/*! 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 34 Write each system of equations i... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Write each system of equations in the form \(\left\\{\begin{array}{l}A x+B y=E \\\ C x+D y=F\end{array}\right.\) and then solve the system. $$\left\\{\begin{array}{c} \frac{x+1}{2}+\frac{y-1}{3}=1 \\ 3 x+y=7 \end{array}\right.$$

Short Answer

Expert verified
The solution to the system of equations is \(x = 1\) and \(y = 4\).

Step by step solution

01

Reformat the First Equation

First, focus on the equation \(\frac{x + 1}{2} + \frac{y - 1}{3} = 1\). Multiply the terms on the left side by the denominator to remove the fractions. This results in \(x + 1 + y - 1 = 2 + 3\). Simplify that to \(x + y = 5\) which is in the form \(Ax + By = E\). An important note is that simplifying the equation helps in reducing errors in future calculations.
02

Reformat the Second Equation

Next, focus on the second equation \(3x + y = 7\). This is already in the standard form, so no changes are necessary. So, \(Cx + Dy = F\), where \(C = 3\), \(D = 1\), and \(F = 7\). This means that the system in standard form is: \(\left\{\begin{array}{l}x + y = 5\ 3x + y = 7\end{array}\right.\)
03

Solving the System

To solve this system, you can subtract the first equation from the second to eliminate \(y\), which results in \(2x = 2\). Solving for \(x\), find that \(x = 1\). Substituting \(x = 1\) into the first equation results in \(1 + y = 5\). Solving for \(y\), you find that \(y = 4\). So, the solution to the system of equations is \(x = 1\) and \(y = 4\).

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 91Ó°ÊÓ!

Key Concepts

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

Linear Equations
Linear equations are fundamental components of algebra that express a relationship in which each term is either a constant or the product of a constant and a single variable. These equations can graphically be represented as straight lines on a coordinate plane, where the slope and y-intercept define the line's direction and position, respectively.

That is, a linear equation in two variables can be written as \( Ax + By = C \), where \( A \), \( B \), and \( C \) are constants. Systems of linear equations, like the one given in the exercise, consist of two or more linear equations with the same variables. Solving such systems is about finding the values of the variables that satisfy all the equations simultaneously.
Algebraic Methods
Algebraic methods involve techniques to manipulate and solve equations or systems of equations. One of these methods, used in the step by step solution, includes clearing fractions by multiplying both sides of the equation by a common denominator. Doing this simplifies the equation and makes it easier to work with.

Another foundational technique involves rearranging terms to isolate variables, such as adding or subtracting terms on both sides of an equation, or using the distributive property to expand expressions. The step by step solution began by applying this method to reformat the given equations into a more manageable form, making the subsequent steps of finding the variable values more straightforward.
Elimination Method
The elimination method is one of several strategies to solve systems of linear equations. It involves combining equations with the goal of eliminating one of the variables. To do this, one can add, subtract, or sometimes multiply the equations by certain coefficients.

In the given solution, the elimination method is used after reformulating the equations into the standard form. By subtracting the first equation from the second, the variable \( y \) is eliminated, allowing to directly solve for \( x \). Once \( x \) is known, it is substituted back into one of the original equations to find the value of \( y \). The elimination method is favored by many due to its straightforward process and ability to effectively deal with systems that have multiple equations.
Fractions in Linear Equations
Dealing with fractions in linear equations can make the process of solving the equations more challenging. However, a solid approach is to eliminate the fractions as an initial step by finding a common denominator and multiplying all terms of the equation by it.

This approach simplifies the equation to a format without fractions, making it easier to apply further algebraic methods. As demonstrated in the exercise's solution, multiplying by the least common denominator cleared the fractions and reformatted the first equation, transforming it into a more standard form. It's crucial for students to master this fraction clearing technique as it is widely applicable and helps reduce calculation errors.

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

A couple has \(\$ 10,000\) to invest for their child's wedding. Their accountant recommends placing at least \(\$ 6000\) in a high-yield investment and no more than \(\$ 4000\) in a low-yield investment. (a) Use \(x\) to denote the amount of money placed into the high-yield investment. Use \(y\) to denote the amount of money placed into the low-yield investment. Write a system of linear inequalities that describes the possible amounts the couple could invest in each type of venture. (b) Graph the region that represents all possible amounts the couple could put into each investment if they wish to follow the accountant's advice.

A farmer has 90 acres available for planting corn and soybeans. The cost of seed per acre is \(\$ 4\) for corn and \(\$ 6\) for soybeans. To harvest the crops, the farmer will need to hire some temporary help. It will cost the farmer \(\$ 20\) per acre to harvest the corn and \(\$ 10\) per acre to harvest the soybeans. The farmer has \(\$ 480\) available for seed and \(\$ 1400\) available for labor. His profit is \(\$ 120\) per acre of corn and \(\$ 150\) per acre of soybeans. How many acres of each crop should the farmer plant to maximize the profit?

An airline charges 380 dollar for a round-trip flight from New York to Los Angeles if the ticket is purchased at least 7 days in advance of travel. Otherwise, the price is 700 dollar . If a total of 80 tickets are purchased at a total cost of 39,040 dollar, find the number of tickets sold at each price.

For the given matrices \(A, B,\) and \(C,\) evaluate the indicated expression. $$\begin{aligned}&A=\left[\begin{array}{rr}3 & -8 \\\2 & 4\end{array}\right] ; \quad B=\left[\begin{array}{rr}-6 & 0 \\\0 & -6\end{array}\right] ; \quad C=\left[\begin{array}{rr}3 & 5 \\\\-2 & 6\end{array}\right]\\\&(A+2 B) C\end{aligned}$$

Consider the following system of equations. $$\left\\{\begin{aligned} x+y &=3 \\\\-2 x-2 y &=-6 \\\\-x-y &=-3 \end{aligned}\right.$$ Use Gauss-Jordan elimination to show that this system Thas infinitely many solutions. Interpret your answer in merms of the graphs of the given equations.

See all solutions

Recommended explanations on Math Textbooks

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.