Problem 15 In Exercises \(11-18,\) solve ea... [FREE SOLUTION] (2024)

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Chapter 4: Problem 15

In Exercises \(11-18,\) solve each proportion. $$\frac{y}{15}=\frac{20}{6}$$

Short Answer

Expert verified

y = 50

Step by step solution

01

Identify the Proportion

The first step is to identify the proportion given in the problem. The proportion is: \[\frac{y}{15}=\frac{20}{6}\]

02

Cross-Multiply

To solve the proportion, we will use cross-multiplication. Cross-multiplying the terms, we get: \[6y = 15 \cdot 20\]

03

Simplify the Multiplication

Now, perform the multiplication on the right side of the equation: \[6y = 300\]

04

Solve for y

Next, isolate the variable \(y\) by dividing both sides of the equation by 6: \[y = \frac{300}{6} = 50\]

05

Verify the Solution

Finally, verify the solution by substituting \(y = 50\) back into the original proportion to ensure both sides are equal:\[\frac{50}{15} = \frac{20}{6}\]Simplify both fractions:\[\frac{50}{15} = \frac{10}{3}\quad\text{and}\quad\frac{20}{6} = \frac{10}{3}\]Both sides are equal, so \(y = 50\) is correct.

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cross-Multiplication

To solve proportions like \(\frac{y}{15} = \frac{20}{6}\), we apply a handy technique called cross-multiplication. This method entails multiplying the numerator of one fraction by the denominator of the other fraction. This step is crucial because it transforms the proportion into a more straightforward equation.
For instance, here we multiply \y\ by 6 and 15 by 20, forming the new equation: \[6y = 15 \times 20\]
Through cross-multiplying, we eliminate the fractions making it easier to solve for the variable.

Simplifying Fractions

Upon performing the cross-multiplication, we're left with a new equation \[6y = 300.\]
The right-side multiplication is an example of simplifying fractions. This step involves completing the arithmetic operations indicated on one side of the equation.
It's a great idea to always break down these operations into simpler, manageable parts, ensuring accuracy and ease of understanding.
The multiplication of 15 and 20 simplifies further to 300, making our equation straightforward.

Variable Isolation

The next key step is isolating the variable we need to solve for, which in this exercise is \y.\
After simplifying the fractions to get \[6y = 300\], our goal now is to have \y\ on one side of the equation by itself.
To isolate \y\, we divide both sides of the equation by 6, the coefficient of \y.\
This step looks like this: \[y = \frac{300}{6}\]
Performing the division, we get:
\[y = 50.\]
The variable \y\ is now isolated, and we have our solution!

Verify Solution

After calculating the value of the variable, it is vital to verify the solution. This ensures the steps we followed are correct and the solution is accurate.
We substitute \y = 50\ back into the original proportion:
\(\frac{50}{15} = \frac{20}{6}\)
We then simplify both fractions. Dividing 50 by 15, we get \[ \frac{50}{15} = \frac{10}{3}\]
Similarly, reducing \(\frac{20}{6} = \frac{10}{3}\).
Since the simplified forms of both fractions match, we've confirmed that our solution \y = 50\ is indeed correct.

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Problem 15 In Exercises \(11-18,\) solve ea... [FREE SOLUTION] (31)

Most popular questions from this chapter

Set up an equation or inequality and solve the problem. Be sure to indicateclearly what quantity your variable represents. Round to the nearest tenthwhere necessary. If David walks at the rate of \(5 \mathrm{mph}\) and jogs at the rate of \(9\mathrm{mph}\) and it takes 2 hours to cover a distance of 16 miles, how muchtime was spent jogging?In each of the following exercises, perform the indicated operations. Expressyour answer as a single fraction reduced to lowest terms. $$\frac{11 c}{60 a^{2} b}-\frac{2 c}{45 a b^{3}}$$Solve the equations and inequalities. $$\frac{w+3}{4}+1=\frac{w+4}{3}$$If the exercise is an equation, solve it; if not, perform the indicatedoperations and express your answer as a single fraction. $$\frac{x}{3}+\frac{x}{2}+\frac{x}{5}=62$$Solve each of the following problems algebraically. Be sure to label what thevariable represents. An air freight service charges a \(\$ 10\) pick-up charge plus a \(\$ 4\) perpound delivery charge. If a package is picked up and delivered for a totalcost of \(\$ 28,\) how many pounds did the package weigh?
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Problem 15 In Exercises \(11-18,\) solve ea... [FREE SOLUTION] (2024)

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