You can solve this problem through factoring.
First, you have the equation,
Then, you can factor the numerator.
You can cancel out the x-6 in both the numerator and the denominator because they would equal to just 1.
You are left with
The function is removable noncontinuous at x=6 because if you plug in 6 in x-6, your denominator would be undefined.
Answer:
Given System of equation:
x-y =6 .....,[1]
2x-3z = 16 ......[2]
2y+z = 4 .......[3]
Rewrite the equation [1] as
y = x - 6 .......[4]
Substitute the value of [4] in [3], we get
Using distributive property on LHS ( i.e, )
then, we have
2x - 12 +z =4
Add 12 to both sides of an equation:
2x-12+z+12=4+12
Simplify:
2x +z = 16 .......[5]
On substituting equation [2] in [5] we get;
2x+z=2x -3z
or
z = -3z
Add 3z both sides of an equation:
z+3z = -3z+3z
4z = 0
Simplify:
z = 0
Substitute the value of z = 0 in [2] to solve for x;
or
2x = 16
Divide by 2 both sides of an equation:
Simplify:
x= 8
Substitute the value of x =8 in equation [4] to solve for y;
y = 8-6 = 2
or
y = 2
Therefore, the solution for the given system of equation is; x = 8 , y = 2 and z =0
