A line is parameterized by
\[\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \end{pmatrix} + t \begin{pmatrix} -1 \\ 5 \end{pmatrix}.\]A second line is parameterized by
\[\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 7 \end{pmatrix} + u \begin{pmatrix} -1 \\ 4 \end{pmatrix}.\]Find the point where the lines intersect.
For the first line,
\[\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \end{pmatrix} + t \begin{pmatrix} -1 \\ 5 \end{pmatrix} = \begin{pmatrix} 2 - t \\ 3 + 5t \end{pmatrix}.\]For the second line,
\[\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 7 \end{pmatrix} + u \begin{pmatrix} -1 \\ 4 \end{pmatrix} = \begin{pmatrix} -u 
\\ 7 + 4u \end{pmatrix}.\]Hence, $2 - t = -u$ and $3 + 5t = 7 + 4u.$  Solving, we find $t = -4$ and $u = -6,$ so
\[\begin{pmatrix} x \\ y \end{pmatrix} = \boxed{\begin{pmatrix} 6 \\ -17 \end{pmatrix}}.\]
\begin{pmatrix}6\-17\end{pmatrix}
Level 2
Precalculus
Problem: A line is parameterized by
\[\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \end{pmatrix} + t \begin{pmatrix} -1 \\ 5 \end{pmatrix}.\]A second line is parameterized by
\[\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 7 \end{pmatrix} + u \begin{pmatrix} -1 \\ 4 \end{pmatrix}.\]Find the point where the lines intersect.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the intersection of the two parameterized lines, we need to set their equations equal to each other. The first line is given by the equation:

x = 2 - t
y = 3 + 5t

The second line is given by:

x = -u
y = 7 + 4u

Setting the x-coordinates equal gives us the equation:

2 - t = -u  (1)

Setting the y-coordinates equal gives us:

3 + 5t = 7 + 4u  (2)

From equation (1), we can express u in terms of t:

u = - (2 - t) = t - 2  (3)

Substituting equation (3) into equation (2) allows us to solve for t:

3 + 5t = 7 + 4(t - 2)

Expanding the right side:

3 + 5t = 7 + 4t - 8
3 + 5t = 4t - 1

Rearranging gives:

5t - 4t = -1 - 3
t = -4

Now substituting t = -4 back into equation (3) to find u:

u = - (2 - (-4)) = - (2 + 4) = -6

Now we can find the intersection point by substituting t = -4 into the first line's parameterization:

x = 2 - (-4) = 2 + 4 = 6
y = 3 + 5(-4) = 3 - 20 = -17

Thus, the point of intersection of the two lines is:

\begin{pmatrix} 6 \\ -17 \end{pmatrix}<|END_OF_LATENT|> The final answer is $\begin{pmatrix}6\-17\end{pmatrix}$.