/*! 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 25 In Exercises \(19-30,\) solve ea... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 25

In Exercises \(19-30,\) solve each system by the addition method. $$ \left\\{\begin{array}{l} 4 x+3 y=15 \\ 2 x-5 y=1 \end{array}\right. $$

Short Answer

Expert verified
The solution to the system of equations is x = 3 and y = 1.

Step by step solution

01

Plan the elimination

First, plan to eliminate one of the variables when the two equations are added together. If 3y from Equation 1 and -5y from Equation 2 could be made equal but opposite, they would cancel out. This can be achieved by multiplying Equation 1 by 5 and Equation 2 by 3.
02

Multiply the equations

Next, multiply Equation 1 by 5 and Equation 2 by 3 to make the coefficients in front of y equal and opposite. This gives: 20x + 15y = 75 and 6x - 15y = 3.
03

Add the two equations

Following on from Step 2, add the two changed equations together. This means 20x and 6x are added together and +15y and -15y are also added together, achieving an equation of: 26x = 78.
04

Solve for x

For solving the value of x, divide both sides of the equation 26x = 78 by 26, leading to: x = 3.
05

Substitute x = 3 into Equation 1

Now that the value of x is known (x = 3), substitute this into the first original equation so that the value can be found for y. Substituting x = 3 into Equation 1 results in: 4*3 + 3y = 15, which simplifies to: 3y = 15 - 12, or: 3y = 3.
06

Solve for y

After substituting x into the first equation, solve the equation for y. This involves dividing both sides by 3 to get: 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.

Addition Method
The addition method helps you solve systems of linear equations effectively. The idea is to organize and manipulate the equations so that one variable gets eliminated when you add them.
For the given system:
  • Start by aligning the equations.
  • Notice that if you add them as they are, no variable will cancel out.
  • So, adjust the equations by multiplication to set them up for elimination.
This adjustment makes the addition method so powerful. It prepares the equations for seamless elimination of one unknown, making it simpler to find the solution.
Elimination Technique
The elimination technique in algebra involves removing one variable from a system of equations. This is done by adding or subtracting equations after aligning their terms properly.
Here's how it works:
  • You need to make the coefficients of one variable opposite in both equations.
  • In this case, multiply Equation 1 by 5 and Equation 2 by 3 to balance the y terms to +15y and -15y.
  • Add these adjusted equations; the y terms will cancel each other, leaving a single-variable equation to solve.
This method highlights the core advantage of elimination: reducing the system to a simpler, single-variable equation.
Solving Linear Equations
Solving linear equations is a fundamental skill in algebra. Once you eliminate one variable, you're left with a linear equation in the form ax = b.
Here's the approach:
  • Simplify the equation if necessary, ensuring only one variable is present.
  • Solve for that variable by isolating it on one side of the equation. For instance, divide both sides by the coefficient of the variable.
In our example, after eliminating y, you're left with the equation 26x = 78. Then simply divide by 26 to get x = 3. Remembering these steps helps with clarity and efficiency in finding the solutions.
Algebraic Methods
Algebraic methods provide various techniques for tackling equations, and systems of equations, including the addition method and elimination technique.
They are vital because they offer:
  • The capacity to handle complex systems in a step-by-step manner.
  • Flexibility, as you can choose the most convenient approach for the problem at hand.
  • A solid foundation for more advanced mathematics.
Understanding these methods builds confidence and competence, enabling easy navigation through algebraic challenges from solving basic linear equations to more intricate system resolutions.

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

write the partial fraction decomposition of each rational expression. $$ \frac{9 x+21}{x^{2}+2 x-15} $$

Members of the group should interview a business executive who is in charge of deciding the product mix for a business. How are production policy decisions made? Are other nethods used in conjunction with linear programming? What are these methods? What sort of academic background, garticularly in mathematics, does this executive have? Present group report addressing these questions, emphasizing the ole of linear programming for the business.

In this exercise, we refer to annual spending per person in 1980 . The combined spending on housing, vehicles/gas, and health care was 7073 dollar. The difference between spending on housing and spending on vehicles/gas was 1247 dollar. The difference between spending on housing and spending on health care was 1466 dollar. Find the average per-person spending on housing, vehicles/gas, and health care in 1980.

write the partial fraction decomposition of each rational expression. $$ \frac{4 x^{2}-5 x-15}{x(x+1)(x-5)} $$

Suppose that you inherit 10,000 dollar. The will states how you must invest the money. Some (or all) of the money must be invested in stocks and bonds. The requirements are that at least 3000 dollar be invested in bonds, with expected returns of 0.08 dollar per dollar, and at least 2000 dollar be invested in stocks, with expected returns of 0.12 dollar per dollar. Because the stocks are medium risk, the final stipulation requires that the investment in bonds should never be less than the investment in stocks. How should the money be invested so as to maximize your expected returns?
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Geometry

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.