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Chapter 3: Problem 60

Solve and check each equation. $$\frac{3}{5}-\frac{2}{3} k=1$$

Short Answer

Expert verified
The solution for the given equation is \(k = -\frac{3}{5}\). To check the correctness of the solution, plug the value of k back into the original equation: \(\frac{3}{5} - \frac{2}{3}\left(-\frac{3}{5}\right) = 1\). This simplifies to \(\frac{3}{5}+\frac{2}{5}=\frac{5}{5}=1\), confirming that the solution is correct.

Step by step solution

01

Find the common denominator

To eliminate the fractions, we need to find the common denominator of 5 and 3, which is 15. We will multiply both sides of the equation by 15 to clear out the fractions.
02

Multiply both sides of the equation by the common denominator

Multiplying both sides of the equation by 15, we get: \(15(\frac{3}{5} - \frac{2}{3}k) = 15(1)\)
03

Distribute the 15 to the terms inside the parentheses

Now, distribute 15 to both terms inside the parentheses: \(\frac{15\times3}{5} - \frac{15\times2}{3}k = 15\)
04

Simplify and write the new equation without fractions

After simplifying the fractions and cancelling out any common factors, we get: \(9 - 10k = 15\)
05

Solve for k

Now we need to isolate k on one side of the equation. First, add 10k to both sides: \(9 = 15 + 10k\) Now, subtract 15 from both sides: \(-6 = 10k\) Finally, divide both sides by 10 to find the value of k: \(k = -\frac{3}{5}\)
06

Check the solution

To check the correctness of the solution, plug the value of k back into the original equation: \(\frac{3}{5} - \frac{2}{3}\left(-\frac{3}{5}\right) = 1\) Simplify the equation to check if the left-hand side equals the right-hand side: \(\frac{3}{5} + \frac{2}{3}\times\frac{3}{5} = \frac{3}{5}+\frac{6}{15}=\frac{3}{5}+\frac{2}{5}=\frac{5}{5}=1\) As the left-hand side equals the right-hand side, our solution is correct. The solution for the given equation is \(k = -\frac{3}{5}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solving Linear Equations
When solving linear equations, the aim is to find the value of the unknown variable that makes the equation true. In our exercise, we need to solve the equation \(\frac{3}{5} - \frac{2}{3}k = 1\). This requires us to isolate the variable \(k\) on one side of the equation.
  • Begin by aiming to eliminate any fractions to simplify the process.
  • Use operations like addition, subtraction, multiplication, or division to manipulate the equation.
In this case, finding a common denominator and multiplying through helps eliminate the fractions. This sets up the path to solve for \(k\) by straightforward algebraic manipulation. Remember:
  • When you perform an operation on one side of the equation, you must do the same on the other side to keep the equation balanced.
  • Check the solution by substituting it back into the original equation to see if it holds true.
Fractions in Equations
Fractions in equations can seem difficult, but are quite manageable with a little practice. They simply express a part of a whole and, in linear equations, often appear when variables are divided by a number, like \(\frac{3}{5} - \frac{2}{3}k\).
  • The main goal is to remove fractions to simplify the equation.
  • Multiply every term by the least common denominator to clear the fractions.
Visualize the process like cleaning the slate; you turn a complex fraction-laden equation into a simpler, cleaner equation without fractions, thereby easing the subsequent steps to solve for the unknown variable.
Common Denominator
The common denominator is a powerful tool used to eliminate fractions in equations by transforming different denominators into a single, shared one. For the equation \(\frac{3}{5} - \frac{2}{3}k = 1\), the denominators are 5 and 3.
  • Find the least common multiple (LCM) of these denominators; 15, in this case.
  • Multiply each term by this common denominator.
By multiplying through, you remove the fractions completely, reducing the complexity of the equation. This allows you to convert the equation into an easier format and proceed with arithmetic operations to isolate the variable, leading to the solution.

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