/*! 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 29 Solve each system by the additio... [FREE SOLUTION] | 91Ó°ÊÓ

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Solve each system by the addition method. Be sure to check all proposed solutions. \(\left\\{\begin{array}{r}x+2 y=2 \\ -4 x+3 y=25\end{array}\right.\)

Short Answer

Expert verified
The solution to the system of equations is \(x = -4\) , \(y = 3\)

Step by step solution

01

- Multiply one or both equations

First, we need to prepare the equations, so that when we add them together, one of the variables is eliminated. Looking at the two original equations: \(x+2y = 2\) and \(-4x+3y = 25\), to eliminate \(x\), let's multiply the first equation by 4: \(4x + 8y = 8\)
02

- Add equations to eliminate a variable

We then add this new equation to the second one, which leads to: \(4x - 4x + 8y + 3y = 8 + 25\), simplifying this gives: \(11y = 33\)
03

- Solve for Y

Solving for \(y\), we get \(y = 33 / 11 = 3\)
04

- Substitute and solve for X

Substitute \(y = 3\) into the first equation: \(x + 2*3 = 2\), which gives \(x = -4\)
05

- Verify the solution

Lastly, we check the solution \((-4, 3)\) by substituting these values back into the original equations: 1st equation : \(-4 + 2*3 = 2\), which is true and 2nd equation: \(-4*(-4) + 3*3 = 25\), also true. Therefore, the solution \(-4, 3\) is correct.

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Key Concepts

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

Simultaneous Equations
Simultaneous equations are a set of equations containing multiple variables that are solved together. The term 'simultaneous' indicates that the equations are interdependent; the solution must work for each equation at the same time. In the given exercise, we tackle a system of two linear equations with two unknown variables, denoted as \(x\) and \(y\).
This particular system is represented as \(x + 2y = 2\) and \( -4x + 3y = 25\). Solving such equations requires finding the value for \(x\) and \(y\) that makes both equations true simultaneously. One prevalent method to solve these equations is the addition method, also known as the elimination method, which we'll delve into further in the next sections. The beauty of simultaneous equations lies in their versatility; they can describe real-world scenarios ranging from business to physics, making them a foundational concept in algebra.
Algebraic Methods
Algebraic methods provide a toolkit for solving equations and uncovering the relationships between variables. In the realm of simultaneous equations, these methods include substitution, graphing, and elimination. The addition method falls under elimination strategies. It entails manipulating the equations in a system to cancel out one variable, paving the way to solve for the other.
In our exercise, we multiplied the first equation by a factor that would allow the coefficients of \(x\) in both equations to be equal and opposite in sign. This multiplication set the stage for us to add the equations together, leading to the elimination of one variable. Algebraic methods are a demonstration of strategic problem-solving, enabling students to approach linear systems with precision and logical sequencing. By mastering algebraic techniques, one can tackle a wide range of problems in mathematics and related disciplines.
Variable Elimination
Variable elimination is the heart of solving systems by the addition method. It's a systematic process to remove one variable in order to simplify the equations to the point where they can be solved sequentially. In our exercise, we used variable elimination to proficiently remove \(x\) from the equations by ensuring the coefficients of \(x\) were numerically equal but had opposite signs. We achieved this by multiplying the first equation.
Once this alignment was accomplished, adding the equations together made \(x\) disappear, leaving a simpler equation with only \(y\) to solve. After finding the value of \(y\), we back-substituted it into one of the original equations to find the value of \(x\). Variable elimination is particularly powerful as it allows for handling more complex systems as well, where more variables are involved. It's a technique that rewards careful planning and attention to the structure of the equations at hand.

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Most popular questions from this chapter

The data can be modeled by $$ f(x)=956 x+3176 \text { and } g(x)=3904 e^{0.134 x} \text {, } $$ in which \(f(x)\) and \(g(x)\) represent the average cost of room and board at public four-year colleges in the school year ending \(x\) years after 2010. Use these functions to solve Exercises 33-34. Where necessary, round answers to the nearest whole dollar. a. According to the linear model, what was the average cost of room and board at public four-year colleges for the school year ending in 2015 ? b. According to the exponential model, what was the average cost of room and board at public four-year colleges for the school year ending in 2015 ? c. Which function is a better model for the data for the school year ending in \(2015 ?\)

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