You can write three equations in the numbers of nickels (n), dime (d), and quarters (q).
n + d + q = 23 . . . . . . . there are 23 coins total
0n +d -q = 2 . . . . . . . . .there are 2 more dimes than quarters
5n +10d +25q = 250 . .the total value is $2.50
The collection includes 11 nickels, 7 dimes, and 5 quarters.
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I used the matrix function of my calculator to solve these equations. You can find q by subtracting from the last equation five times the sum of the first two equations.
(5n +10d +25q) -5((n +d +q) +(d -q)) = (250) -5(23 +2)
25q = 125 . . . . . . . simplify
q = 5
From the second equation,
d = q +2 = 7
And from the first,
n = 23 -5 -7 = 11
Answer:
There are 111 blue bags
Step-by-step explanation:
Let
x ----> the number of black bags
y ---> the number of blue bags
we know that
isolate the variable y
-----> equation A
----> equation B
substitute equation B in equation A
therefore
There are 111 blue bags
<span>x+(2x-4)=11
Answer is x=5.</span>
Answer:
most is 160 hope I helped
