Question

The cost of a gym membership is a onetime fee of \(\$ 140,\) plus a monthly fee of \(\$ 40 .\) Brendan wrote a \(\$ 500\) check to pay his gym membership for a certain number of months, including the onetime fee. How many months of membership did he pay for? F. 3 G. 4 H. 9 J. 12 K. 13

Short answer

Answer: 9 months

Step-by-step solution

Let x be the number of months Brendan paid for his gym membership. #Step 2: Write down the equation#

The total cost of the gym membership is equal to the one-time fee plus the monthly fee times the number of months, which is also equal to the amount Brendan paid for the membership. So, the equation can be written as: \(140 + 40x = 500\) #Step 3: Solve the equation for x#

We will isolate x by first subtracting 140 from both sides of the equation: \(40x = 360\) Next, we will divide both sides of the equation by 40: \(x = \dfrac{360}{40}\) \(x = 9\) #Step 4: Check the answer against the given options and state the conclusion#

x = 9 matches option H, so Brendan paid for 9 months of gym membership. The correct answer is H. 9.

Key concepts

Algebra

Algebra is a branch of mathematics dealing with symbols and the rules for manipulating these symbols. It is a unifying thread of almost all of mathematics and includes everything from solving elementary equations to studying abstractions such as groups, rings, and fields. The power of algebra lies in its ability to generalize mathematical concepts, thus making it possible to deal with abstract concepts and apply them to solve real-world problems.

For instance, in our exercise, algebra is used to represent the cost of a gym membership and the amount Brendan paid with variables and constants. The one-time fee (), monthly fee (), and total amount () are all accounted for using algebraic expressions. This allows us to solve for the unknown quantity, which, in this case, is the number of months of membership paid for by Brendan. By understanding and manipulating these algebraic expressions, we can find concrete answers to theoretical problems.

Solving Linear Equations

Solving linear equations is fundamental in algebra. A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and a single variable. These equations can be as simple as \( x + 2 = 5 \) or more complex, involving multiple terms and variables.

In the context of ACT Math problems, we often deal with linear equations in one variable. The goal is to isolate the variable on one side of the equation to find its value. The steps involved typically include simplifying the equation, using inverse operations to cancel out terms, and isolating the variable of interest. As seen in the provided exercise:

• We start by setting up the equation \(140 + 40x = 500\).
• We simplify by isolating the variable \(x\).
• Finally, we solve for \(x\) getting \(x = 9\), denoting the number of months Brendan paid for his gym membership.

Word Problems

Word problems are mathematical problems presented in a narrative format, often involving real-life scenarios. They require comprehension and translation of text into mathematical expressions, a crucial skill for both ACT Math tests and real-world problem-solving. While they can seem daunting due to their wordy nature, breaking down a word problem into understandable parts is a key strategy.

To solve word problems effectively, one must:

• Read the problem thoroughly.
• Identify and highlight the key pieces of information.
• Translate the words into an algebraic equation.
• Solve the equation using appropriate mathematical techniques.
• Verify that the solution makes sense within the context of the problem.

In our example, identifying the one-time fee, the monthly fee, and the total amount Brendan wrote on the check was crucial. Translating these into the equation \(140 + 40x = 500\) and solving for the number of months he could pay for was the challenge that we had successfully tackled using algebra.