/*! 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 65 Suppose you are solving the syst... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose you are solving the system $$ \left\\{\begin{array}{l} {3 x+8 y=-5} \\ {2 x-4 y=3} \end{array}\right. $$ You decide to use the addition method by multiplying both sides of the second equation by \(2 .\) In which of the following was the multiplication performed correctly? Explain. a. \(4 x-8 y=3\) b. \(4 x-8 y=6\)

Short Answer

Expert verified
b. \(4x - 8y = 6\) is multiplied correctly.

Step by step solution

01

Write Down the Original System of Equations

The system of equations given is: 1. \(3x + 8y = -5\) 2. \(2x - 4y = 3\)
02

Identify the Multiplication Factor

To use the addition (or elimination) method, we multiply the second equation by 2 as per the problem instruction: \(2(x) - 2(4y) = 2(3)\)
03

Perform Multiplication on Each Term of the Second Equation

Multiply each term of the second equation by 2:- The term \(2x\) becomes \(4x\) - The term \(-4y\) becomes \(-8y\) - The term \(3\) (right side) becomes \(6\) Thus, the new equation is \(4x - 8y = 6\).
04

Compare the Options

The correct equation from the multiplication should be \(4x - 8y = 6\). Compare this with the provided options: - Option a: \(4x - 8y = 3\) (Incorrect) - Option b: \(4x - 8y = 6\) (Correct)

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Key Concepts

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

Addition Method
The addition method, often referred to as the elimination method, is a popular technique for solving systems of linear equations. This method involves strategically adding or subtracting equations to eliminate one of the variables, making it simpler to find the values of the remaining variable.
For example, consider a system of two equations with two variables. The goal is to find a way to combine these equations so that one of the variables cancels out. This method is particularly useful when the coefficients of a variable in each equation are evenly divisible or can be easily adjusted through multiplication to become equal.
By eliminating a variable, we transform a system with two variables into one with a single variable, which is much easier to solve. The simplicity of this approach makes it quite effective for students learning algebra.
Multiplication Factor
The multiplication factor is key in the addition method. It allows manipulation of one or both equations so that when they are combined, one variable is easily eliminated.
In the given exercise, the second equation was multiplied by a factor of 2. This means multiplying every term in that equation by 2 to obtain a new equation. This aligns the coefficients of one of the variables in both equations.
Using a multiplication factor can make the resulting coefficients equal but opposite in sign, facilitating the elimination of one variable when the equations are added together. This strategic adjustment simplifies the system, making it easier to solve.
Elimination Method
Elimination method is essentially synonymous with the addition method. It focuses on removing one variable by adding or subtracting equations. By doing this, you isolate one variable on one side of the equation.
In the example given, after using the multiplication factor, the "eliminated" variable was successfully setup. This is because multiplying the second equation by 2 resulted in the variables being set up to cancel each other out when the equations were added together.
Elimination method remains effective due to its ability to produce simple equations that are much easier to solve, while reducing errors that often come with substitution or other methods.
Solving Systems of Equations
Solving systems of equations involves finding values for the variables that satisfy all of the equations simultaneously. This means both or all equations hold true at the same time for the solution found.
The process begins with setting up the equations properly, identifying coefficients that can easily be manipulated, and applying methods such as addition or elimination. Each approach has its benefits depending on the coefficients and arrangement of equations.
In a classroom setting, students often practice solving systems of equations using different methods to see which is most efficient for different types of problems. Mastery of these techniques ensures flexibility and deeper understanding when encountering complex systems. By practicing both addition and elimination methods, students are equipped to handle a variety of systems with skill and confidence.

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Most popular questions from this chapter

Solve each system of equations by the addition method. If a system contains fractions or decimals, you may want to first clear each equation of fractions or decimals. See Examples 2 through 6 $$ \left\\{\begin{array}{l} {3 x+y=-11} \\ {6 x-2 y=-2} \end{array}\right. $$

Without graphing, decide. See Examples 7 and \(8 .\) a. Are the graphs of the equations identical lines, parallel lines, or lines intersecting at a single point? b. How many solutions does the system have? $$ \left\\{\begin{array}{l} {2 x+y=0} \\ {y=-2 x+1} \end{array}\right. $$

Solve each system of linear equations by graphing. See Examples 3 through 6 $$ \left\\{\begin{array}{l} {x-2 y=-6} \\ {-2 x+4 y=12} \end{array}\right. $$

Macadamia nuts cost an astounding 16.50 dollars per pound, but research by an independent firm says that mixed nuts sell better if macadamias are included. The standard mix costs 9.25 dollars per pound. Find how many pounds of macadamias and how many pounds of the standard mix should be combined to produce 40 pounds that will cost 10 dollars per pound. Find the amounts to the nearest tenth of a pound.

Dave and Sandy Hartranft are frequent flyers with Delta Airlines. They often fly from Philadelphia to Chicago, a distance of 780 miles. On one particular trip they fly into the wind, and the flight takes 2 hours. The return trip, with the wind behind them, only takes \(1 \frac{1}{2}\) hours. If the wind speed is the same on each trip. find the speed of the wind and find the speed of the plane in still air.

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