Answer:
This is a linear function.
You can do substitution 2x-2=x^2 -x -6; isolate all terms on one side 0= x^2 -x-2x -6+2; combine like terms x^2 -3x -4=0; factor the quadratic (x-4)(x+1)=0; each term is zero x-4=0 so x=4 and x+1=0 so x=-1. Now, y=2•4-2=6 and y=2•(-1) -2= -4 ; solutions for the system are ( 4,6) and ( -1, -4)
If you have multiple equations with multiple variables, you can either do clever substitutions, or turn it into a matrix on which you can perform linear combinations or multiplications (Gauss elimination)
1 1 1 1
2 1 -1 8
1 -1 1 -5
(note how the above 3 rows represent the 3 equations, just got rid of the variables, plus sign and equals sign)
subtract row1 from row3, that eliminates x and z from row 3.
1 1 1 1
2 1 -1 8
0 -2 0 -6
divide row3 by -2, that will give y a factor of 1
1 1 1 1
2 1 -1 8
0 1 0 3
The last row now says y=3
Answer: R=V
Step-by-step explanation: :p
