Problem 78 - Elementary Algebra

Question

Solve and round off the solutions to the nearest hundredth. $$ (3 x-1)(x+4)=2 x(x+6)-(x-3) $$

Step-by-Step Solution

Answer

The solutions are approximately \( x = 2.65 \) and \( x = -2.65 \).

Step 1: Expand Both Sides

The given equation is \( (3x-1)(x+4) = 2x(x+6) - (x-3) \). First, expand each side separately. Start with the left-hand side: \( (3x-1)(x+4) = 3x^2 + 12x - x - 4 = 3x^2 + 11x - 4 \).Now, expand the right-hand side: \( 2x(x+6) - (x-3) = 2x^2 + 12x - x + 3 = 2x^2 + 11x + 3 \).

Step 2: Move All Terms to One Side

Subtract \( (2x^2 + 11x + 3) \) from both sides of the equation:\( 3x^2 + 11x - 4 - (2x^2 + 11x + 3) = 0 \).Simplify this to get: \( 3x^2 + 11x - 4 - 2x^2 - 11x - 3 = 0 \).

Step 3: Simplify the Equation

Combine like terms:\( 3x^2 - 2x^2 + 11x - 11x - 4 - 3 = 0 \).This simplifies to:\( x^2 - 7 = 0 \).

Step 4: Solve for x

Add 7 to both sides:\( x^2 = 7 \).Take the square root of both sides:\( x = \pm \sqrt{7} \).

Step 5: Round to the Nearest Hundredth

Calculate \( \sqrt{7} \approx 2.645751311 \), therefore:\( x \approx \pm 2.65 \) when rounded to the nearest hundredth.

Key Concepts

Algebraic Expansion

Algebraic expansion is a technique used to simplify expressions and make solving equations more manageable. It involves applying the distributive property to expand products into sums of terms. In the provided exercise, you have two sides of an equation that need expansion.

Let's break down the left-hand side, \( (3x-1)(x+4) \):

• Multiply each term in the first binomial by each term in the second binomial.
• This results in four separate terms: \( 3x \times x = 3x^2 \), \( 3x \times 4 = 12x \), \( -1 \times x = -x \), and \( -1 \times 4 = -4 \).
• Combine these terms to simplify to \( 3x^2 + 12x - x - 4 \).

Next, the right-hand side, \( 2x(x+6) - (x-3) \):

• Expand \( 2x(x+6) \) to get \( 2x^2 + 12x \).
• Distribute the negative sign in \( -(x-3) \) to get \( -x + 3 \).
• Combine to simplify to \( 2x^2 + 11x + 3 \).

The goal of algebraic expansion is to completely eliminate parenthesis so you can work with finished polynomials. It prepares you to move on to solving the equation by putting all terms in a standard form.

Equation Solving

Equation solving is about finding the value of the variable that satisfies the equation. Once you've expanded both sides of the equation, you can begin solving it.

The next step involves rearranging the equation to have all terms on one side, aiming to set the entire equation to zero. This often requires moving terms across the equals sign and ensures that you can solve the equation by isolating variable terms:

• After expansion, move all terms from the right-hand side to the left by subtracting the entire expression: \( 3x^2 + 11x - 4 - (2x^2 + 11x + 3) = 0 \).
• This simplifies to \( x^2 - 7 = 0 \).

Now, with \( x^2 - 7 = 0 \), you're a step closer to finding the solution:

• Add 7 to both sides to isolate the squared term: \( x^2 = 7 \).
• Next, find the square root of both sides to solve for \( x \): \( x = \pm \sqrt{7} \).

Equation solving involves these steps of rearranging and isolating terms to make finding unknowns straightforward. It’s crucial to systematically apply algebraic rules for accuracy.

Rounding Numbers

Rounding numbers is an essential mathematical skill, particularly when your final solutions are not exact numbers but decimals extending infinitely. It makes solutions more manageable and easier to interpret by reducing the number of decimal places.

For the provided exercise, rounding is the final step. After calculating \( \sqrt{7} \), which is approximately 2.645751311, apply rounding rules:

• Identify the place value to which you need to round; here, it's the nearest hundredth.
• The hundredth place is the second digit to the right of the decimal point, where we have the digit 4.
• Look at the next digit (5), which tells you whether to round up or stay; since it's 5, you round up.

Thus, round \( 2.645751311 \) to \( 2.65 \). Rounding not only simplifies results but also maintains the required precision according to the problem's instructions. It is a good practice to know the rules for different rounding scenarios, whether it's tenths, hundredths, or other decimal places.