Problem 54 - Intermediate Algebra

Question

Solve each proportion. $$ \frac{2 y+6}{3}=\frac{4 y-16}{5} $$

Step-by-Step Solution

Answer

The solution is \( y = 39 \).

Step 1: Cross-Multiply

Start by cross-multiplying the two fractions. This involves multiplying the numerator of the first fraction by the denominator of the second fraction, and the numerator of the second fraction by the denominator of the first fraction. This gives:\[ 5(2y + 6) = 3(4y - 16) \]

Step 2: Distribute Multiplication

Now, distribute the multiplication over the terms in each parenthesis:\[ 5 \cdot 2y + 5 \cdot 6 = 3 \cdot 4y - 3 \cdot 16 \]Simplifying each term yields:\[ 10y + 30 = 12y - 48 \]

Step 3: Move Terms Involving y

To isolate terms with \(y\) on one side, subtract \(10y\) from both sides:\[ 30 = 2y - 48 \]

Step 4: Move Constant Terms

Add 48 to both sides to move constants to one side:\[ 30 + 48 = 2y \]Simplifying gives:\[ 78 = 2y \]

Step 5: Solve for y

Divide both sides by 2 to solve for \(y\):\[ y = \frac{78}{2} \]This simplifies to:\[ y = 39 \]

Step 6: Check the Solution

Substitute \(y = 39\) back into the original equation to verify the solution:\[ \frac{2(39) + 6}{3} = \frac{4(39) - 16}{5} \]This gives:\[ \frac{78 + 6}{3} = \frac{156 - 16}{5} \]Simplifying both sides yields:\[ \frac{84}{3} = \frac{140}{5} \]Both sides simplify further to 28, confirming the solution is correct.

Key Concepts

Cross-Multiplication in Proportions

Cross-multiplication is a fundamental concept to solve equations involving two equal fractions, known as proportions. When we have a proportion such as \( \frac{2y+6}{3} = \frac{4y-16}{5} \), cross-multiplication helps us to eliminate the fractions. This is done by multiplying the numerator of one fraction by the denominator of the other fraction and setting the two products as equal.
Here's how it works:

• Multiply \(2y + 6\) by 5 (the denominator of the second fraction).
• Multiply \(4y - 16\) by 3 (the denominator of the first fraction).

The resulting equation, \(5(2y + 6) = 3(4y - 16)\), is without fractions, simplifying the solving process for the unknown variable \(y\). This method of cross-multiplication is especially useful because it transforms a proportion problem into a simpler linear equation.

Distributive Property in Algebra

The distributive property is crucial in simplifying expressions, especially when handling equations obtained from cross-multiplication. For our problem \( 5(2y + 6) = 3(4y - 16) \), you must distribute the multiplication across each term inside the parentheses.
Applying the distributive property in this example involves:

• Multiplying 5 by both \(2y\) and 6 to get \(10y + 30\).
• Multiplying 3 by both \(4y\) and -16, resulting in \(12y - 48\).

After distribution, we convert the equation into \(10y + 30 = 12y - 48\). The distributive property simplifies complex equations by breaking them into manageable parts, making it easier to find solutions.

Solving Algebraic Equations

Once the expression is simplified using distribution, solving the algebraic equation involves organizing and isolating the variable. From \(10y + 30 = 12y - 48\), the steps to find \(y\) include:

• First, move all terms involving \(y\) to one side. You subtract \(10y\) from both sides, giving \(30 = 2y - 48\).
• Next, move constant terms to the opposite side. Adding 48 to both sides results in \(78 = 2y\).

Finally, divide by 2 to isolate \(y\), leading to \(y = 39\). These systematic steps in solving algebraic equations help in managing and simplifying the problem to reach the desired value.

Verifying Solutions in Proportions

After determining a solution in a math problem, verifying your result is essential to ensure accuracy. Substituting back \(y = 39\) into the original proportion \( \frac{2y+6}{3} = \frac{4y-16}{5} \) lets you check correctness:

• Replace every \(y\) with 39, resulting in \(\frac{2(39) + 6}{3} = \frac{4(39) - 16}{5}\).
• Calculate: \(\frac{84}{3} = \frac{140}{5}\).

Both sides simplify to 28, confirming the equality and validating that \(y = 39\) is correct. This verification step assures the solution satisfies the original equation, avoiding mistakes in problem-solving.