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Solving a System of Linear Equations In Exercises \(19 - 24\) , solve the system of linear equations and check any solutions algebraically. $$\left\\{ \begin{aligned} - 2 x + 3 y & = 10 \\ x + y & = 0 \end{aligned} \right.$$

Short Answer

Expert verified
The solution to the system of equations is \(x = -2\) and \(y = 2\).

Step by step solution

01

Isolate One Variable from an Equation

For simplicity, choose equation \(x + y = 0\). Isolating y would give \(y = -x\).
02

Substitution into the Other Equation

Substitute \(y = -x\) into the first equation. This would result into \(-2x + 3(-x) = 10\), simplifying this further, gives \( -5x = 10\).
03

Solve for x

Now solve for x by dividing each side by -5, results in \(x = -2\).
04

Substitute x into the Isolated Equation

Substitute \(x = -2\) into the first isolated equation \(y = -x\), which results in \(y = 2\).
05

Check the Solution

Verify these solutions by substituting them into the original equations. For \(-2x + 3y = 10\), substituting results in \( -2*(-2) + 3*2 = 10\), which is correct as \(4 + 6 = 10\). For \(x + y = 0\), substituting gives \(-2 + 2 = 0\), which is also correct; thus the solutions are valid.

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Key Concepts

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

Substitution Method
The Substitution Method is a strategy used to solve systems of linear equations where one variable is expressed in terms of the other variable, and then substituted into another equation. This process reduces the system of equations into a single variable equation, which we can solve using standard algebraic techniques.

For example, given the equations \(-2x + 3y = 10\) and \(x + y = 0\), we start by isolating a variable in the simpler equation, preferably where the coefficient is 1. In this case, from \(x + y = 0\), we can derive \(y = -x\). Next, we substitute the expression for \(y\) back into the other equation. This eliminates \(y\) and gives us a linear equation with one variable, \(x\), that we can solve.

Utilizing the substitution method effectively requires careful algebraic manipulation and the ability to isolate variables accurately. This method is particularly useful for systems with equations that are easy to manipulate into \(y = ...\) or \(x = ...\) form.
Algebraic Solutions
Algebraic solutions involve finding the set of values for variables that satisfy all equations within a system. To reach the algebraic solution, you may employ various methods such as substitution, elimination, or graphing, but all seek the same end result: the exact numerical values that make all equations true simultaneously.

Once the value for one variable is found using a chosen method, it’s plugged back into one of the original equations to find the other variable. After finding the values, it’s crucial to verify them by substituting back into each original equation to ensure they resolve correctly. Algebraic verification is a step that cannot be overlooked as it confirms the correctness of the solution.
Linear Equation Systems
A linear equation system is composed of two or more linear equations involving the same set of variables. The solution to the system typically consists of the values of the variables that satisfy all the equations simultaneously.

These systems can represent numerous real-world scenarios, such as supply-demand in economics, or intersections between geometric figures in coordinate geometry. Understanding how to navigate these systems is fundamental to algebra. Every system has three possible types of solutions:
  • A single, unique solution suggesting the lines intersect at one point.
  • No solution if the lines are parallel and never intersect.
  • Infinite solutions if the equations represent the same line.
Identifying the nature of a linear equation system allows for a more targeted approach when solving them.
Isolating Variables
Isolating variables is a foundational skill in algebra, forming the bedrock of solving equations and systems of equations. This process involves manipulating the equation to get one variable alone on one side of the equal sign.

For instance, if we're given \(x + y = 0\), we can isolate \(y\) by subtracting \(x\) from both sides to get \(y = -x\). This approach simplifies complex problems and reveals relationships between variables. When dealing with linear systems, being able to isolate a variable efficiently can make more advanced techniques, like substitution or elimination, much more accessible.

Isolation is more than just a mechanical skill; it involves knowing which variable to isolate and how it makes subsequent steps easier, thus reflecting a deeper understanding of algebraic structure.

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Most popular questions from this chapter

Optimal Revenue An accounting firm has 780 hours of staff time and 272 hours of reviewing time available each week. The firm charges \(\$ 1600\) for an audit and \(\$ 250\) for a tax return. Each audit requires 60 hours of staff time and 16 hours of review time. Each tax return requires 10 hours of staff time and 4 hours of review time. What numbers of audits and tax returns will yield an optimal revenue? What is the optimal revenue?

Finding Minimum and Maximum Values, find the minimum and maximum values of the objective function and where they occur, subject to the constraints \(x \geq 0, y \geq 0,3 x+y \leq 15\) and \(4 x+3 y \leq 30 .\) $$ z=5 x+y $$

Fitting a Line to Data To find the least squares regression line \(y=a x+b\) for a set of points \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right)\) you can solve the following system for \(a\) and \(b\) . $$ \left\\{\begin{array}{c}{n b+\left(\sum_{i=1}^{n} x_{i}\right) a=\left(\sum_{i=1}^{n} y_{i}\right)} \\ {\left(\sum_{i=1}^{n} x_{i}\right) b+\left(\sum_{i=1}^{n} x_{i}^{2}\right) a=\left(\sum_{i=1}^{n} x_{i} y_{i}\right)}\end{array}\right. $$ In Exercises 55 and \(56,\) the sums have been evaluated. Solve the given system for \(a\) and \(b\) to find the least squares regression line for the points. Use a graphing utility to confirm the result. $$ \left\\{\begin{aligned} 5 b+10 a &=20.2 \\ 10 b+30 a &=50.1 \end{aligned}\right. $$

Solving a Linear Programming Problem, sketch the region determined by the constraints. Then find the minimum and maximum values of the objective function (if possible) and where they occur, subject to the indicated constraints. $$ \begin{array}{c}{\text { Objective function: }} \\ {z=4 x+5 y} \\ {\text { Constraints: }} \\ {x \geq 0} \\ {y \geq 0} \\ {x+y \geq 8} \\ {3 x+5 y \geq 30}\end{array} $$

Finding Minimum and Maximum Values, find the minimum and maximum values of the objective function and where they occur, subject to the constraints \(x \geq 0, y \geq 0,3 x+y \leq 15\) and \(4 x+3 y \leq 30 .\) $$ z=2 x+y $$

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