# Use an algebraic approach to solve the problem.?

Sarah has a collection of nickels, dimes, and quarters worth \$15.2. She has 8 more dimes than nickels and twice as many quarters as dimes. How many coins of each kind does she have?

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Sarah has a collection of nickels, dimes, and quarters worth \$15.2. She has 8 more dimes than nickels and twice as many quarters as dimes. How many coins of each kind does she have?

n= nickels       'say'

d= dimes       'say'

q= quarters     'say'

Given that

⇒ 5n + 10d + 25q = 1520 ----------(1)

n = d - 8

q = 2d

substitute n = d - 8 , q = 2d in the equation (1)

Now we get

5(d - 8) + 10d + 25(2d) = 1520

Simplify

5d - 40 + 10d + 50d = 1520

5d -40 + 10d + 50d +40 = 1520 + 40

Simplify

65d = 1560

Divide each side by 65

65d / 65 = 1560 / 65

Simplify

d = 24

substitute d = 24 in the n = d - 8 and q = 2d

Now we get

n = 24 -8 = 16

q = 2(24) = 48

There fore  d = 24 , n = 16 and q = 48

Let number of nickels be x.

Number of dimes = x + 8.

Number of quarters = 2(x + 8).

The equation will be 5x + 10(x + 8) + 25{ 2(x + 8) } = 1520.

5x + 10(x + 8) + 25{ 2x + 16 } = 1520

5x + 10x + 80 + 50x + 400 = 1520

65x + 480 = 1520

65x = 1520 - 480

65x = 1040

x = 1040/65 = 16.

Therefore number of nickels = 16.

Therefore number of dimes = x + 8 = 16 + 8 = 24.

Therefore number of quarters = 2(x + 8) = 2(16 + 8) = 2(24) = 48.