araceli asked in Science & MathematicsMathematics · 6 months ago

# Math help?

Solve the problem by setting up and solving a system of three linear equations in three variables.

A box contains \$7.00 in nickels, dimes, and quarters. There are 41 coins in all, and the sum of the numbers of nickels and dimes is 3 less than the number of quarters. How many coins of each kind are there?

nickels?

dimes ?

quarters?

Relevance
• Mike G
Lv 7
6 months ago

N+D+Q = 41

N+D = Q-3

Q-3+Q = 41

Q = 22

N+D = 19

5N+10D+25Q = 700

5N+10(19-N) = 150

-5N = -40

N = 8

Nickels = 8

Dimes = 11

Quarters = 22

• 6 months ago

n + d + q = 41

n + d + 3 = q

5n + 10d + 25q = 700

n + d = q – 3

n + d + q = 41

(q – 3) + q = 41

2q = 41 + 3 = 44

q = 22

5n + 10d + 25q = 700

5n + 10d + 25•22 = 700

5n + 10d + 550 = 700

5n + 10d = 150

n + 2d = 30

n + d = q – 3

n + d = 22 – 3

n + d = 19

n + 2d = 30

n + 2d = 30

–n – d = –19

d = 11

n = 19 – d = 8

so we have

q = 22

d = 11

n = 8

check

n + d + q = 41

8 + 11 + 22 = 41 ok

n + d + 3 = q

8 + 11 + 3 = 22 ok

5n + 10d + 25q = 700

40 + 110 + 550 = 700 ok