Solving Proportions Homework Solutions And Explanations

This article delves into the fascinating world of proportions and how to solve for unknown variables within them. We will tackle three distinct proportion problems, providing step-by-step solutions and clear explanations to enhance your understanding. Mastering proportions is crucial in various mathematical and real-world applications, from scaling recipes to understanding financial ratios. We'll break down each problem, highlighting the key concepts and techniques involved. So, let's embark on this mathematical journey and conquer these proportion challenges!

Question 1: Solving for 'n' in a Proportion

Our first challenge involves solving for the variable 'n' in the proportion:

rac{2}{n+1}=rac{9}{6}

To solve this proportion, we employ the fundamental principle of cross-multiplication. This principle states that if rac{a}{b} = rac{c}{d}, then $ad = bc$. Applying this to our problem, we multiply the numerator of the first fraction by the denominator of the second fraction and vice-versa. This gives us:

$2 * 6 = 9 * (n + 1)$

This simplifies to:

$12 = 9(n + 1)$

Now, we need to isolate 'n'. We begin by distributing the 9 on the right side of the equation:

$12 = 9n + 9$

Next, we subtract 9 from both sides of the equation to further isolate the term containing 'n':

$12 - 9 = 9n + 9 - 9$

This simplifies to:

$3 = 9n$

Finally, to solve for 'n', we divide both sides of the equation by 9:

rac{3}{9} = rac{9n}{9}

This results in:

n = rac{1}{3}

Therefore, the solution to the first proportion is n = rac{1}{3}. This means that when 'n' is equal to one-third, the two fractions in the original proportion are equivalent. To verify our solution, we can substitute rac{1}{3} back into the original equation:

rac{2}{rac{1}{3} + 1} = rac{9}{6}

Simplifying the left side, we get:

rac{2}{rac{4}{3}} = rac{9}{6}

Which further simplifies to:

rac{2 * 3}{4} = rac{9}{6}

rac{6}{4} = rac{9}{6}

Both fractions simplify to rac{3}{2}, confirming that our solution for 'n' is correct. Understanding the process of solving proportions is essential for many mathematical applications, and this example provides a solid foundation for tackling more complex problems. The ability to manipulate equations and isolate variables is a key skill in algebra and beyond.

Question 2: Solving for 'r' with Variables in Numerators

Our second question presents a proportion where the variable 'r' appears in both numerators:

rac{r}{5}=rac{r-8}{10}

Similar to the first question, we utilize the principle of cross-multiplication to solve for 'r'. Cross-multiplying gives us:

$10 * r = 5 * (r - 8)$

This simplifies to:

$10r = 5(r - 8)$

Now, we distribute the 5 on the right side of the equation:

$10r = 5r - 40$

To isolate 'r', we subtract 5r from both sides of the equation:

$10r - 5r = 5r - 40 - 5r$

This simplifies to:

$5r = -40$

Finally, we divide both sides of the equation by 5 to solve for 'r':

rac{5r}{5} = rac{-40}{5}

This results in:

$r = -8$

Therefore, the solution to the second proportion is $r = -8$. Substituting this value back into the original equation to verify our solution:

rac{-8}{5} = rac{-8 - 8}{10}

Simplifying the right side, we get:

rac{-8}{5} = rac{-16}{10}

Both fractions simplify to rac{-8}{5}, confirming our solution. This example highlights how to handle proportions with variables in multiple terms. The crucial step is to carefully distribute and combine like terms to isolate the variable. This process is fundamental to solving various algebraic equations and is a valuable skill in mathematics.

Question 3: Solving for 'x' with Variables in Both Numerators and Denominators

Our third and final question presents a slightly more complex proportion, with the variable 'x' appearing in both numerators and denominators:

rac{9}{5}=rac{x-10}{x+4}

Again, we begin by applying the principle of cross-multiplication:

$9 * (x + 4) = 5 * (x - 10)$

This simplifies to:

$9(x + 4) = 5(x - 10)$

Now, we distribute on both sides of the equation:

$9x + 36 = 5x - 50$

To isolate 'x', we first subtract 5x from both sides:

$9x + 36 - 5x = 5x - 50 - 5x$

This simplifies to:

$4x + 36 = -50$

Next, we subtract 36 from both sides:

$4x + 36 - 36 = -50 - 36$

This simplifies to:

$4x = -86$

Finally, we divide both sides by 4 to solve for 'x':

rac{4x}{4} = rac{-86}{4}

This results in:

x = -rac{43}{2} or $x = -21.5$

Therefore, the solution to the third proportion is x = -rac{43}{2}. To verify this solution, we substitute it back into the original equation:

rac{9}{5} = rac{-rac{43}{2} - 10}{-rac{43}{2} + 4}

Simplifying the right side, we need to find a common denominator:

rac{9}{5} = rac{-rac{43}{2} - rac{20}{2}}{-rac{43}{2} + rac{8}{2}}

This becomes:

rac{9}{5} = rac{-rac{63}{2}}{-rac{35}{2}}

Dividing by a fraction is the same as multiplying by its reciprocal:

rac{9}{5} = -rac{63}{2} * -rac{2}{35}

Simplifying, we get:

rac{9}{5} = rac{63}{35}

Both fractions simplify to rac{9}{5}, confirming our solution. This final example showcases the importance of careful distribution, combining like terms, and handling fractions when solving proportions. The ability to solve complex proportions like this demonstrates a strong understanding of algebraic principles.

In conclusion, we have successfully solved three proportion problems, each presenting unique challenges. We've utilized the principle of cross-multiplication, distribution, and simplification to isolate the variables and find their values. Mastering these techniques is crucial for success in algebra and other mathematical disciplines. Remember to always verify your solutions by substituting them back into the original equation. Practice is key to solidifying your understanding of proportions and improving your problem-solving skills. Keep exploring the fascinating world of mathematics, and you'll be amazed at what you can achieve!