Problem 13 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Solve the given equation. $$ \frac{3}{5}(k+1)=\frac{1}{4}(2 k+3) $$

Step-by-Step Solution

Answer

The short answer based on the given step-by-step solution is: \(k = \frac{3}{2}\).

Step 1: Write down the given equation

The given equation is \[ \frac{3}{5}(k+1)=\frac{1}{4}(2 k+3) \]

Step 2: Clear the fractions by multiplying both sides of the equation by the least common multiple (LCM) of the denominators

The least common multiple of the denominators 5 and 4 is 20. We will multiply both sides of the equation by 20 to eliminate the fractions. \[ 20 \times \frac{3}{5}(k+1) = 20 \times \frac{1}{4}(2 k+3) \]

Step 3: Simplify the equation

After multiplying, we simplify the equation: \[ \frac{20 \times 3}{5}(k+1) = \frac{20 \times 1}{4}(2 k+3) \] \[ 12(k+1) = 5(2 k+3) \]

Step 4: Distribute and combine like terms

Distribute the numbers on both sides of the equation: \[ 12k + 12 = 10k + 15 \]

Step 5: Move all terms with the variable to one side of the equation

Subtract 10k from both sides of the equation: \[ 12k - 10k + 12 = 15 \] This simplifies to: \[ 2k + 12 = 15 \]

Step 6: Solve for the variable

To solve for `k`, first subtract 12 from both sides of the equation: \[ 2k = 3 \] Now, divide both sides by 2: \[ k = \frac{3}{2} \] The solution for the equation is \(k = \frac{3}{2}\).

Key Concepts

Understanding Linear Equations

Linear equations are like simple math stories that help us find unknown values. They often look like a line in a graph and usually have a form where you solve for a variable, like \(x\) or \(k\). This equation involves finding the value of \(k\) that makes both sides equal.
Understanding linear equations is important because they help you solve real-life problems like budgeting or calculating distances. In this exercise, we'll go through a clear and simple problem to solve a linear equation.

• They have no exponents higher than one.
• All operations are something a middle school math student would know—like addition or multiplication.

Getting comfortable with linear equations means getting comfortable with a vital part of algebra.

Fraction Clearing Made Simple

Fractions in equations can make the math look complicated, but we have a neat trick to handle them. We can "clear" fractions by finding something called the Least Common Multiple (LCM).
In the given equation, there are two fractions with denominators 5 and 4. The LCM of these numbers is 20. By multiplying each term in the equation by 20, we make the fractions disappear:
\[20 \times \frac{3}{5}(k+1) = 20 \times \frac{1}{4}(2 k+3)\]
This becomes an equation without fractions, making it easier to solve:
\[12(k+1) = 5(2k+3)\]
This step helps us focus on solving the problem without getting bogged down by fractions. It simplifies the pathway to finding the solution.

Variable Isolation Insight

Isolating the variable means getting \(k\) all alone on one side of the equation. This makes it clear what value makes the equation true.
After clearing fractions, we distribute the numbers:
\[12k + 12 = 10k + 15\]
Now, to isolate \(k\), we'll move all terms containing \(k\) to one side. Subtract 10k from both sides:

• This leaves us with \(2k + 12 = 15\).

Next, get rid of the constant by subtracting 12 from both sides, giving:
\[2k = 3\]
Finally, divide by 2 to isolate \(k\):
\[k = \frac{3}{2}\]
Breaking it down piece by piece makes the process straightforward.

Equation Solving Steps Explained

Solving equations involves clearly outlined steps that lead to a solution. By following these steps methodically, you can solve almost any linear equation. Here's a quick recap of our process:

• **Write down the equation**: Start with what's given.
• **Clear fractions**: Use the LCM to get rid of fractions, making the equation simpler.
• **Distribute and combine like terms**: Simplify each side of the equation.
• **Isolate the variable**: Move everything around to have the variable on one side alone.
• **Solve for the variable**: Final math operations give you the answer.

Understanding these steps not only helps in math exams but in everyday problem-solving situations as well. Practice makes these steps feel more natural.