You are given 144 in one five and ten dollar bills There are

You are given $144 in one, five, and ten dollar bills. There are 35 bills. How many bills of each type are there? 3. Suppose that in question # 2 there are two more ten-dollar bills than five-dollar bills. How many bills of each type are there?

Solution

Let x denote the number of one dollar bills, y denotes the number of five dollar bills and z denotes the number of 10 dollar bills

We are given x+y+z= 35 and also x+5y+10z = 144

Subtracting these two will give: -4y-9z = - 109 => 4y+9z=109 => y=(109-9z)/4

And hence x+(109-9z)/4 +z = 35

We have one independent variable here and two dependent variables

x+ (109-5z)/4 = 35 => x= 35-(109-5z)/4 = (31+5z)/4

So we get {x,y,z} = {31+5z/4, (109-9z)/4, z}

One solution of this system will be : 36/4, 100/4, 1 = { 9, 25, 1}

So 9 one dollar bills, 25 five dollar bills and 1 ten dollar bill

#3 Now we are given x+y+z = 35, x+5y+10z = 144, z = y+2

So we get 4y+9z=109

Now plug in z=y+2 => 4y+9y+18 = 109 => 13y = 91 => y=7

And so z= 9

x= 35-7-9 = 35-16 = 19

So we get 19 one dollar bills, 7 five dollar bills, and 9 ten dollar bills