Problem 201 - Elementary Algebra

Question

Two containers of gasoline hold a total of fifty gallons. The big container can hold ten gallons less than twice the small container. How many gallons does each container hold?

Step-by-Step Solution

Answer

The small container holds 20 gallons, and the big container holds 30 gallons.

Step 1: Define the Variables

Let the amount of gasoline the small container can hold be represented by the variable \( x \). Let the amount of gasoline the big container can hold be represented by the variable \( y \).

Step 2: Set Up the Equations

According to the problem, the total amount of gasoline is fifty gallons, so we have the equation: \( x + y = 50 \). Additionally, it is given that the big container can hold ten gallons less than twice the small container, resulting in the equation: \( y = 2x - 10 \).

Step 3: Substitute the Expression

Substitute the expression for \( y \) from the second equation into the first equation: \( x + (2x - 10) = 50 \).

Step 4: Combine Like Terms

Combine like terms in the equation: \( 3x - 10 = 50 \).

Step 5: Solve for x

Add 10 to both sides of the equation: \( 3x = 60 \). Then, divide both sides by 3: \( x = 20 \).

Step 6: Solve for y

Use the value of \( x \) to find \( y \) using the equation \( y = 2x - 10 \): \( y = 2(20) - 10 \). Therefore, \( y = 40 - 10 \). Thus, \( y = 30 \).

Step 7: Verify the Solution

Check that the solutions meet the original conditions: \( x + y = 20 + 30 = 50 \) gallons, which is correct. Also, \( y = 2x - 10 = 2(20) - 10 = 30 \) gallons, which is correct.

Key Concepts

Variable Substitution

Variable substitution is a powerful technique in solving equations. It involves replacing a variable with its equivalent value or expression. In our problem, we had two variables: the amount of gasoline the small container can hold (\(x\)) and the amount in the big container (=y)). The problem gave us a relationship: the total gasoline is fifty gallons and the big container holds ten gallons less than twice the small container.

By substituting the expression for =y=2x−10) into the equation =x+y=50), we simplified our complex system into a single equation: =x+(2x−10)=50.). This is key in simplifying and solving the problem more efficiently. Variable substitution helps to break down complex relationships step-by-step, making it easier to manage and solve. It is useful in linear equations, quadratic equations, and many other algebra problems.

Linear Equations

Linear equations are mathematical statements that show the equality of two expressions with one or more variables. These equations form straight lines when plotted on a graph. In our example, the problem provides us with two linear equations:

ewline=x+y=50
ewline=y=2x−10

To solve these equations, we looked for the intersection point of the lines represented by these equations, but here we used algebraic methods directly. The equations were manipulated and combined to isolate variables and find their values. Solving linear equations often involves steps like:

• Combining like terms
• Isolating a variable
• Using substitution

These are fundamental tools in algebra critical for solving real-world problems just like in our exercise.

Problem-Solving Steps

Solving problems involving systems of equations requires a clear approach. In this exercise, we followed several well-defined steps:

• Define the Variables: Represent the unknowns with variables (=x) and =y)).
  
• Set Up the Equations: From the problem's information, create your system of equations.
  
• Substitute and Simplify: Use variable substitution to replace one variable with an equivalent expression.
  
• Solve for the Variable: Combine like terms and solve for your variable.
  
• Find the Remaining Variable: Use your obtained value to find the other variable.
  
• Verify the Solution: Check both equations to ensure the solutions satisfy the original problem conditions.
  

Following these steps systematically ensures we don't miss key details and leads us to the correct solution efficiently. This structured approach applies broadly across different math problems and is useful for developing good problem-solving habits.