Hi, I was just wondering how you guys would solve this problem. I got it wrong because I didn't use

Kaplan's n-variable, n-equations rule. Which basically states that if for every variable there is a linear equation , then it is sufficient.

At a certain restaurant, the total price for one hamburger and one drink is $3.50. How much does one drink cost?

Statement 1- If the price of the drink were half of its present price, the price of a drink would be 1/5 of the price of a hamburger.

Statement 2- The cost of one hamburger and two drinks is $4.50.

- I picked B, because I thought the problem couldn't be solved from statement one. But to some things up, I was just wondering if you guys apply the n-variable, n-equations rule to statement 1 and 2 and move on knowing you got the problem correctly.

, meaning that no equation can be derived with the help of others or by arithmetic operation (multiplication, addition).

\(x+y=2\) and \(3x+3y=6\) --> we do have two linear equations and two variables but we can not solve for \(x\) or \(y\) as the second equation is just the first one multiplied by 3 (basically we have only one distinct equation);

\(x+y=1\), \(y+z=2\) and \(x+2y+z=3\) --> we have 3 linear equations and 3 variables but we can not solve for \(x\), \(y\) or \(z\) as the third equation can be derived with the help of first two if we sum them (basically we have only two distinct equation);

Given: \(h+d=3.5\). Question: \(d=?\)

(1) \(\frac{d}{2}=\frac{h}{5}\). Two

equations for two variables, hence sufficient.

(2) \(h+2d=4.5\). Two

equations for two variables, hence sufficient.

Answer: D.