Problem 72
Question
\(t-6 t=-13+t-3 t\)
Step-by-Step Solution
Verified Answer
The solution is \(t = \frac{13}{3}\).
1Step 1: Combine Like Terms on Each Side
First, combine the like terms on both sides of the equation. Start with the left side: \(t - 6t\) combines to \(-5t\). On the right side, \(t - 3t\) combines to \(-2t\). This reduces the equation to \(-5t = -13 - 2t\).
2Step 2: Move Variables to One Side
The next step is to get all the variable terms on one side. Add \(2t\) to both sides to move \(-2t\) from the right side to the left side. This results in \(-5t + 2t = -13\), which simplifies to \(-3t = -13\).
3Step 3: Solve for the Variable
Now that we have \(-3t = -13\), isolate \(t\) by dividing both sides by -3. This gives us \(t = \frac{-13}{-3}\), which simplifies to \(t = \frac{13}{3}\).
Key Concepts
Combining Like TermsIsolating the VariableEquivalent Equations
Combining Like Terms
When solving a linear equation, one essential technique is the process of combining like terms. Like terms are terms that have the same variable raised to the same power. In the original exercise, the terms to focus on are those containing the variable \(t\) on both sides of the equation:
\[ t - 6t = -13 + t - 3t \]
To simplify the left side, you subtract \(6t\) from \(t\), resulting in \(-5t\). Similarly, on the right side, subtract \(3t\) from \(t\) to get \(-2t\). Now, your simplified equation looks like this:
\[ -5t = -13 - 2t \]
Combining like terms helps reduce equations to simpler forms, making it easier to move on to the next solution steps.
\[ t - 6t = -13 + t - 3t \]
To simplify the left side, you subtract \(6t\) from \(t\), resulting in \(-5t\). Similarly, on the right side, subtract \(3t\) from \(t\) to get \(-2t\). Now, your simplified equation looks like this:
\[ -5t = -13 - 2t \]
Combining like terms helps reduce equations to simpler forms, making it easier to move on to the next solution steps.
Isolating the Variable
Isolating the variable is a key step in solving equations as it helps find the value of the unknown. Once we have the simplified equation,
\[ -5t = -13 - 2t \]
we need to get all terms with the variable \(t\) on one side of the equation. This is achieved by adding \(2t\) to both sides, moving any additional terms across the equal sign. The equation becomes:
\[ -5t + 2t = -13 \]
This results in:
\[ -3t = -13 \]
By performing arithmetic operations that "undo" each other, we ensure that the variable remains by itself on one side of the equation—hence, isolated.
\[ -5t = -13 - 2t \]
we need to get all terms with the variable \(t\) on one side of the equation. This is achieved by adding \(2t\) to both sides, moving any additional terms across the equal sign. The equation becomes:
\[ -5t + 2t = -13 \]
This results in:
\[ -3t = -13 \]
By performing arithmetic operations that "undo" each other, we ensure that the variable remains by itself on one side of the equation—hence, isolated.
Equivalent Equations
While solving linear equations, generating equivalent equations is crucial. These are equations that have the same solutions. As we performed operations like combining like terms and isolating the variable, each new equation we derived was equivalent to the original.
For instance, we transformed:
\[ t - 6t = -13 + t - 3t \]
to:
\[ -3t = -13 \]
Each step involved creating equations that retained the same solution. Applying an operation to both sides, such as adding, subtracting, or dividing by the same number, maintains the equivalence. In our final step, solving \(-3t = -13\) involved dividing by -3 on both sides:
\[ t = \frac{13}{3} \]
This ensures that the process is logical and the solution accurate across all transformations.
For instance, we transformed:
\[ t - 6t = -13 + t - 3t \]
to:
\[ -3t = -13 \]
Each step involved creating equations that retained the same solution. Applying an operation to both sides, such as adding, subtracting, or dividing by the same number, maintains the equivalence. In our final step, solving \(-3t = -13\) involved dividing by -3 on both sides:
\[ t = \frac{13}{3} \]
This ensures that the process is logical and the solution accurate across all transformations.
Other exercises in this chapter
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