Problem 78 - Algebra and Trigonometry

Question

Find the solution of the equation rounded to two decimals. \(3.95-x=2.32 x+2.00\)

Step-by-Step Solution

Answer

The solution is approximately x = 0.59.

Step 1: Isolate x

Start by bringing all the terms with \(x\) to one side of the equation. Add \(x\) to both sides to group terms with \(x\) together: \[3.95 = 2.32x + 2.00 + x\]This simplifies to \[3.95 = 3.32x + 2.00\]

Step 2: Simplify the equation

Subtract 2.00 from both sides to get rid of the constant term on the side with \(x\):\[3.95 - 2.00 = 3.32x\]This results in:\[1.95 = 3.32x\]

Step 3: Solve for x

Divide both sides by 3.32 to solve for \(x\):\[x = \frac{1.95}{3.32}\]Calculate the division to find \(x\):\[x \approx 0.587\]

Step 4: Round to Two Decimal Places

Round the solution \(x \approx 0.587\) to two decimal places:\[x \approx 0.59\]

Key Concepts

Variable Isolation

In algebra, variable isolation is a foundational skill. It means rearranging an equation to have the variable on one side of the equation and constants on the other. For example, consider the equation given in the exercise, \(3.95 - x = 2.32x + 2.00\). To isolate \(x\), you need all the \(x\)-terms on one side. To accomplish this, start by adding \(x\) to both sides. This step adjusts the equation as \(3.95 = 2.32x + 2.00 + x\). Notice how the \(x\) initiates grouping responsibility on one side. Simplifying the terms \(2.32x + x\) gives \(3.32x\), leading to the simplified appearance: \(3.95 = 3.32x + 2.00\). By strategically handling the variable \(x\) and associated terms, variable isolation is achieved successfully.

Equation Solving

Equation solving involves the process of finding the value of the variable that makes the equation true. Once you have isolated the variable as further demonstrated with \(3.95 = 3.32x + 2.00\), the next target is simplifying further to pinpoint \(x\). Begin by subtracting 2.00 from both sides: \(3.95 - 2.00 = 3.32x\). This transformation leads to \(1.95 = 3.32x\). Finally, solve for \(x\) by dividing both sides by 3.32, yielding \(x = \frac{1.95}{3.32}\). Solving provides \(x \approx 0.587\). Through these strategic actions via addition, subtraction, and division, equation solving clears pathways to the variable's exact value. Patience and accuracy are key in these steps.

Rounding Numbers

After solving the equation and obtaining the variable value as a decimal, often you need to round it to a specified degree of precision. In this example, \(x \approx 0.587\) needs rounding to two decimal places. For rounding, observe the digit immediately after the second decimal place. If it is 5 or greater, increase the second decimal digit by one. Otherwise, leave it unchanged. Here, since the third place is 7 (greater than 5), \(0.587\) rounds to \(0.59\). Rounding numbers is essential for simplification in reporting and can often enhance clarity and communicate precision in your results.