Problem 73 - Algebra and Trigonometry

Question

\(71-78\) Find the solution of the equation correct to two decimals. $$ 2.15 x-4.63=x+1.19 $$

Step-by-Step Solution

Answer

The solution is approximately 5.06.

Step 1: Identify like terms

Identify like terms in the equation to isolate the variable terms on one side of the equation and constant terms on the other. The equation is \(2.15x - 4.63 = x + 1.19\). The terms \(2.15x\) and \(x\) are like terms (involving the variable \(x\)), and \(-4.63\) and \(1.19\) are constant terms.

Step 2: Move variable terms to one side

Subtract \(x\) from both sides of the equation to combine variable terms. \(2.15x - x - 4.63 = 1.19\). Simplifying gives: \(1.15x - 4.63 = 1.19\).

Step 3: Isolate the variable term

To isolate the \(x\) term, add \(4.63\) to both sides: \(1.15x - 4.63 + 4.63 = 1.19 + 4.63\). This simplifies to: \(1.15x = 5.82\).

Step 4: Solve for x

Divide both sides of the equation by \(1.15\) to solve for \(x\): \(x = \frac{5.82}{1.15}\). Calculating gives: \(x \approx 5.06\).

Step 5: Round the solution

Since the problem requests the answer correct to two decimal places, the value from the previous calculation is already in this form: Thus, the solution is \(x \approx 5.06\).

Key Concepts

Isolating Variables

When working with linear equations, a crucial step is isolating the variable. This means rearranging the equation such that the variable you're solving for, commonly denoted as \( x \), is on one side of the equation by itself. Here is how you can achieve this:

• First, look at the equation and identify where the variable appears. Sometimes it appears more than once and may be combined into one term.
• Move all the terms containing the variable to one side of the equation and the constant terms to the other.
• To shift terms across the equation, we often use addition or subtraction. You should do the same operation on both sides of the equation to maintain equality.

For example, in the equation \(2.15x - 4.63 = x + 1.19\), isolating the variable involves combining like terms and moving them to one side. You end up with an equation that makes it easy to solve.

Like Terms

Understanding the concept of like terms is essential when dealing with linear equations. Like terms are terms that contain the same variable raised to the same power. They are the portions of an equation that can be combined algebraically.

• For example, in the equation \(2.15x - 4.63 = x + 1.19\), the terms \(2.15x\) and \(x\) are like terms because they both contain the variable \(x\).
• Meanwhile, \(-4.63\) and \(1.19\) are constants and are considered like terms because they contain no variables.

Combining like terms typically means simplifying the equation by subtraction or addition. In our equation, subtracting \(x\) from both sides allowed us to simplify \(2.15x - x\) to \(1.15x\). This kind of simplification helps you move closer to isolating the variable.

Decimal Approximation

In solving equations, decimal approximation is often needed, especially when a problem asks for rounding to a certain number of decimal places. This is particularly true in real-world problems where precise measurements are not always possible.

• For the given problem, once the calculation \(x = \frac{5.82}{1.15}\) is performed, we get an approximate value for \(x\).
• The instruction specifies rounding to two decimal places, which is crucial as it aligns with many practical applications requiring precision.
• Thus, in this example, \(x\approx 5.06\) serves as a decimal approximation of the exact value.

Round accurately by inspecting the third decimal place. If it is 5 or greater, round the second decimal place up. This ensures the approximation is as close to the true value as possible.

Solving for x

The ultimate goal of many equations is to solve for \(x\), the variable of interest. Solving for \(x\) means finally determining its numerical value that satisfies the equation.

• After simplifying and isolating \(x\) using the steps of combining like terms and isolating variables, you're often left with an equation that looks like \(ax = b\).
• In this form, solve by dividing both sides by \(a\), giving \(x = \frac{b}{a}\).
• This division step leads to the solution for \(x\), as seen in the problem where \(x = \frac{5.82}{1.15}\).

Do not forget to perform division carefully, maintaining attention to decimals if necessary. Once solved, check your result by substituting back into the original equation to ensure it holds true, which helps to verify your solution's accuracy.