Problem 9 - Algebra and Trigonometry with Analytic Geometry

Question

Solve the equation. $$0.3(3+2 x)+1.2 x=3.2$$

Step-by-Step Solution

Answer

The solution is approximately \(x = 1.28\).

Step 1: Distribute the 0.3

Distribute 0.3 to each term inside the parentheses. The equation becomes: \(0.3 \times 3 + 0.3 \times 2x = 0.9 + 0.6x\). This simplifies the equation to: \(0.9 + 0.6x + 1.2x = 3.2\).

Step 2: Combine like terms

Combine the terms involving \(x\). So, \(0.6x + 1.2x\) becomes \(1.8x\). Substitute this back into the equation: \(0.9 + 1.8x = 3.2\).

Step 3: Isolate the term with variable

Subtract 0.9 from both sides of the equation to isolate the term with \(x\): \(1.8x = 3.2 - 0.9\). This simplifies to \(1.8x = 2.3\).

Step 4: Solve for x

Divide both sides of the equation by 1.8 to solve for \(x\). So, \(x = \frac{2.3}{1.8}\). Compute the division to find \(x = 1.2777...\) (repeating).

Key Concepts

Distributive Property

The distributive property is an essential concept when solving equations, especially when dealing with parentheses. The property allows you to multiply a sum or difference within the parentheses by a number outside of them. This helps to eliminate parentheses and simplify the equation. For example, in the equation \(0.3(3 + 2x)\), the distributive property is used to multiply \(0.3\) with each term inside the parentheses. This results in two separate terms: \(0.3 \times 3\) and \(0.3 \times 2x\). After performing the calculations, you'll have a simplified form as \(0.9 + 0.6x\). Distributing correctly is crucial as errors here can lead to incorrect solutions later in the process. Remember, break down each term separately and always double-check your multiplication to avoid mistakes.

Combining Like Terms

Once the distributive property has been applied, and the equation is free from parentheses, the next step often involves **combining like terms**. Like terms are terms that have the same variable raised to the same power. In our exercise, we have the terms \(0.6x\) and \(1.2x\). These terms are 'like' because they both contain the variable \(x\). By adding them together, we simplify the expression: \(0.6x + 1.2x = 1.8x\).

• Combining like terms reduces the complexity of the equation.
• It helps to consolidate similar elements, making it easier to solve the equation next.

Make sure to only combine terms that are precisely like each other. Also, handle coefficients accurately to avoid errors.

Isolating Variables

The goal when solving linear equations is often to isolate the variable on one side of the equation. In our example, we want to get \(x\) by itself. To do this, you need to perform operations that will move other numbers to the opposite side of the equation from the variable. Here, we start with \(1.8x = 3.2 - 0.9\).
Subtract \(0.9\) from \(3.2\) to further simplify the equation to \(1.8x = 2.3\).

• Remember that whatever operation you apply to one side, you must apply to the other side too, to keep the equation balanced.
• Keep the variable positive, often making the equation easier to solve.

By isolating the variable, you're setting yourself up for the final step, which is solving for the variable itself. It is like peeling away layers until you're left with a straightforward equation involving the variable and a constant.

Division in Equations

When it comes to actually solving the equation for the variable, division plays a crucial role if the variable is multiplied by a coefficient. After isolating the variable in the equation \(1.8x = 2.3\), you need to divide both sides by \(1.8\) to solve for \(x\). This means you are performing the operation \(x = \frac{2.3}{1.8}\).
Perform the division to find the exact value of \(x\), which is approximately \(1.2777\) when rounded to four decimal places.

• Always divide by the coefficient of the variable to complete the solution process.
• Keep track of decimal places, especially if a precise answer is required.
• Check your work by plugging the value of \(x\) back into the original equation.

This step finalizes the solving process by giving you the value for \(x\), confirming how other steps prepared the equation for this conclusive operation.