Set the two expressions on line $m$ equal to each other and solve the resulting linear equation.

At point $P$, think of $y$ as the sum of the acute angles that lines $m$ and $n$ make with a horizontal line parallel to $r$ and $s$.

In Desmos, enter y = x + 10 and y = 3x - 20. Click the intersection point and note the $x$-coordinate (call it $t$).

Create a value for that coordinate, for example t = 15 (use the intersection’s $x$-value). Then enter a = t + 10 and b = 2t + 5 to compute the two angle measures.

Enter a + b. This sum is the measure of $y$.

Along transversal $m$, the angle labeled $(x+10)^\circ$ at line $r$ and the angle labeled $(3x-20)^\circ$ at line $s$ are alternate interior angles, so they are equal:

Substitute $x=15$:

• $(x+10)^\circ = (15+10)^\circ = 25^\circ$
• $(2x+5)^\circ = (2\cdot 15+5)^\circ = 35^\circ$

At point $P$, angle $y$ is formed by ray $PB$ (on line $n$) and ray $PC$ (on line $m$) on the right side of the intersection.

Ray $PB$ makes a $35^\circ$ angle with the horizontal parallel line, and ray $PC$ makes a $25^\circ$ angle with the horizontal parallel line, on opposite sides of that horizontal. Therefore,

So, the value of $y$ is $60^\circ$.