Problem 24 - Algebra 1

Question

Solve the proportion. Check for extraneous solutions. $$\frac{r+4}{3}=\frac{r}{5}$$

Step-by-Step Solution

Answer

Consequently, the solution to the proportion is \(r = -10\).

Step 1: Cross Multiply

To solve the proportion, that is \(\frac{r+4}{3}=\frac{r}{5}\), use the cross multiplication method. Multiply \(3*r\) and \(5*(r+4)\), which results in \(3r = 5r + 20\).

Step 2: Solve the equation

Subtract \(5r\) on both sides of \(3r = 5r + 20\), get \(-2r = 20\). Then, divide both sides by -2 to solve for 'r'. Consequently, \(r = -10\).

Step 3: Verify the Solution

Replace 'r' by -10 into the original equation: Check if \(\frac{-10+4}{3} = \frac{-10}{5}\). By simplifying both sides, we get \(-2 = -2\), which means that the two sides of the original equation are equal, therefore -10 is not an extraneous solution.

Key Concepts

Understanding Cross Multiplication in Proportions

Cross multiplication is like a magic trick when dealing with proportions. It helps us solve equations quickly and efficiently. A proportion is an equation stating that two ratios are equal. For example, if you have \( \frac{a}{b} = \frac{c}{d} \), you can use cross multiplication. Here's how it works:

• Multiply the numerator of the first ratio by the denominator of the second ratio.
• Do the same with the other numerator and denominator pair.

So it looks like: \( a \times d = b \times c \).
In the given exercise, \( \frac{r+4}{3} = \frac{r}{5} \), applying cross multiplication means calculating \( 3 \times r = 5 \times (r+4) \). This simplifies the problem by turning a proportion into an equation that can be more easily solved.

Solving Equations Step by Step

Once you have used cross multiplication, the next step is to solve the equation you've created. Let's take the equation from our example: \( 3r = 5r + 20 \). The goal is to get 'r' all alone on one side. Follow these steps:

• Subtract \( 5r \) from both sides to isolate 'r'. This gives you \( 3r - 5r = 20 \).
• Simplify to get \(-2r = 20\).
• To solve for 'r', divide both sides by the coefficient of 'r', which is -2.

After dividing, you find \( r = -10 \). Making sure each step is clear and simple will help you solve such equations successfully.

Identifying Extraneous Solutions

Extraneous solutions are results that come up during the solving process but don't actually work when substituted back into the original equation. This happens because certain operations, like squaring or multiplying, can introduce solutions that weren't there originally.
In our example, you found \( r = -10 \). To determine if it's extraneous, substitute \( r \) back into the original proportion: \( \frac{-10+4}{3} = \frac{-10}{5} \). Simplifying both sides gives: \( \frac{-6}{3} = \frac{-10}{5} \), or \( -2 = -2 \).
Since both sides match perfectly, \( r = -10 \) is a valid solution and not extraneous. Checking your work is crucial because it confirms that the solution solves the actual problem without being misled by extraneous results.