Problem 15
Question
Solve the equations. $$ 4 t+2(t+1)-5 t=13 $$
Step-by-Step Solution
Verified Answer
Question: Solve the equation: \(4 t+2(t+1)-5 t=13\)
Answer: \(t = 11\)
1Step 1: Distribute the coefficient inside the parentheses
First, we distribute the coefficient '2' inside the parentheses:
$$
4 t+2(t+1)-5 t=13
$$
$$
4t + 2t + 2 - 5t = 13
$$
2Step 2: Combine like terms on both sides of the equation
Now, combine the like terms (all the terms containing 't'):
$$
(4t - 5t + 2t) + 2 = 13
$$
$$
t + 2 = 13
$$
3Step 3: Isolate the variable 't'
In this step, we want to isolate 't' on one side of the equation. To do this, subtract '2' from both sides of the equation:
$$
t + 2 - 2 = 13 - 2
$$
$$
t = 11
$$
So, the solution to the equation is \(t = 11\).
Key Concepts
Distributive PropertyCombining Like TermsIsolation of Variable
Distributive Property
The distributive property is a very useful tool in algebra that allows us to simplify expressions by multiplying a single term across terms inside parentheses. In our original exercise, we see it applied in the equation:
\[4t + 2(t+1) - 5t = 13\]
The term \(2(t+1)\) shows us a common case where the distributive property helps. Here's how we can apply it:
\[4t + 2t + 2 - 5t = 13\]
By doing this, we've simplified the expression and prepared it for the next steps. Using the distributive property is essential when dealing with equations that include parentheses, as it helps to break down and work through the equation systematically.
\[4t + 2(t+1) - 5t = 13\]
The term \(2(t+1)\) shows us a common case where the distributive property helps. Here's how we can apply it:
- Multiply the coefficient \(2\) with each term inside the parentheses \((t+1)\).
- This results in \(2 \cdot t + 2 \cdot 1\), which simplifies to \(2t + 2\).
\[4t + 2t + 2 - 5t = 13\]
By doing this, we've simplified the expression and prepared it for the next steps. Using the distributive property is essential when dealing with equations that include parentheses, as it helps to break down and work through the equation systematically.
Combining Like Terms
Once the distributive property has been applied, the next step is to combine like terms. This step simplifies the equation further making it easier to solve. In the equation:
\[4t + 2t + 2 - 5t = 13\]
We can identify the 'like terms,' which are all the terms with the variable \(t\):
\[t + 2 = 13\]
This combination process is crucial as it makes the equation much simpler, revealing the structure needed for further solutions.
\[4t + 2t + 2 - 5t = 13\]
We can identify the 'like terms,' which are all the terms with the variable \(t\):
- \(4t\)
- \(2t\)
- \(-5t\)
- Start by summing \(4t\) and \(2t\) to get \(6t\).
- Next, subtract \(5t\) from \(6t\) to arrive at \(t\).
\[t + 2 = 13\]
This combination process is crucial as it makes the equation much simpler, revealing the structure needed for further solutions.
Isolation of Variable
The final step in solving the equation is isolating the variable, which involves getting the variable by itself on one side of the equation. After combining like terms, our equation is:
\[t + 2 = 13\]
To isolate \(t\), we need to remove the constant term (\(+2\)) from the left side. This is achieved by performing the same arithmetic operation on both sides of the equation:
\[t = 11\]
Now, the variable \(t\) is isolated, revealing the solution. Isolating the variable is one of the most important steps in solving equations, as it underscores the transformation of the equation into its simplest possible form.
\[t + 2 = 13\]
To isolate \(t\), we need to remove the constant term (\(+2\)) from the left side. This is achieved by performing the same arithmetic operation on both sides of the equation:
- Subtract \(2\) from both sides of the equation.
- This operation looks like: \[t + 2 - 2 = 13 - 2\].
\[t = 11\]
Now, the variable \(t\) is isolated, revealing the solution. Isolating the variable is one of the most important steps in solving equations, as it underscores the transformation of the equation into its simplest possible form.
Other exercises in this chapter
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