For real numbers $t,$ the point of intersection of the lines $x + 2y = 7t + 3$ and $x - y = 2t - 2$ is plotted.  All the plotted points lie on a line.  Find the slope of this line.
I want to find the slope of the line that passes through all the points of intersection of the two given lines. To do that, I need to find a general expression for the coordinates of the point of intersection in terms of $t.$ I can use either of the given equations to express $x$ in terms of $t$ and $y.$ For example, from the second equation, I get $x = 2t - 2 + y.$ I can substitute this into the first equation and solve for $y$ in terms of $t.$ This gives me $2t - 2 + y + 2y = 7t + 3,$ or $3y = 5t + 5.$ Dividing by 3, I get $y = \frac{5}{3}t + \frac{5}{3}.$ Then, using the expression for $x$ in terms of $t$ and $y,$ I get $x = 2t - 2 + \frac{5}{3}t + \frac{5}{3}.$ Simplifying, I get $x = \frac{11}{3}t - \frac{1}{3}.$ So, the point of intersection has coordinates $(\frac{11}{3}t - \frac{1}{3}, \frac{5}{3}t + \frac{5}{3}).$ To find the slope of the line that passes through all these points, I need to find the ratio of the change in $y$ to the change in $x$ as $t$ varies. This is the same as the derivative of $y$ with respect to $x.$ I can use the chain rule to find this derivative. I get $\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}.$ To find $\frac{dy}{dt},$ I just need to differentiate the expression for $y$ in terms of $t.$ I get $\frac{dy}{dt} = \frac{5}{3}.$ To find $\frac{dt}{dx},$ I need to differentiate the expression for $x$ in terms of $t$ and then take the reciprocal. I get $\frac{dx}{dt} = \frac{11}{3}$ and $\frac{dt}{dx} = \frac{3}{11}.$ Therefore, the slope of the line is $\frac{dy}{dx} = \frac{5}{3} \cdot \frac{3}{11} = \frac{5}{11}.$ 