Problem 60 - College Algebra with Corequisite Support

Question

For the following exercises, use a graphing calculator to solve for \(x\). Round the answers to the nearest hundredth. \(0.5(12.3)^{2}-48 x=\frac{3}{5}\)

Step-by-Step Solution

Answer

The value of \(x\) is approximately 1.56.

Step 1: Simplify the Left Side

First, simplify the expression on the left side of the equation. Calculate \(0.5 \times (12.3)^2\). \( (12.3)^2 = 151.29 \) Then, \(0.5 \times 151.29 = 75.645\). So the equation becomes: \( 75.645 - 48x = \frac{3}{5} \).

Step 2: Convert the Right Side to Decimal

Convert the fraction on the right side to a decimal for easier computation. \( \frac{3}{5} = 0.6 \). Thus, the equation is now: \( 75.645 - 48x = 0.6 \).

Step 3: Isolate x

Subtract 75.645 from both sides to isolate the \(-48x\) term. \( -48x = 0.6 - 75.645 \) \( -48x = -75.045 \).

Step 4: Solve for x

Divide both sides of the equation by \(-48\) to solve for \(x\). \( x = \frac{-75.045}{-48} \) \( x = 1.5634375 \). Rounding to the nearest hundredth gives us: \(x = 1.56\).

Key Concepts

Solving Equations

Solving equations involves finding the value of the unknown variable, which often requires manipulating the equation until the variable is isolated. In the problem given, you begin by simplifying and rearranging the original equation.
Here are some key steps for solving equations:

• Start by simplifying each side of the equation. This can include expanding brackets, simplifying terms, or converting fractions to decimals for easier calculations.
• Next, work to isolate the variable. You can do this by removing any constants on the same side of the equation as the variable through addition or subtraction.
• Finally, divide or multiply to solve for the variable. If the variable has a coefficient, divide both sides by that number to find its value.

Using these steps, you can methodically solve many algebraic equations. It can be helpful, as in this exercise, to use a graphing calculator to check your work and ensure accuracy with complex calculations.

Algebra Problems

Algebra problems may look challenging at first, but they are simply puzzles waiting to be solved. At their core, these problems are all about relationships between numbers and symbols.
Algebra problems often require tasks such as:

• Simplification: Breaking down complex expressions into simpler ones, sometimes requiring the use of distributive properties or combining like terms.
• Substituting values: Replacing a variable with a specific value to assist with the solution of the equation.
• Solving inequalities: Understanding how to find the range of values a variable can take rather than a single answer.

A graphing calculator can be a powerful tool when solving more complex algebra problems, as it can help visualize the equation, make quick calculations, and confirm whether your solutions meet every condition of the problem. As always, practice with a range of algebra problems will help you become more comfortable and confident in finding solutions efficiently.

Rounding Numbers

Rounding numbers is a valuable skill in calculations, used to simplify numbers while maintaining their approximate value. It is especially useful when dealing with lengthy decimals or messy fractions.
Here’s how you can round numbers to the nearest hundredth:

• Identify the place value to which you need to round. In the case of the hundredths place, it's the second digit to the right of the decimal point.
• Look at the next digit (thousandths place). If this digit is 5 or greater, round the hundredths place up by one. If it's less than 5, don't change the hundredths digit.

For example, in the calculation from the given exercise, the number 1.5634375 rounds to 1.56 when keeping value close to the actual number. Rounding is often required in problems that specify an answer to a particular decimal place, and practicing this with different numbers can help reinforce the concept in your mind.