Two lines defined by the equations $y = mx + 4$ and $y = 3x + b$, where $m$ and $b$ are constants, intersect at the point $(6, 10)$. What is the value of $b + m$?
Since the two lines intersect at $(6,10)$, each must pass through that point.  So, we can substitute this point for $x$ and $y$ in the two given equations, then solve for $m$ and $b$.  In the first equation we find: \begin{align*}
10&=m(6)+4\\
\Rightarrow\qquad 6&=6m\\
\Rightarrow\qquad 1&=m
\end{align*} Using the same method in the second equation gives: \begin{align*}
10&=3(6)+b\\
\Rightarrow\qquad 10&=18+b\\
\Rightarrow\qquad -8&=b
\end{align*} The value of $b+m$ is equal to $1+(-8)=\boxed{-7}$.
-7
Level 3
Algebra
Problem: Two lines defined by the equations $y = mx + 4$ and $y = 3x + b$, where $m$ and $b$ are constants, intersect at the point $(6, 10)$. What is the value of $b + m$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the values of $m$ and $b$, we start by substituting the coordinates of the intersection point $(6, 10)$ into both line equations. 

For the first line $y = mx + 4$, substituting $x = 6$ and $y = 10$ gives:
10 = m(6) + 4.
This simplifies to:
10 = 6m + 4.
Subtracting 4 from both sides results in:
6 = 6m.
Dividing both sides by 6 yields:
m = 1.

Next, we substitute the same point into the second line equation $y = 3x + b$. Substituting $x = 6$ and $y = 10$ gives:
10 = 3(6) + b.
This simplifies to:
10 = 18 + b.
Subtracting 18 from both sides results in:
b = 10 - 18,
which simplifies to:
b = -8.

Now, we need to find the value of $b + m$. We have:
b + m = -8 + 1.
Calculating this gives:
b + m = -7.

Thus, the final answer is -7.<|END_OF_LATENT|> The final answer is $-7$.