/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 47 Solve each equation. $$ \fra... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 47

Solve each equation. $$ \frac{3 y-2}{5}=\frac{6 y-5}{11} $$

Short Answer

Expert verified
y = -1

Step by step solution

01

Cross Multiply

Multiply both sides of the equation by the denominators to eliminate the fractions. This gives you: [ 11(3y - 2) = 5(6y - 5) ]
02

Distribute

Distribute the numbers outside the parentheses on both sides of the equation: [ 33y - 22 = 30y - 25 ]
03

Gather Like Terms

Subtract 30y from both sides to get all the variable terms on one side: [ 33y - 30y - 22 = -25 ] This simplifies to: [ 3y - 22 = -25 ]
04

Solve for y

Add 22 to both sides to isolate the variable term: [ 3y - 22 + 22 = -25 + 22 ] This simplifies to: [ 3y = -3 ] Finally, divide by 3 to solve for y: [ y = -1 ]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Cross Multiplication
When solving rational equations, the first step is often cross multiplication. This involves multiplying the numerator of each fraction by the denominator of the other fraction. This helps eliminate the fractions and simplifies the equation.
In the given exercise, the equation is:
\(\frac{3 y - 2}{5} = \frac{6 y - 5}{11}\)
To cross-multiply, we take the numerator of the left fraction (\(3y - 2\)) and multiply it by the denominator of the right fraction (11). Similarly, we multiply the numerator of the right fraction (\(6y - 5\)) by the denominator of the left fraction (5).
This leads us to:
\[ 11(3y - 2) = 5(6y - 5) \]
Distribution
Distribution is the next step after cross multiplication. It's essential for eliminating parentheses and combining terms. In our example, we need to distribute the numbers outside the parentheses across the terms within the parentheses.
Specifically:
  • Distribute 11 across \(3y - 2\): \[ 11 \times 3y - 11 \times 2 = 33y - 22 \]
  • Distribute 5 across \(6y - 5\): \[ 5 \times 6y - 5 \times 5 = 30y - 25 \]
So, after distributing, the equation looks like this:
\[ 33y - 22 = 30y - 25 \]
Combining Like Terms
Combining like terms means bringing together similar variable terms or constant terms. This often simplifies the equation greatly, making it easier to solve.
In the given equation:
\[ 33y - 22 = 30y - 25 \]
We want to get all the y terms on one side and constant terms on the other.
  • Subtract \(30y\) from both sides to get: \(33y - 30y - 22 = -25\)
  • This simplifies to: \(3y - 22 = -25\)
Isolating Variables
The final step involves isolating the variable, which means getting the variable alone on one side of the equation. This helps determine its value.
In the simplified equation:
\[ 3y - 22 = -25 \]
We need to isolate y. Start by adding 22 to both sides:
  • Add 22 to both sides: \(3y - 22 + 22 = -25 + 22\)
  • This simplifies to: \(3y = -3\)
Now, divide both sides by 3 to solve for y:
\( y = \frac{-3}{3} = -1 \)
Thus, the solution to the equation is:
\( y = -1 \)

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Solve each formula for the specified variable. \(S=\frac{1}{2}(a+b+c)\) for \(b\)

Find the value of the remaining variable in each formula. Use 3.14 as an approximation for \(\pi(p i) .\) \(\mathscr{A}=\frac{1}{2} b h(\) area of a triangle \() ; \quad b=8, h=16\)

Graph the solution set, and write it using interval notation \(1 \leq 3+\frac{2}{3} p \leq 7\)

In one day, a store sold \(\frac{2}{3}\) as many DVDs as Blu-ray discs. The total number of DVDs and Blu-ray discs sold that day was \(280 .\) How many DVDs were sold?

Solve each inequality. Graph the solution set, and write it using interval notation. \(-4<-2 x<12\)
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.