Solve 16-3(x-7)=-14: Step-by-Step Answer and Guide

Solving the linear equation 16-3(x-7)=-14 is a fundamental exercise in algebra that tests your ability to manage distribution, negative signs, and variable isolation. The process requires a systematic approach to simplify the expression on the left and isolate the variable x to determine its specific value. This walkthrough breaks down each mathematical step to ensure clarity and understanding, transforming a seemingly complex problem into a straightforward calculation. By following these logical stages, you can confidently arrive at the correct solution without relying on guesswork.

Initial Equation Analysis

The primary goal is to find the value of x that satisfies the equality 16-3(x-7)=-14. At first glance, the presence of parentheses and the negative coefficient attached to the group might seem intimidating. However, the standard procedure involves addressing the parentheses first to remove the grouping symbols. This initial step is crucial for revealing the individual terms that make up the equation, allowing for easier manipulation and combination of like terms as the solving process progresses.

Step 1: Distributing the Negative Three

To eliminate the parentheses, we distribute the -3 across the terms inside (x - 7). This means multiplying -3 by x, which gives us -3x, and then multiplying -3 by -7, which results in +21 due to the rule that a negative times a negative yields a positive. The equation now transforms from 16 - 3(x - 7) = -14 into 16 - 3x + 21 = -14. This expansion is the essential bridge that moves us from a compact form to a more accessible linear format.

a black and white photo with numbers written in the shape of a christmas tree on it
a black and white photo with numbers written in the shape of a christmas tree on it

Step 2: Combining Like Terms

Next, we combine the constant terms on the left side of the equation. We have the 16 and the newly distributed +21, which are like terms because they are both numbers without variables. Adding 16 and 21 results in 37, significantly simplifying the expression. The equation is now reduced to the much cleaner form of 37 - 3x = -14, making it much easier to isolate the variable term.

Isolating the Variable Term

The core of solving for x revolves around getting the term with the variable (in this case, -3x) alone on one side of the equation. To move the constant 37 to the right side, we subtract 37 from both sides of the equality. This action maintains the balance of the equation. The calculation on the right side involves subtracting 37 from -14, which is equivalent to adding the negative of 37 to -14, resulting in -51. The equation is now simplified to -3x = -51.

Final Calculation for X

With the variable term isolated as -3x, the final step is to determine the value of x itself. Since x is being multiplied by -3, we must perform the inverse operation by dividing both sides of the equation by -3. Dividing -51 by -3 yields a positive result, as the negatives cancel out. The calculation 51 divided by 3 results in the integer 17, giving us the solution x = 17.

the back side of a computer screen with numbers and letters on it, all in white
the back side of a computer screen with numbers and letters on it, all in white

Verification of the Solution

To ensure the accuracy of our answer, we substitute x = 17 back into the original equation: 16 - 3(17 - 7). First, we solve the expression inside the parentheses, which is 10. Multiplying 3 by 10 gives 30. The expression becomes 16 - 30, which equals -14. Since the left side of the equation matches the right side exactly, we have confirmed that x = 17 is indeed the correct and valid solution.

Summary of Results

Working through the problem 16-3(x-7)=-14 methodically involves distributing, combining like terms, and isolating the variable. We discovered that the variable x holds the value of 17, a result that holds true under strict verification. Understanding these steps provides a reliable framework for tackling any similar linear equation, ensuring accuracy and confidence in algebraic problem-solving.

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