Apile of coins, consisting of quarters and half dollars, is worth $11.75. if there are 2 more quarters than half dollars, how many of each are there?

The pile contains 17 quarters and 15 half-dollars.

Let x = the number of quarters and y = the number of half-dollars.

We have two equations:

(1) $0.25x + $0.50y  = $11.75

(2) x = y +2

Substitute the value of x from Equation (2) into Equation (1).

0.25(y+2) + 0.50y = 11.75

0.25y + 0.50 + 0.50y = 11.75

0.75y = 11.75 – 0.50 = 11.25

y = 11.25/0.75 = 15

Substitute the value of y in Equation (2).

x = 15 + 2 = 17

The pile contains 17 quarters and 15 half-dollars.

Check: 17×$0.25 + 15×$0.50 = $4.25 + $7.50 = $11.75.

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x = 17

step-by-step explanation:

in the given figure, two triangles are inscribed in a circle where ∠bae ≅ ∠dae which makes the side be and de equal so we can write it as:

3x - 24 = x + 10

solving this equation for x to get:

3x - x = 10 + 24

2x = 34

x = 34/2

x = 17

therefore, the value of x = 17.