Answer:
8 one-dollar bills
3 five-dollar bills
2 ten-dollar bills
Step-by-step explanation:
Let x = # of one-dollar bills, y = # of five-dollar bills, and z = # of ten-dollar bills. Total amount in the wallet is $43, so the first equation would be 1x + 5y + 10z = 43. Next, there are 4 times as many one-dollar bills as ten-dollar bills, so x = 4z. There are 13 bills in total, so x + y + z = 13
x + 5y + 10z = 43
x = 4z
x + y + z = 13
x + 5y + 10z = 43
x + 0y - 4z = 0
x + y + z = 13
5y + 14z = 43
-y - 5z = -13
5y + 14z = 43
-5y - 25z = -65
-11z = -22
z = 2
x = 4z
x = 4*2 = 8
x + y + z = 13
8 + y + 2 = 13
10 + y = 13
y = 3
If you have at least as many equations as your have unknown variables, the 'system' is solvable (unless the equations are copies of eachother).
In this case, isolate one letter and plug it in the other. I'm going for the 2a in the bottom one (it doesn't matter)
2a - 4b = 12 => (divide by two and move the b's to the other side)
a = 6 + 2b. (plug this one into the top equation)
4(6+2b) + 6b = 10 => 24+8b+6b = 10 => 24 + 14b = 10 => 14b = -14 => b=-1
a = 6 + -2 = 4.
So a=4 and b=-1.
F(6) = 7(6) - 4
f(6) = 42 - 4
f(6) = 38
Hope this helps!
im just gonna take the points for falling for that.
7.6 x 10^2 that's the answer
