Question 282766
Your answer is that the number of hole punchers is 5.


You can't find it directly because you have 2 equations in 3 unknowns which means that one of the unknowns can only be found in relation to one of the other unknowns.


To find your solution, I did the following:


The two equations you have to solve simultaneously are:


x + y + z = 100
.5x + 3y + 10z = 100


Multiply the second equation by 2 to get:


x + y + z = 100
x + 6y + 20z = 200


Subtract first equation from second equation to get:


5y + 19z = 100


Solve y in terms of z to get:


y = (100 - 19z)/5


This is the same as:


y = 100/5 - 19z/5


y will not be an integer unless z = 5 or a multiple of 5 which means that the smallest z can be is 5.


Sounds like she could buy either 5 or 10 hole punchers and y would be an integer.


If z were 10, however, then she could not have bought any x or y because 10 * 10 = 100 which means zero money for x and y.


z would have to be equal to 5.


Here's how the equations work out.


x + y + 5 = 100
.5x + 3y + 50 = 100


Subtract 5 from both sides of the first equation and subtract 50 from both sides of the second equation to get:


x + y = 95
.5x + 3y = 50


Multiply the second equation by 2 to get


x + y = 95
x + 6y = 100


Subtract the first equation from the second equation to get:


5y = 5


Divide both sides of the equation by 5 to get:


y = 1


Use the first equation to solve for x to get:


x = 94


Your values for x,y,z are 94,1,5.


Plug them into your original equations to get:


x + y + z = 100 becomes 94 + 1 + 5 = 100 which is true.
.5x + 3y + 10z = 100 becomes .5*94 + 3*1 + 10*5 = 47 + 3 + 50 = 100 which is true.


Both original equations are true when you use the values for x,y,z of 94,1,5, so those values are good.


Your answer is:


She bought 5 hole punchers.


that would be selection E.