Problem 33 - Algebra for College Students

Question

For Problems \(31-44\), solve each equation for the indicated variable. $$ \frac{-2}{x-4}=\frac{5}{y-1} \text { for } y $$

Step-by-Step Solution

Answer

The solution for \( y \) is \( y = -\frac{5}{2}x + 11 \).

Step 1: Cross-multiply to clear fractions

To solve for \( y \), we'll start by cross-multiplying to eliminate the fractions. Multiply both sides of the equation by the products of the denominators: \((x-4)(y-1)\). This yields: \[-2(y - 1) = 5(x - 4)\].

Step 2: Distribute the terms

Now distribute the terms on both sides of the equation:On the left: \(-2(y - 1) = -2y + 2\).On the right: \(5(x - 4) = 5x - 20\).This gives us the equation:\(-2y + 2 = 5x - 20\).

Step 3: Solve for \( y \)

To isolate \( y \), first subtract 2 from both sides:\(-2y = 5x - 20 - 2\)which simplifies to:\(-2y = 5x - 22\).Now divide every term by -2 to solve for \( y \):\[y = -\frac{5x - 22}{2}\]which can be rearranged to:\[y = -\frac{5}{2}x + 11\].

Key Concepts

Cross-Multiplication

Solving equations with fractions on both sides can be tricky. One powerful tool to handle these kinds of problems is cross-multiplication. By using cross-multiplication, we can efficiently get rid of fractions from equations. Here's how it works.

• We take the numerator of one fraction and multiply it by the denominator of the other fraction.
• Repeat this process with the remaining numerator and denominator.

This approach transforms an equation with fractions into a standard equation without fractions, making it easier to solve. For example, in our exercise, we started with:\[\frac{-2}{x-4} = \frac{5}{y-1}\] By cross-multiplying, we got: \[-2(y - 1) = 5(x - 4)\] Now, we are ready to simplify further. Cross-multiplication can really simplify an equation by removing the complexity of fractions.

Isolating Variables

A key strategy in solving equations is isolating the variable you want to solve for. When you isolate a variable, you make it the subject of the formula. This means getting the variable on one side of the equation and everything else on the other. Here's the process:

• First, perform operations that will simplify the equation, such as distributing terms.
• Next, add or subtract terms on both sides of the equation to position the variable on one side.
• Then, if needed, multiply or divide both sides to leave your desired variable by itself.

This strategy allows you to directly determine the value of the variable. In our example, once we had:\[-2(y - 1) = 5(x - 4)\] We isolated \( y \) by first distributing, then subtracting 2, leading to:\[-2y = 5x - 22\]Finally, dividing by \(-2\) resulted in:\[y = -\frac{5x - 22}{2}\] Isolating variables is a fundamental skill that simplifies the equation and helps one find the solution more straightforwardly.

Algebraic Manipulation

Algebraic manipulation involves performing various operations on an equation to simplify or rearrange it. This is essential when solving equations to clarify or break down complex expressions into simpler forms. Some common techniques include:

• Distributing multiplication over addition or subtraction, known as the distributive property.
• Combining like terms to simplify expressions.
• Using inverse operations to undo addition, subtraction, multiplication, or division performed on a variable.

In our example, after cross-multiplying:\[-2(y - 1) = 5(x - 4)\],we used the distributive property:\[-2y + 2 = 5x - 20\].Later, we subtracted and rearranged terms to further simplify:\[-2y = 5x - 22\].Finally, we divided by \(-2\) to express \(y\) in terms of \(x\). Each step in algebraic manipulation helps transform the equation and bring us closer to solving it by making adjustments that lead to a clear solution for the variable. By understanding these core operations, you can more easily solve a wide range of algebraic equations.