Y = mx+b
-1 = 13/4 (-1) +b
-4/4 + 13/4 = b
9/4 = b
525/1000 = 105/200 = 21/40
<span>There are 4 long rows and 5 short rows in the theater. x represents the number of chairs in each long row and y represents the number of chairs in each short row.
So, total number of chairs in 4 long rows= 4x
Total number of chairs in 5 short rows = 5y
Total number of chairs in the theater on a normal day = 4x + 5y
When 2 chairs are added to each long row, the number of chairs will change to (x+2).
So, total number of chairs in 4 long rows will be = 4(x+2)
When 3 chairs are added to each short row, the number of chairs will change to (y+3)
So, total number of chairs in 5 short rows will be = 5(y+3)
Thus, total number of chairs in the theater in rush day = 4(x+2) + 5(y+3)
= 4x + 8 + 5y + 15
= 4x + 5y + 23
Thus we can say the number of chairs increase by 23 as compared to a normal day.
</span>
ANSWER: 32 five-dollar bills
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EXPLANATION:
Let x be number of $5 bills
Let y be number of $10 bills
Since we have total of 38 bills, we must have the sum of x and y be 38
x + y = 38 (I)
Since the total amount deposited is $220, we must have the sum of 5x and 10y be 220 (x and y are just the "number of" their respective bills, so we multiply them by their value to get the total value):
5x + 10y = 220 (II)
System of equations:

Divide both sides of equation (II) by 5 so our numbers become smaller

Rearrange (I) to solve for y so that we can substitute into (II)

Substituting this into equation (II) for the y:

We have 32 five-dollar bills
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If we want to finish off the question, use y = 38 - x to figure out number of $10 bills

32 five-dollar bills and 6 ten-dollar bills
Combine the two equations in the right amounts to eliminate y :
-2 (2x + 3y) + (3x + 6y) = -2 (17) + 30
-4x - 6y + 3x + 6y = -34 + 30
-x = -4
x = 4
Solve for y :
2x + 3y = 17
8 + 3y = 17
3y = 9
y = 3