Answer:
There are 12 bills of 10$ and 11 bills of 20$
Step-by-step explanation:
Let
x ------> number of 10$ bills
y -----> number of 20$ bills
we know that
x+y=23 -----> x=23-y -----> equation A
10x+20y=340 ----> equation B
substitute equation A in equation B and solve for y
10(23-y)+20y=340
230-10y+20y=340
10y=340-230
y=110/10=11
Find the value of x
x=23-y ----> x=23-11=12
therefore
There are 12 bills of 10$ and 11 bills of 20$
