Write as a Vector Equality

Problem

{[(x+2)/6−(y+6)/3+z/2=0],[(x+1)/2+(y−1)/2−z/4=6],[(x−5)/4+(y+1)/3+(z−2)/2=83/12])

Solution

1. Clear the denominators by multiplying each equation by its least common multiple (6, 4, and 12 respectively).

6*((x+2)/6−(y+6)/3+z/2)=6*(0)⇒(x+2)−2*(y+6)+3*z=0

4*((x+1)/2+(y−1)/2−z/4)=4*(6)⇒2*(x+1)+2*(y−1)−z=24

12*((x−5)/4+(y+1)/3+(z−2)/2)=12*(83/12)⇒3*(x−5)+4*(y+1)+6*(z−2)=83

2. Simplify and rearrange each equation into the standard form a*x+b*y+c*z=d

x+2−2*y−12+3*z=0⇒x−2*y+3*z=10

2*x+2+2*y−2−z=24⇒2*x+2*y−z=24

3*x−15+4*y+4+6*z−12=83⇒3*x+4*y+6*z=106

3. Extract the coefficients to form the matrix A and the constants to form the vector b

A=[[1,−2,3],[2,2,−1],[3,4,6]]

b=[[10],[24],[106]]

4. Construct the vector equality by expressing the system as A*x=b where x is the vector of variables.

[[1,−2,3],[2,2,−1],[3,4,6]]*[[x],[y],[z]]=[[10],[24],[106]]

Final Answer

[[1,−2,3],[2,2,−1],[3,4,6]]*[[x],[y],[z]]=[[10],[24],[106]]

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