5y+z=-26

Answer 1

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.