Question

If you were to use the substitution method to solve the following system, choose the new equation after the expression equivalent to x from the first equation is substituted into the second equation.
x – 2y = 2
x + 3y = 17

Answer 1

So, solve the first equation for x:

x = 2 + 2y

Now, substitute that equation into the second equation for x...

x + 3y = 17
(2+2y) + 3y = 17
2 + 5y = 17
5y = 17 - 2 = 15
y = 3

Substitute that back into any of our equations to solve for x. I'll use x = 2 + 2y

x = 2 + 2(3) = 8.

So, x = 8, y = 3 is our solution. Make sense?

Answer 2

You can also solve this problem using linear algebra
\[\begin{bmatrix}
1 & -2\\
1 & 3
\end{bmatrix}\begin{bmatrix}
x\\
y
\end{bmatrix}=\begin{bmatrix}
2\\
17
\end{bmatrix}\]Matrix determinant
\[det=\frac{1}{5}\]Matrix minor
\[\begin{bmatrix}
3 & 1\\
-2 & 1
\end{bmatrix}\]Matrix cofactor
\[\begin{bmatrix}
3 & -1\\
2 & 1
\end{bmatrix}\]Matrix transpose
\[\begin{bmatrix}
3 & 2\\
-1 & 1
\end{bmatrix}\]Putting it all together
\[\begin{bmatrix}
x\\
y
\end{bmatrix}=\frac{1}{5}\begin{bmatrix}
3 & 2\\
-1 & 1
\end{bmatrix}\begin{bmatrix}
2\\
17
\end{bmatrix}=\frac{1}{5}\begin{bmatrix}
40\\
15
\end{bmatrix}=\begin{bmatrix}
8\\
3
\end{bmatrix}\]Therefore,
\[x=8, y=3\]Sorry, I'm bored waiting for class. lol