Let $t$ be TNYWR.

In the diagram, the number that goes in each unshaded box above the bottom row is the sum of the numbers in the two unshaded boxes immediately below to the left and to the right. For example, $23=9+14$. What is the value of $x-y$ ?

![](https://cdn.mathpix.com/cropped/2024_04_30_4efeb798298dfec26b46g-3.jpg?height=352&width=1258&top_left_y=1559&top_left_x=455)
Using the rule that the sum of two entries in consecutive boxes in one row is the entry between these boxes in the row above, the entries in the second row from the bottom should be $x+8, x+y, y+9,23$.

Thus, the first entry in the third row from the bottom should be $2 x+y+8$ and the last entry in this row should be $y+32$.

Also, the two entries in the second row from the top should be $5 x+4 y+8$ and $3 x+4 y+32$.

![](https://cdn.mathpix.com/cropped/2024_04_30_75c1cf70fbe3735bfd92g-18.jpg?height=357&width=1258&top_left_y=1258&top_left_x=539)

Now the given entry $3(x+y)$ is the sum of the two entries below it.

Thus, $3 x+3 y=(x+y)+(y+9)$, which gives $2 x+y=9$.

Also, the top entry $t$ is the sum of the two entries below it.

Thus, $t=(5 x+4 y+8)+(3 x+4 y+32)$, which gives $8 x+8 y=t-40$ or $x+y=\frac{1}{8} t-5$.

Since $2 x+y=9$ and $x+y=\frac{1}{8} t-5$, then $x=(2 x+y)-(x+y)=9-\left(\frac{1}{8} t-5\right)=14-\frac{1}{8} t$.

Since $x+y=\frac{1}{8} t-5$ and $x=14-\frac{1}{8} t$, then $y=(x+y)-x=\frac{1}{4} t-19$.

Thus, $x-y=\left(14-\frac{1}{8} t\right)-\left(\frac{1}{4} t-19\right)=33-\frac{3}{8} t$.

Since the answer to (b) is 32 , then $t=32$ and so $x-y=33-\frac{3}{8} t=21$.

ANSWER: $15,32,21$