Sam has 3 types of toys. teddy bears cost $10, toy cars cost $7, and legos cost $13. the total cost of his 15 toys is $171. there are five times as many legos as there are teddy bears. how many toys of each type are there?

pls

2 bears, 3 toy cars, and 10 legos

Step-by-step explanation:

We have 2 things going on here:  the NUMBER if toys and the COST of the toys.  They are not similar so they cannot be combined.  We have to have 1 equation represent each.

First, the NUMBER of toys:

We are told that he has 15 toys altogether, and they are bears, cars, and legos.  b = bears, c = cars, l = legos, ok?  The equation for the number of toys is:

b + c + l = 15.  But we are told then that the number of legos is 5 times more than bears, so l = 5b.  We make the replacement:

b + c + 5b = 15 and

6b + c = 15

Next we deal with the COST.  If each bear costs $10, we represent that as 10b; if each car costs $7, we represent that as 7c; if each set of legos costs $13, we represent that as 13(5b) = 65b.  So the equation for the cost is

10b + 7c + 13(5b) = 171 and

75b + 7c = 171

Now we have 2 equations with only 2 unknowns so we solve this system using the method of elimination/addition.  We can cancel out the c terms if we multiply the first equation by -7 to give us the new system:

75b + 7c = 171

-42b - 7c = -105

The c's are gone, leaving us with

33b = 66 and b = 2.

Now sub that in to folve for c:

6(2) + c = 15 and

c = 3.

Sub both of those in to find l:

2 + 3 + l = 15 and

l = 10