Answer:
![\left[\begin{array}{ccc}3&-5  &|12\\4&-2  &|15\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%26-5%20%20%26%7C12%5C%5C4%26-2%20%20%26%7C15%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
When making a matrix of two equations with the variables x and y, the result will be a matrix with three columns:
- a column for the values of x in each equation
- a column for the values of y in each equation
- a column for the independent values of each equation
since our system of equations is:

we can see that the value for x in the first equation is 3 and in the second equation is 4, thus the first column will have the numbers 3 and 4:
![\left[\begin{array}{ccc}3&&\\4&&\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%26%26%5C%5C4%26%26%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Now for the values of y we hvae -5 in the first equation and -2 in the second equation, we update the matrix with another column with the values of -5 and -2:
![\left[\begin{array}{ccc}3&-5&\\4&-2&\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%26-5%26%5C%5C4%26-2%26%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Finally, the last column is the independent values of each equation (or the results) in the first equation that number is 12 and in the second equation is 15, thus the matrix is:
![\left[\begin{array}{ccc}3&-5&12\\4&-2&15\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%26-5%2612%5C%5C4%26-2%2615%5C%5C%5Cend%7Barray%7D%5Cright%5D)
usually there is a line separating the columns for the values of x and y, and the independent values: 
![\left[\begin{array}{ccc}3&-5  &|12\\4&-2  &|15\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%26-5%20%20%26%7C12%5C%5C4%26-2%20%20%26%7C15%5C%5C%5Cend%7Barray%7D%5Cright%5D)
this is the matrix of the system of equations
 
        
             
        
        
        
Elimination:
7x - 3y = 20
5x + 3y = 16
(add)
12x = 36
÷ 12
x = 3
(5 × 3) + 3y = 16
15 + 3y = 16
- 15
3y = 1
÷ 3
y = 1/3
Substitution:
5x + 3y = 16
- 3y
5x = 16 - 3y
÷ 5
x = 3.2 - 0.6y
5(3.2 - 0.6y) + 3y = 16
16 - 3y + 3y = 16
16  = 16
- 16
6y = 0
÷ 6
y = 0
Sorry the substitution messed up for some reason, I'll fix it after I've answered the other question
        
             
        
        
        
Every linear graph is just a straight line, so if there are any curves or unnatural shapes, than you know it is not linear. As for equations, if it can be shaped into y=mx+b where m and b are numbers, then it will be a linear equation.
        
             
        
        
        
Answer:
<em>Both 16 and 1 are squares. This suggests using the formula for the difference of squares. Answer link.</em>
Step-by-step explanation:
Every time you have a difference in the format of
A2−B2 you can factor as (A−B)(A+B).
Then you have to identify if your quantity is the difference of two squared quantities.
It is. In fact we can write it as:
16x2−1=(4x)2−12
Then we can apply our rule
