Question 56·Medium·Linear Equations in Two Variables
A community center sold regular memberships and premium memberships, collecting a total of $2,050. The relationship is modeled by the equation
How many dollars more does a premium membership cost than a regular membership?
For linear word problems that give an equation like , first match each variable to what it counts (items, memberships, hours, etc.) and then interpret each coefficient as the price or rate per item. Before jumping into solving for or , check what the question actually asks—often it only wants a comparison like a difference between prices, which you can get directly by subtracting the relevant coefficients. This saves time and avoids unnecessary algebra.
Hints
Identify what the variables mean
Look at the problem statement: what do and stand for in terms of memberships?
Focus on the coefficients (the numbers in front of the variables)
In the expression , think about what the and represent in the context of selling memberships and collecting money.
Translate the question into a math operation
Once you know the cost of a regular membership and the cost of a premium membership, what operation should you use to find how many dollars more one costs than the other?
Desmos Guide
Use Desmos to confirm the price difference
In Desmos, type 75 - 50 and look at the output. That value is how many dollars more a premium membership costs than a regular membership.
Step-by-step Explanation
Interpret the equation in context
The equation is .
- is the number of regular memberships.
- is the number of premium memberships.
- means each regular membership costs $50.
- means each premium membership costs $75.
So the price of a regular membership is $50, and the price of a premium membership is $75.
Find how much more a premium membership costs
The question asks: How many dollars more does a premium membership cost than a regular membership?
That is:
- (price of premium) (price of regular)
Compute this difference: .
So a premium membership costs $25 more than a regular membership.