Problem 72 - Algebra and Trigonometry

Question

\(71-78\) Find the solution of the equation correct to two decimals. $$ 8.36-0.95 x=9.97 $$

Step-by-Step Solution

Answer

The solution is approximately \(-1.69\).

Step 1: Isolate the Variable Term

First, we need to isolate the term containing the variable \(x\). To do this, subtract \(8.36\) from both sides of the equation: \ 8.36 - 0.95x - 8.36 = 9.97 - 8.36 \ This simplifies to: \ -0.95x = 1.61

Step 2: Solve for the Variable

Now, we need to solve for \(x\). Divide both sides of the equation by \(-0.95\): \ \(-0.95x = 1.61\) \ \(x = \frac{1.61}{-0.95}\) \ Calculating this division gives \(x \approx -1.6947368421\).

Step 3: Round to Two Decimal Places

Round the solution to two decimal places. The number we have is approximately \(-1.6947368421\), so rounding to two decimal places gives \(-1.69\).

Key Concepts

Decimal Approximation

When working with solutions like the one given in the exercise, complete accuracy is often not feasible due to the nature of decimal numbers. Decimal approximation comes into play to provide a manageable number that is close to the exact solution. In our case, the solution to the equation was calculated to be approximately \[ x \approx -1.6947368421 \]This long decimal reflects greater precision, but it can be cumbersome in practice. In real-world applications, we use an approximation to simplify our calculations or present results more clearly. Approximating decimals involves reducing the number of digits beyond the decimal point while maintaining a value that's acceptably close to the actual figure. Thus,- A shorter, approximate result is often easier to communicate.- It enables simpler calculations in further steps.Approximation helps ensure the result is both understandable and practically usable, especially when using decimals.

Rounding Numbers

Rounding is the process of eliminating decimal places to achieve a specified level of precision. This operation helps make numbers simpler and more understandable, often without significantly affecting the accuracy of calculations. Rounding involves looking at the digit immediately after the desired decimal place and adjusting accordingly.In the given exercise, after determining the approximate solution \[ x \approx -1.6947368421 \]we need to round it to two decimal places. You focus on the third decimal digit, which is 4 in this case:- If this intermediate digit is less than 5, you round down, keeping the last desired decimal unchanged.- If it's 5 or more, you round up, increasing the last desired digit by one.Here, since the intermediate number is 4, we round down, resulting in:\[ x \approx -1.69 \]This rounded figure ensures clarity and precision for practical use in further calculations or presentations.

Isolating Variables

Isolating the variable in a linear equation is a fundamental skill in algebra. It involves manipulating the equation so that the variable stands alone on one side, typically the left. This step is crucial because it simplifies the process of solving the equation.To isolate the variable, follow these general steps:- Identify the term containing the variable.- Move other terms to the opposite side of the equation by using operations like addition or subtraction.- In the given exercise, subtract the constant 8.36 from both sides to get:\[ -0.95x = 1.61 \]At this stage, the term -0.95x is isolated. Finish by dividing by the coefficient of x, which is -0.95 in this example:\[ x = \frac{1.61}{-0.95} \]This technique simplifies solving the equation, allowing you to find the value of the variable efficiently.