Question 1188270
you need to introduce 2 variables.
x is the total amount of his money.
c is the cost of one pack of cards.


he spent 3/10 of his money on 11 packs of cards.


3/10 of his money = 3/10 * x.
total cost for the cards = 11 * c.


equation is 3/10 * x = 11 * c.


with the rest of his money plus another 20 dollars, he could buy 29 more cards.


rest of his money is 7/10 * x + 10.
total cost for the cards is 29 * c.


equation is 7/10 * x + 20 = 29 * c.


you have two equations that need to be solved simultaneously.


they are:


3/10 * x = 11 * c.
7/10 * x + 20 = 29 * c


multiply both sides of the first equation by 7 and multiply both sides of the second equation by 3 to get:


21/10 * x = 77 * c
21/10 * x + 60 = 87 * c


subtract the first equation from the second to get:


60 = 10 * c


solve for c to get:


c = 60/10 = 6.


the cost of 1 pack of cards is 6 dollars.


now that you know the cost for 1 pack of cards, you can find out much money he had to start with.


3/10 * x = 11 * c = 11 * 6 = 66 dollars.


solve for x to get:


x = 66 * 10 / 3 = 220 dollars.


your soluion is that the cost of one pack of cards is 6 dollars and the total amount of money that he had to start with was 220 dollars.


i'm assuming dollars, but you can replace that with any other denomination that you are dealing with.


to confirm, replace the original equations with 220 for x and 6.00 for c.


3/10 * x = 11 * c becomes 3/10 * 220 = 11 * 6 which becomes 66 = 66.
7/10 * x + 20 = 29 * c becomes 7/10 * 220 + 20 = 29 * 6 which becomes 174 = 174.


both equations are confirmed to be good when x = 220 and c = 6.


your solution is:


he has 220 dollars to start with and the cost of one pack of cards was 6 dollars.