BTGmoderatorDC wrote:A man spends $48 to buy 6 hamburgers and 8 colas for his office workers. The next day, he buys 5 hamburgers and 4 colas and spends $32. Assuming the prices of hamburgers and colas remain constant, what is the price of one hamburger and one cola?

A. $6
B. $7
C. $8
D. $9
E. $10

If 6h+8c = 48 and 5h+4c = 32, what is the value of h+c?
In

In the blue equation, the coefficient for h is LESS than the coefficient for c.
In the red equation, the coefficient for h is GREATER than the coefficient for c.

Given this situation, we can use the following approach to determine the value of h+c:
1. Alter the equation(s) so that the difference between the coefficients in each equation is THE SAME
2. Add the two equations
3. Divide as necessary to determine the value of h+c

In the blue equation, the difference between the coefficients = 8-6 = 2.
In the red equation, the difference between the coefficients = 5-4 = 1.
If we divide the blue equation by 2, the difference between the coefficients will decrease to 1:
6h+8c = 48 --> 3h+4c = 24 --> 4-3 = 1

Adding together 3h+4c = 24 and 5h+4c = 32, we get:
8h+8c = 56
Dividing both sides by 8, we get:
h+c = 7

The correct answer is B.

Let the price of one hamburger be \(x\) and that of cola is \(y\)

\(6x+8y=48 \Rightarrow 3x+4y=24\qquad (1)\)
\(5x+4y=32\qquad (2)\)

Subtract equation \((1)\) from \((2)\) to get \(2x=8\) or \(x=4\).

Substitute the value of \(x\) in any equation to get \(y=3\).

Hence, \(x+y=4+3=7\Rightarrow \) __B__.