Problem 61
Question
Solve the equation. Round your answer to two decimal places. $$\frac{x}{3.25}+1=2.08$$
Step-by-Step Solution
Verified Answer
The solution to the equation is \(x = 3.51\).
1Step 1: Simplify the equation
Initially, the equation is \(\frac{x}{3.25} + 1 = 2.08\). To simplify it, subtract 1 from both sides of the equation to get rid of the '+1' on the left side so the equation becomes: \(\frac{x}{3.25} = 2.08 - 1\). This simplifies to \(\frac{x}{3.25} = 1.08\).
2Step 2: Getting rid of fractions
To eliminate the fraction, multiply both sides of the equation by 3.25: \(x = 1.08 \times 3.25\).
3Step 3: Solve for x
To find the value of \(x\), calculate \(1.08 \times 3.25\) to get \(x = 3.51\).
Key Concepts
Understanding AlgebraSimplifying EquationsHandling Fractions in Equations
Understanding Algebra
Algebra is one of the fundamental branches of mathematics, and its main goal is to solve for unknown variables. In an equation like \( \frac{x}{3.25} +1 = 2.08 \), \( x \) represents the unknown that we are tasked to find. To approach algebra problems effectively:
Algebra can include operations like addition, subtraction, multiplication, or division to "balance" the equation on both sides. This balance is critical, as it ensures whatever you do to one side of the equation is equally done to the other side to maintain equality.
By mastering these techniques, you gain a powerful toolset for solving a variety of mathematical problems.
- Identify the variable you need to solve.
- Understand the equation structure.
- Apply algebraic operations to isolate the variable.
Algebra can include operations like addition, subtraction, multiplication, or division to "balance" the equation on both sides. This balance is critical, as it ensures whatever you do to one side of the equation is equally done to the other side to maintain equality.
By mastering these techniques, you gain a powerful toolset for solving a variety of mathematical problems.
Simplifying Equations
Equation simplification is the process of altering the equation to a simpler form while retaining equality. The initial equation \( \frac{x}{3.25} + 1 = 2.08 \) required simplification to correctly solve for \( x \).
Here are steps and tips for simplifying equations:
In our example, we subtracted 1 from both sides to eliminate the constant on the left side. This led to: \( \frac{x}{3.25} = 1.08 \). Simplifying equations makes it easier to isolate the variable. It's an essential step before solving equations, especially in algebra.
Here are steps and tips for simplifying equations:
- Identify any additional or subtracted constants and remove them by performing the reverse operation (e.g., if a number is added, subtract it from both sides).
- Simplify complicated expressions to a more manageable form.
In our example, we subtracted 1 from both sides to eliminate the constant on the left side. This led to: \( \frac{x}{3.25} = 1.08 \). Simplifying equations makes it easier to isolate the variable. It's an essential step before solving equations, especially in algebra.
Handling Fractions in Equations
Fractions in equations might seem tricky at first, but with systematic steps, they can be simplified. In our example, the fraction \( \frac{x}{3.25} \) can be intimidating, but here are some strategies to handle them:
In this instance, multiplying both sides by 3.25 converted the equation from a fractional form to a simple multiplication: \( x = 1.08 \times 3.25 \).
Understanding how to work with fractions in equations opens the door to solving many complex algebraic problems efficiently.
- To remove the fraction, multiply both sides of the equation by the denominator. This action will "cancel" out the denominator, leaving the numerator or variable isolated.
- Make sure you apply the multiplication to both sides to maintain the balance of the equation.
In this instance, multiplying both sides by 3.25 converted the equation from a fractional form to a simple multiplication: \( x = 1.08 \times 3.25 \).
Understanding how to work with fractions in equations opens the door to solving many complex algebraic problems efficiently.
Other exercises in this chapter
Problem 61
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