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Chapter 3: Problem 2

Solve each equation and check your answer. $$-6 t+1=16$$

Short Answer

Expert verified
\(t = -\frac{5}{2}\)

Step by step solution

01

Subtract 1 from both sides of the equation

To isolate the term with the variable, subtract 1 from both sides: \(-6t+1-1 = 16-1\) This simplifies to: \(-6t = 15\)
02

Divide both sides of the equation by -6

To isolate the variable t, divide both sides by -6: \(\frac{-6t}{-6} = \frac{15}{-6}\) This simplifies to: \(t = -\frac{5}{2}\)
03

Check the answer

To check the answer, substitute the value of t back into the original equation: \(-6\left(-\frac{5}{2}\right) + 1 \stackrel{?}{=} 16\) Calculate the left-hand side of the equation: \(5(3) + 1 = 15 + 1 = 16\) Since the left-hand side is equal to the right-hand side (16 = 16), the solution \(t = -\frac{5}{2}\) is correct.

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Key Concepts

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

Equation Isolation
Before solving a linear equation, the first step is focusing on isolating the term that contains the variable. This means you need to get all the variable-related terms on one side of the equation. In our original exercise, we begin with:
  • \(-6t+1=16\)
To isolate the term with \(t\), subtract 1 from both sides. This removes the constant from the left side, helping in isolating \(-6t\):
  • \(-6t + 1 - 1 = 16 - 1\)
By performing this operation, the equation simplifies to:
  • \(-6t = 15\)
Now, all terms involving \(t\) are on one side, paving the way for the next step.
Variable Manipulation
Once the variable term is isolated, you'll need to solve for the variable itself through manipulation. In this context, manipulation involves performing operations to get the variable by itself. Continuing with our example:
  • Start from the simplified equation: \(-6t = 15\)
To solve for \(t\), divide both sides of the equation by \(-6\):
  • \(\frac{-6t}{-6} = \frac{15}{-6}\)
By doing this, the equation becomes:
  • \(t = -\frac{5}{2}\)
This division isolates \(t\), providing the solution to the equation. Always remember to use inverse operations (like dividing when there's multiplication) to isolate the variable.
Checking Solutions
After finding a solution, it's crucial to check if it's correct by substituting it back into the original equation. This ensures the operations performed were accurate. For our exercise, plug the solution \(t = -\frac{5}{2}\) back into the original equation:
  • \(-6\left(-\frac{5}{2}\right) + 1\)
First, compute the multiplication:
  • \(5\cdot3 = 15\)
Add 1 to the result:
  • \(15 + 1 = 16\)
Both sides of the equation equal 16, confirming the solution is correct. This step verifies that the solved value for \(t\) satisfies the original equation, giving confidence in the accuracy of the solution.

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