Second term + Fourth term: $(a - d) + (a + d) = 2a = 10$ - RoadRUNNER Motorcycle Touring & Travel Magazine
Understanding the Second and Fourth Terms in Linear Simplification: Solving $(a - d) + (a + d) = 2a = 10$ with Clear Steps
Understanding the Second and Fourth Terms in Linear Simplification: Solving $(a - d) + (a + d) = 2a = 10$ with Clear Steps
Mastering algebra often begins with recognizing patterns in equations—especially those involving variables and constants. One classic example is the expression $(a - d) + (a + d) = 2a = 10$, which highlights the importance of term pairing and simplification. In this article, we explore its mathematical meaning, breaking down how the second and fourth terms combine and why the final result of $2a = 10$ leads directly to powerful simplifications in solving equations.
Understanding the Context
The Equation: $(a - d) + (a + d) = 2a = 10$
Consider the equation
$$(a - d) + (a + d) = 10.$$
This sum combines two binomials involving the variables $a$ and $d$. The key insight lies in observing how the terms relate to one another:
- The first term: $a - d$
- The second term: $a + d$
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Key Insights
Now notice that the $-d$ and $+d$ are opposites. When added, these terms cancel out:
$$(a - d) + (a + d) = a - d + a + d = 2a.$$
Thus, the sum simplifies neatly to $2a$, eliminating the variables $d$, resulting in $2a = 10$.
Why the Second and Fourth Terms Matter
In algebra, pairing like terms is essential. Here, the variables $d$ appear as opposites across the two expressions:
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- In $(a - d)$, $d$ is subtracted.
- In $(a + d)$, $d$ is added.
When combined, $ -d + d = 0 $, removing $d$ entirely from the expression. This cancellation is a core principle in solving equations and simplifying expressions:
> Rule: Opposite variables cancel when added together.
Thus, the selective pairing of $ -d $ and $ +d $ directly leads to the simplified form $2a$, a foundational step toward solving for the variable.
Solving for $a$: From $2a = 10$
With the simplified equation
$$2a = 10,$$
we solve for $a$ by isolating the variable:
- Divide both sides by 2:
$$a = rac{10}{2} = 5.$$
So, the value of $a$ is $5$. This demonstrates how understanding term relationships—particularly cancellation—supports efficient problem-solving.
