Need help with an algebra problem.

Part of the proceeds from a garage sale was $340 worth of $10 and $20 bills. If there were 4 more $10 bills than $20 bills, find the number of each denomination.

let $20 denomination be x, denomination of $10 is x+4

20 * x + 10 * (x +4) = 340

30x + 40 = 340

x = 10

there are 10 $20 bills and 14 $10 bills---answer

let 'a' represent $10 and 'b' represent $20

340 = 10a + 20b

a = b + 4

340 = 10(b + 4) + 20b
340 = 10b + 40 + 20b
300 = 30b
b = 10

a = b + 4
a = 10 + 4
a = 14

check

10(14) + 20(10) = 140 + 200 = 340 (correct!)

10x+20y=340
10(y+4)+20y=340
30y=300
y=10 number of $20 bills
x=14 number of $10 bills
God bless you.

N = $10 bills
T = $20 bills
10N+20T = 340
N+2T=34
Also N=T+4
3T+4 = 34
T = 10
N = 14

10x + 20y = 340

x = y + 4

10 (y + 4) + 20y = 340
10y + 40 + 20y = 340
30y = 300
y = 10

x = 10 + 4 = 14

answer would be 14 pcs $10 bills & 10 pcs $20 bills

14 $10 bills
10 $20 bills

14 $10 bills and 10 $20 bills

10twentys and 14tens