Question

Use Cramer's rule to solve the system of equations.
6x=2-y
3x+1=2y

Answer 1

6x+y=2
3x-2y=-1
\[\left[\begin{matrix}6 & 1 \\ 3 & -2\end{matrix}\right]\left(\begin{matrix}x \\ y\end{matrix}\right)=\left(\begin{matrix}2 \\ -1\end{matrix}\right)\]

Answer 2

\[Ax=b \therefore x=A ^{-1}b\]

Answer 3

it's supposed to be an ordered pair involving 2 fractions

Answer 4

yh, let me land first..chillax

Answer 5

lol, we need to know if the matrix has a unique solution or not, we check know this if det A=0

Answer 6

Hi, across!

Answer 7

det A is not equal to zero, it's -15 so we find it's unique solution

Answer 8

\[A ^{-1}\times b=1/15\left[\begin{matrix}2 & 1 \\ 3 & -6\end{matrix}\right]\left(\begin{matrix}2 \\ -1\end{matrix}\right)\]

Answer 9

You are given

\(6x=2-y\),
\(3x+1=2y\).

But we re-write it as

\(6x+y=2\),
\(3x-2y=-1\).

We then have that\[A=\begin{bmatrix}6&1\\3&-2\end{bmatrix},\]\[\vec{x}=\begin{bmatrix}x\\y\end{bmatrix},\text{ and}\]\[b=\begin{bmatrix}2\\-1\end{bmatrix}.\]It follows that \(\det(A)=-15\). Then\[A_1=\begin{bmatrix}2&1\\-1&-2\end{bmatrix},\]\[A_2=\begin{bmatrix}6&2\\3&-1\end{bmatrix},\]\(\det(A_1)=-3\), and \(\det(A_2)=-12\). Finally,\[x=\frac{\det(A_1)}{\det(A)}=\frac{-3}{-15}=\frac{1}{5},\text{ and}\]\[y=\frac{\det(A_2)}{\det(A)}=\frac{-12}{-15}=\frac{12}{15}.\]Those are your solutions.

Answer 10

That's what I was looking for! Thanks!

Answer 11

so we should have
\[\left(\begin{matrix}x \\ y\end{matrix}\right)=\left(\begin{matrix}1/5 \\ 12/15\end{matrix}\right)\]