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Please help this is algebra!

Please help this is algebra!-example-1

1 Answer

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Answer:

7 nickels

20 dimes

5 quarters

Explanation:

n=nickels. d=dimes. q=quarters

n+d+q=32

0.05n+0.1d+0.25q=3.60

d=8+n+q

Solve for number of dimes:

d-8=8-8+n+q

d-8=n+q

d-d-8=n+q-d

-8=n+q-d

32=n+q+d

- (-8=n+q-d)

32 - (-8) = n-n + q-q + d-(-d)

32 + 8 = d + d

40 = 2d

d = 20 => simplify

20=8+n+q ==> substitute 20 for d (d=8+n+q)

20-8 = 8-8+n+q

n+q = 12

0.05n+0.1(20)+0.25q=3.60 ==> substitute 20 for d

0.05n + 2 + 0.25q = 3.60

(0.05n + 2 + 0.25q = 3.60)*100 ->remove the decimals by multiplying by 100

5n + 200 + 25q = 360

5n + 200 - 200 + 25q = 360 - 200

5n + 25q = 160

Solve for number of quarters:

(5n + 25q)/5 = 160/5 ==> simplify the equation

n + 5q = 32

- (n + q = 12)

n-n + 5q-q = 32-12

0 + 4q = 20

4q = 20

q = 5 ==> simplify

Solve for number of nickels:

n+20+5=32 ==> plug in 20 for d and 5 for q (n+d+q=32)

n+25=32

n+25-25=32-25 ==> solve for n

n=7 ==> simplify

n=7: 7 nickels

d = 20: 20 dimes

q = 5: 5 quarters

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