Problem 32 - Algebra Form and Function

Question

Put the equation in standard form. \(5 b(y+b x+2)=4 b(4-x+2 b),\) with \(b\) constant

Step-by-Step Solution

Answer

Question: Rewrite the equation in its standard form: (5b)(y + bx + 2) = (4b)(4 - x + 2b) Answer: x = \(\frac{-16b + 5by - 10b - 8b^2}{5b^2 - 4b}\)

Step 1: Distribute the terms by using the distributive property with constant\(b\)

Begin by distributing \(5b\) and \(4b\) through their respective parentheses: \((5b)\cdot(y+bx+2) = (4b)\cdot(4-x+2b).\) Multiplying, we have \(5by+5b^2x+10b = 16b - 4bx + 8b^2.\)

Step 2: Group the terms for the variable \(x\)

Now, let's isolate the terms containing the variable \(x\): \(5b^2x - 4bx = -16b + 5by - 10b - 8b^2.\)

Step 3: Factor out x

Next, we can factor out x from the left side of the equation: \(x(5b^2 - 4b) = -16b + 5by - 10b - 8b^2.\)

Step 4: Solve for \(x\) by dividing both sides by \((5b^2 - 4b)\)

Now, we can solve for \(x\) by dividing both sides of the equation by \((5b^2 - 4b)\): \(x = \frac{-16b + 5by - 10b - 8b^2}{5b^2 - 4b}.\) Now the equation is in standard form.

Key Concepts

Distributive Property

The distributive property is a fundamental concept in algebra that helps us to simplify expressions and solve equations more easily. It allows us to "distribute" a factor outside a set of parentheses to each term within the parentheses. In the given equation

• Multiply the constant \( 5b \) by each term inside its parentheses: \( (y + bx + 2) \). This results in: \( 5b(y + bx + 2) = 5by + 5b^2x + 10b \).
• Similarly, multiply \( 4b \) by each term inside its parentheses: \( (4 - x + 2b) \). This results in: \( 4b(4 - x + 2b) = 16b - 4bx + 8b^2 \).

After applying the distributive property, both sides of the equation become simpler expressions, which we can further manipulate to isolate variables or terms. It helps us rearrange and clarify the expression for the next steps of solving the equation.

Factoring

Factoring is the process of breaking down an expression into a product of its simpler terms or components. It can help simplify equations and find solutions for unknown variables by making an equation easier to handle.

In the provided exercise, factoring is used when we reach the step of rearranging the equation to isolate the variable \( x \). We group the terms containing \( x \) which results in:

\( 5b^2x - 4bx \).

Next, we factor out the common factor, \( x \), from these terms. This gives us:

• \( x(5b^2 - 4b) \). This essentially pulls \( x \) out of the product and allows us to isolate or solve for it more easily.

Factoring out \( x \) is crucial as it simplifies the left side of the equation, making it straightforward to further isolate \( x \) or to observe any other needed connections or transformations in the equation.

Equation Manipulation

Equation manipulation involves rearranging and transforming equations to isolate variables, which is a key aspect of solving equations. It often includes using algebraic rules and properties such as the distributive property, as well as factoring, to simplify and handle equations effectively.

In this step-by-step solution, after applying the distributive property and factoring \( x \), we come to the manipulation of isolating \( x \) directly. The steps include:

• Arranging terms around \( x \) gives us \( x(5b^2 - 4b) = -16b + 5by - 10b - 8b^2 \).
• The next goal is to solve for \( x \), which translates to dividing both sides by \( (5b^2 - 4b) \).
• This division isolates \( x \), resulting in the equation: \( x = \frac{-16b + 5by - 10b - 8b^2}{5b^2 - 4b} \).

By strategically manipulating the equation, we turn it into a standard form where \( x \) is expressed in terms of other variables and constants, clarifying its relationship within the expression. Such manipulations are at the core of solving algebraic equations efficiently.