Solve $3.5(2h+4.5)=57.75$: A Step-by-Step Guide

When faced with an algebraic equation like $3.5(2h+4.5)=57.75$, it's natural to wonder about the solution to the equation. Our goal today is to break down this problem step-by-step, making it easy to understand and apply the principles of solving for an unknown variable. We'll navigate through the distributive property and isolation of the variable 'h' to arrive at the correct answer, rounding to the nearest tenth if needed. This process isn't just about getting a number; it's about understanding how to manipulate equations to find unknown values, a fundamental skill in mathematics.

Understanding the Equation and Our Goal

The equation $3.5(2h+4.5)=57.75$ presents us with a common algebraic challenge. Here, '$h$' is our unknown variable, and our primary objective is to isolate '$h$' on one side of the equals sign. To do this, we'll employ a series of inverse operations. Think of it like unwrapping a gift; we need to undo each step in reverse order to get to the present inside. In this case, the 'present' is the value of '$h$'. We'll start by simplifying the left side of the equation using the distributive property and then proceed to isolate '$h$' through division and subtraction. It's crucial to remember that whatever operation we perform on one side of the equation, we must perform the exact same operation on the other side to maintain the balance and equality. This principle is the bedrock of solving algebraic equations and ensures that our final answer is accurate and valid. We'll also keep in mind the instruction to round to the nearest tenth if necessary, which means our final answer might be a decimal.

Step 1: Apply the Distributive Property

The first key to solving $3.5(2h+4.5)=57.75$ is to tackle the expression in parentheses. The distributive property states that when a number multiplies a sum or difference, it multiplies each term within the parentheses. So, we'll multiply $3.5$ by both $2h$ and $4.5$. This step transforms the equation from a product outside parentheses into a sum of terms, making it easier to manage.

• $3.5 * 2h = 7h$
• $3.5 * 4.5 = 15.75$

Now, we substitute these results back into the original equation:

$7h + 15.75 = 57.75$

This simplified form brings us closer to isolating '$h$'. It's a common strategy in algebra: expand and simplify to make the path to the solution clearer. By applying the distributive property, we've effectively removed the parentheses and created a linear equation with two terms on the left side. This is a significant step forward in our quest to find the value of '$h$'. Remember, precision in these initial steps is vital, as any error here will propagate through the rest of the calculation. Double-checking the multiplication is always a good idea.

Step 2: Isolate the Variable Term

With the equation now in the form $7h + 15.75 = 57.75$, our next move is to get the term containing '$h$' by itself. Currently, $15.75$ is being added to $7h$. To undo this addition, we perform the inverse operation: subtraction. We will subtract $15.75$ from both sides of the equation to maintain its balance.

• $7h + 15.75 - 15.75 = 57.75 - 15.75$

Performing the subtraction on both sides:

• $7h = 42$

This step is critical because it separates the variable term from any constant terms, bringing us one step closer to solving for '$h$'. We've successfully eliminated the added constant, leaving us with a simpler equation where '$h$' is only multiplied by a coefficient. This process of using inverse operations to move terms around is fundamental to algebraic manipulation. Each step builds upon the last, systematically simplifying the equation until the variable is isolated. Think of it as peeling away layers of complexity to reveal the core unknown.

Step 3: Solve for the Variable 'h'

We are now at the equation $7h = 42$. Here, '$h$' is being multiplied by $7$. To isolate '$h$', we need to perform the inverse operation of multiplication, which is division. We will divide both sides of the equation by $7$.

• $7h / 7 = 42 / 7$

Performing the division:

• $h = 6$

This is our solution! We have successfully isolated '$h$' and found its value. In this particular case, the division resulted in a whole number, so no rounding to the nearest tenth was necessary. This outcome is often satisfying in mathematics, as it represents a clean and exact answer. However, if the division had resulted in a decimal, we would have applied rounding rules at this stage to meet the requirements of the problem. The process of solving equations involves a sequence of logical steps, and reaching this point signifies the successful completion of that sequence. It's a demonstration of how consistent application of mathematical rules leads to a definitive answer.

Verifying the Solution

It's always a good practice in mathematics to verify the solution to ensure accuracy. We found that $h = 6$. Let's plug this value back into the original equation: $3.5(2h+4.5)=57.75$.

• Substitute $h = 6$: $3.5(2(6)+4.5)=57.75$
• Simplify inside the parentheses: $3.5(12+4.5)=57.75$
• Continue simplifying inside the parentheses: $3.5(16.5)=57.75$
• Perform the multiplication: $57.75 = 57.75$

Since both sides of the equation are equal, our solution $h = 6$ is correct. This verification step confirms that our calculations were accurate and that we have indeed found the correct value for '$h$' that satisfies the given equation. It's a confidence booster and a safeguard against potential errors.

Conclusion

We have successfully navigated the steps to find the solution to the equation $3.5(2h+4.5)=57.75$. By applying the distributive property and using inverse operations (subtraction and division), we isolated the variable '$h$' and found that $h = 6$. This process highlights the systematic approach required in algebra to solve for unknown values. Remember that mastering these techniques is essential for tackling more complex mathematical problems.

For further exploration into algebraic equations and problem-solving strategies, you can refer to reliable resources like Khan Academy. They offer a wealth of information and practice problems to hone your mathematical skills.