### Instruction:
In the diagram, points $X$, $Y$ and $Z$ are on the sides of $\triangle UVW$, as shown.  Line segments $UY$, $VZ$ and $WX$ intersect at $P$.  Point $Y$ is on $VW$ such that $VY:YW=4:3$. If $\triangle PYW$ has an area of 30 and $\triangle PZW$ has an area of 35, determine the area of $\triangle UXP$. [asy]
size(6cm);
pair v = (0, 0); pair w = (10, 0); pair u = (3.5, 7);
pair y = 4 * w / 7;
pair x = 56 * u / 140;
pair p = IP(w--x, u--y);
pair z = IP(v--(10 * p), u--w);
draw(u--v--w--cycle);
draw(u--y);draw(x--w);draw(z--v);

label("$U$", u, N);
label("$X$", x, NW);
label("$P$", p, NE + 0.2 * W);
label("$Z$", z, NE);
label("$V$", v, SW);
label("$Y$", y, S);
label("$W$", w, SE);/*
label("$a$", centroid(p, v, y), fontsize(10));
label("$b$", centroid(p, z, u), fontsize(10));
label("$c$", centroid(p, u, x), fontsize(10));
label("$d$", centroid(p, x, v), fontsize(10));
label("$30$", centroid(p, y, w) + 0.2 * W, fontsize(10));
label("$35$", centroid(p, z, w), fontsize(10));*/
[/asy]

### Response:

Recall that if two triangles have their bases along the same straight line and they share a common vertex that is not on this line, then the ratio of their areas is equal to the ratio of the lengths of their bases. We will use this fact extensively throughout the proof.

Let the area of $\triangle PYV$, $\triangle PZU$, $\triangle UXP$, and $\triangle XVP$, be $a$, $b$, $c$, and $d$, respectively. [asy]
size(6cm);
pair v = (0, 0); pair w = (10, 0); pair u = (3.5, 7);
pair y = 4 * w / 7;
pair x = 56 * u / 140;
pair p = IP(w--x, u--y);
pair z = IP(v--(10 * p), u--w);
draw(u--v--w--cycle);
draw(u--y);draw(x--w);draw(z--v);

label("$U$", u, N);
label("$X$", x, NW);
label("$P$", p, NE + 0.2 * W);
label("$Z$", z, NE);
label("$V$", v, SW);
label("$Y$", y, S);
label("$W$", w, SE);
label("$a$", centroid(p, v, y), fontsize(10));
label("$b$", centroid(p, z, u), fontsize(10));
label("$c$", centroid(p, u, x), fontsize(10));
label("$d$", centroid(p, x, v), fontsize(10));
label("$30$", centroid(p, y, w) + 0.2 * W, fontsize(10));
label("$35$", centroid(p, z, w), fontsize(10));
[/asy] Since $$\frac{|\triangle PYV|}{|\triangle PYW|}=\frac{VY}{YW}=\frac{4}{3},$$then $$a = |\triangle PYV|=\frac{4}{3}\times |\triangle PYW|=\frac{4}{3}(30)=40.$$Also, $\frac{|\triangle VZW|}{|\triangle VZU|}=\frac{ZW}{ZU}=\frac{|\triangle PZW|}{|\triangle PZU|}$ or $|\triangle VZW|\times |\triangle PZU| = |\triangle PZW| \times |\triangle VZU|$. Thus, $$\frac{|\triangle VZU|}{|\triangle PZU|}=\frac{|\triangle VZW|}{|\triangle PZW|}=\frac{35+30+40}{35}=\frac{105}{35}=\frac{3}{1}.$$Therefore,  $\frac{|\triangle VZU|}{|\triangle PZU|}=\frac{3}{1}$, or $\frac{b+c+d}{b}=\frac{3}{1}$ or $b+c+d=3b$ and $c+d=2b$.

Next, $$\frac{|\triangle UVY|}{|\triangle UYW|}=\frac{VY}{YW}=\frac{4}{3},$$so $$\frac{40+c+d}{30+35+b}=\frac{4}{3}.$$Since $c+d=2b$, we have $3(40+2b)=4(65+b)$, so $120+6b=260+4b$, then $2b=140$ and $b=70$.

Next, $$\frac{|\triangle UXW|}{|\triangle XVW|}=\frac{UX}{XV}=\frac{|\triangle UXP|}{|\triangle XVP|},$$or $$\frac{35+b+c}{30+a+d}=\frac{c}{d}.$$Since $b=70$ and $a=40$, $\frac{105+c}{70+d}=\frac{c}{d}$, or $d(105+c)=c(70+d)$. Thus, $105d+cd=70c+cd$ or $105d=70c$, and $\frac{d}{c}=\frac{70}{105}=\frac{2}{3}$ or $d=\frac{2}{3}c$.

Since $c+d=2b=2(70)=140$, we have $$c+d=c+\frac{2}{3}c=\frac{5}{3}c=140,$$or $c=\frac{3}{5}(140)=84$. Therefore, the area of $\triangle UXP$ is $84$.