Problem 12 - Algebra 2 and Trigonometry

Question

Solve each proportion for the variable. \(\frac{x-1}{15}=\frac{2}{5}\)

Step-by-Step Solution

Answer

The solution is \(x = 7\).

Step 1: Cross-Multiply

To solve the proportion \(\frac{x-1}{15}=\frac{2}{5}\), start by cross-multiplying the terms. Multiply \((x-1)\) by 5 and 15 by 2, resulting in the equation:\[ 5(x-1) = 2 \times 15 \]

Step 2: Simplify the Right Side

Next, simplify the right side of the equation. Calculate \(2 \times 15\):\[ 5(x-1) = 30 \]

Step 3: Distribute the 5

Distribute the 5 to both terms inside the parenthesis on the left side:\[ 5x - 5 = 30 \]

Step 4: Isolate x

To isolate \(x\), add 5 to both sides of the equation to eliminate the \(-5\):\[ 5x = 35 \]

Step 5: Solve for x

Divide both sides by 5 to solve for \(x\):\[ x = \frac{35}{5} \]Calculate the division:\[ x = 7 \]

Key Concepts

Cross-Multiplication

In proportion problems, one common way to solve is using cross-multiplication. This technique involves multiplying diagonally across the equal sign in a fraction equation. Let's break down the process:
* First, identify the two ratios in the equation. In this example, it is \(\frac{x-1}{15} = \frac{2}{5}\).
* Cross-multiply by multiplying the numerator of the first ratio by the denominator of the second ratio. Then, multiply the numerator of the second ratio by the denominator of the first ratio.
* For our exercise, we multiply \(x-1\) by 5 and 15 by 2, resulting in the equation: \[ 5(x-1) = 2 \times 15 \]
This method converts the proportion into an equation without fractions, making it easier to solve for the variable. Remember, cross-multiplication is only valid when you have one ratio set equal to another ratio.

Solving for Variables

Once we've applied cross-multiplication, the goal is to solve for the variable, in this case, \(x\). Here's how we continue from our proportion:
* After cross-multiplying, simplify the equation if needed. For example, \(2 \times 15\) simplifies to 30, giving us: \[ 5(x-1) = 30 \]
* The next step is to isolate \(x\). Start by distributing or rearranging terms, eventually using addition or subtraction to get \(x\) alone on one side of the equation.
* With our equation, distributing gives us: \[ 5x - 5 = 30 \] Then, add 5 to both sides: \[ 5x = 35 \]
* Finally, divide by the coefficient of \(x\) to solve for it. Here, divide both sides by 5: \[ x = \frac{35}{5} = 7 \]
Step by step, adjusting each part of the equation helps make \(x\) clear and isolate it for the final solution.

Distributive Property

The distributive property is a handy mathematical tool used in solving equations. It helps to clear out parentheses by distributing a multiplier across terms within the parentheses.
* For the equation \(5(x-1) = 30\), apply the distributive property by multiplying 5 with each term within the parentheses: \(5 \times x\) and \(5 \times -1\). Thus, the equation becomes \[ 5x - 5 = 30 \].
* This property is crucial for simplifying equations, especially after cross-multiplication, where parentheses often appear.
* It helps us convert the equation into a simpler form, preparing us to isolate the variable.
The distributive property ensures that every part of an expression is addressed, allowing for further steps in solving complex equations efficiently. Always distribute first before attempting to combine or eliminate any further terms on either side of the equation.