Question

Rewrite the equation so that \(y\) is a function of \(x .\) $$3(x-2 y)=-12(x+2 y)$$

Short answer

\(y = -\frac{5x}{6}\)

Step-by-step solution

Distribute the terms

We begin by distributing the terms which gives us \(3x - 6y = -12x - 24y\)

Move all terms involving y to one side of the equation and x terms to the other side

By adding \(6y\) to both sides and \(12x\) to both sides we get \(15x = -18y\)

Finally, solve for y

Dividing by \(-18\) gives \(y = -\frac{15x}{18} = -\frac{5x}{6}\)

Key concepts

Distributive Property

Understanding the distributive property is key to restructuring algebraic equations and simplifying complex expressions. It operates on the notion that multiplying a single term by a group of terms enclosed in parentheses requires distributing the multiplication to each term within. This property is expressed mathematically as:

\( a(b + c) = ab + ac \).

In the given exercise, we apply this property to expand the equation \(3(x - 2y) = -12(x + 2y)\). Distributing the coefficients 3 and -12 to both terms within their respective parentheses results in:

\(3 \cdot x - 3 \cdot 2y = -12 \cdot x - 12 \cdot 2y\),
which simplifies to \(3x - 6y = -12x - 24y\). Not only does this step make the equation look tamer, it also sets the stage for easier manipulation as we move to the next steps.

Isolating Variables

Isolating a variable is a fundamental technique in solving linear equations, which involves manipulating the equation to express one variable explicitly in terms of the others. This process usually entails moving terms containing the target variable to one side of the equation and everything else to the opposite side.

In our exercise, we aimed to isolate \(y\). To do this, we performed two main actions:

• We added \(6y\) to both sides to eliminate the \(-6y\) on the left, maintaining the equality of the equation.
• We then added \(12x\) to both sides to move the \(x\)-terms to the left side and have only \(y\)-terms on the right side.

This process yielded \(15x = -18y\), a simpler equation with all \(y\)-terms on one side, perfectly lined up for solving for y, the focus of our equation.

Solving for y

Once we have an equation with the variable of interest on one side—such as \(y\) in the given problem—we can solve for that variable. The goal is to adjust the equation so that it reads \(y = \text{something}\).

In the simplified equation \(15x = -18y\), the next logical step is to get \(y\) on its own. We can do this by dividing both sides of the equation by the coefficient of \(y\), which is -18 in this case. This operation gives us the solution for \(y\) in terms of \(x\):

\(y = \frac{15x}{-18}\).
Simplifying the fraction by dividing both numerator and denominator by their greatest common divisor, which is 3, we finally obtain the functional relationship between \(y\) and \(x\):

\(y = -\frac{5x}{6}\).
This not only provides the answer to the original exercise but also illustrates the power of algebraic manipulation in isolating and solving for a specific variable.