Answer:To solve the system of equations using linear combination and substitution, we'll follow these steps:
Step 1: Choose two equations and eliminate one variable.
Let's choose the first and second equations. We can eliminate the variable z by multiplying the first equation by 3 and the second equation by -1, and then adding them together.
3(x+y-3z) = 3(8) -> 3x+3y-9z=24
-1(2x-3y+z) = -1(-6) -> -2x+3y-z=6
Adding these two equations gives us:
(3x+3y-9z) + (-2x+3y-z) = 24+6
Simplifying, we get:
x+6y-10z = 30 (Equation 3)
Step 2: Choose two different equations and eliminate the same variable.
Now, let's choose the first and third equations. We can eliminate the variable x by multiplying the first equation by -3 and the third equation by 1, and then adding them together.
-3(x+y-3z) = -3(8) -> -3x-3y+9z=-24
1(3x+4y-2z) = 1(20) -> 3x+4y-2z=20
Adding these two equations gives us:
(-3x-3y+9z) + (3x+4y-2z) = -24+20
Simplifying, we get:
y+7z = -4 (Equation 4)
Step 3: Solve the system of equations.
We now have two equations (Equation 3 and Equation 4) with two variables (x and y). We can solve this system of equations using substitution.
From Equation 4, we can express y in terms of z:
y = -7z - 4
Substitute this value of y into Equation 3:
x+6(-7z-4)-10z=30
Simplifying, we get:
x-52z-24-10z=30
x-62z=54
x = 62z + 54
So, the solution to the system of equations is:
x = 62z + 54
y = -7z - 4
z = z (z is a free variable)
This means that for any value of z, we can find the corresponding values of x and y using the given equations.