The problem states that Luis bought 5 scones and 2 large coffees for $16.48. Rachel bought 4 scones and 3 large coffees for $15.83. The cost of one scone is given as $1.89. The task is to find the cost of one large coffee.

1. Problem Description

The problem states that Luis bought 5 scones and 2 large coffees for $16.

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8. Rachel bought 4 scones and 3 large coffees for $15.

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3. The cost of one scone is given as $1.

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9. The task is to find the cost of one large coffee.

2. Solution Steps

Let $s$ be the cost of one scone and $c$ be the cost of one large coffee.

We have two equations:

$5s + 2c = 16.48$

$4s + 3c = 15.83$

We are given that $s = 1.89$. Substitute this value into the equations:

$5(1.89) + 2c = 16.48$

$4(1.89) + 3c = 15.83$

Simplify the equations:

$9.45 + 2c = 16.48$

$7.56 + 3c = 15.83$

Solve the first equation for $2c$:

$2c = 16.48 - 9.45$

$2c = 7.03$

$c = 7.03/2 = 3.515$

Solve the second equation for $3c$:

$3c = 15.83 - 7.56$

$3c = 8.27$

$c = 8.27/3 = 2.75666...$

We can solve this system of equations in the following way.

Multiply the first equation by 3 and the second equation by 2 to eliminate $c$.

$3(5s + 2c) = 3(16.48)$

$15s + 6c = 49.44$

$2(4s + 3c) = 2(15.83)$

$8s + 6c = 31.66$

Subtract the second equation from the first:

$(15s + 6c) - (8s + 6c) = 49.44 - 31.66$

$7s = 17.78$

$s = 17.78/7 = 2.54$

Substitute $s = 2.54$ into $5s + 2c = 16.48$:

$5(2.54) + 2c = 16.48$

$12.7 + 2c = 16.48$

$2c = 16.48 - 12.7$

$2c = 3.78$

$c = 3.78/2 = 1.89$

However, we are given that the cost of one scone is $1.

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9. Substitute $s = 1.89$ into the equations:

$5(1.89) + 2c = 16.48$

$4(1.89) + 3c = 15.83$

$9.45 + 2c = 16.48$

$7.56 + 3c = 15.83$

$2c = 16.48 - 9.45 = 7.03$

$c = 7.03 / 2 = 3.515$

$3c = 15.83 - 7.56 = 8.27$

$c = 8.27 / 3 = 2.756666...$

We now recalculate from the start. We have the equation $5s + 2c = 16.48$ and $4s + 3c = 15.83$. $s = 1.89$. Thus $5(1.89) + 2c = 16.48$ so $9.45 + 2c = 16.48$. $2c = 16.48 - 9.45 = 7.03$. $c = \frac{7.03}{2} = 3.515$. Also $4(1.89) + 3c = 15.83$, so $7.56 + 3c = 15.83$, thus $3c = 15.83 - 7.56 = 8.27$. Then $c = \frac{8.27}{3} = 2.7566... \approx 2.76$ (to two decimal places). It appears the problem may have an error. If we average the coffee costs obtained through each equation, we get $(3.515 + 2.7566...)/2 \approx 3.14$.

Let's use substitution or elimination method.

Multiply first equation by 3, and second equation by -2:

$15s + 6c = 49.44$

$-8s - 6c = -31.66$

Add the two equations:

$7s = 17.78$

$s = 2.54$

If we know that s = 1.89, we do

$5(1.89) + 2c = 16.48$

$9.45 + 2c = 16.48$

$2c = 7.03$

$c = 3.515$

3. Final Answer

3.52