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Chapter 7: Problem 12

In Exercises 11 - 20, solve the system by the method of substitution. Check your solution(s) graphically. \( \left\\{\begin{array}{l}x - 4y = -11\\\x + 3y = 3\end{array}\right. \)

Short Answer

Expert verified
The solution to the system equations is \(x = \frac{15}{7}\) and \(y = \frac{-2}{7}\)

Step by step solution

01

Isolate a variable

First, isolate a variable in one of the equations. It is often easiest to choose the variable that ought to be simpler to isolate. In the given case, let's solve the second equation, \(x + 3y = 3\), for \(x\). It will give \(x = 3 - 3y\).
02

Substitute the isolated variable

Substitute \(x = 3 - 3y\) into the first equation, \(x - 4y = -11\). This yields \((3 - 3y) - 4y = -11\), simplify it to resolve for \(y\). After simplifying it we get \(y = \frac{-2}{7}\).
03

Back Substitute y in the equation for x

Substitute \(y = \frac{-2}{7}\) into the equation solved for \(x\), \(x = 3 - 3y\). This yields \(x = 3 - 3(\frac{-2}{7})\) which simplifies down to \(x = \frac{15}{7}\).
04

Checking solutions graphically

To check the solution graphically, graph both equations on the same set of axes. The point of intersection of the two lines represents the solution to the system. If the point (\(\frac{15}{7}\), \(\frac{-2}{7}\)) lies on both lines, then the solution is correct.

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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 straightforward approach used to solve systems of equations. This method involves isolating one variable in one of the equations and then substituting that expression into the other equation. Here's how it works step-by-step:
  • Isolate a Variable: Begin by solving one of the equations for one variable. This means getting the variable by itself on one side of the equation. For example, if the equation is \( x + 3y = 3 \), then solve for \( x \) to get \( x = 3 - 3y \).
  • Substitute Into the Other Equation: Take the expression you found for the isolated variable and substitute it into the other equation. This reduces the system to one equation with one variable. By substituting \( x = 3 - 3y \) into the other equation \( x - 4y = -11 \), you end up with \((3 - 3y) - 4y = -11\).
  • Solve for the Remaining Variable: Simplify the equation to find the value of the second variable. In this case, simplifying yields \(y = \frac{-2}{7}\).
  • Back-Substitute to Find the Other Variable: Use the value of \(y\) to find \(x\) by substituting back into the expression \(x = 3 - 3y\). This gives \(x = \frac{15}{7}\).
This method works well for linear equations and allows us to find the solution easily when substitution is convenient.
Graphical Solutions
Graphical solutions provide a visual method to solve a system of equations. Unlike algebraic approaches, such as substitution, graphical solutions involve plotting both equations on the same set of axes to find their intersection point.
  • Drawing the Lines: Each equation represents a straight line on the graph. For instance, the equation \(x - 4y = -11\) can be rewritten in slope-intercept form \(y = \frac{1}{4}x +\frac{11}{4}\) and graphed. Similarly, plot the second equation in the same fashion.
  • Intersection Point: The solution to the system of equations is the point where the two lines intersect. In our example, the lines intersect at the point \(\left(\frac{15}{7}, \frac{-2}{7}\right)\).
  • Verification: Checking your algebraic solution against the graphical representation confirms accuracy. If the graphical intersection matches your algebraic solution, then your solution is likely correct.
Graphical solutions not only confirm algebraic results but also provide a clear visual understanding of what the solutions represent. Each line and intersection represents real-world phenomena modeled by the equations.
Linear Equations
Linear equations are a foundational concept in algebra, involving relationships that produce straight lines when graphed. They take the general form \(Ax + By = C\) and have distinct characteristics that make them powerful for modeling situations.
  • Standard Form and Slope-Intercept Form: The standard form is useful for easy identification of coefficients and constants. However, converting to slope-intercept form \(y = mx + b\) often makes graphing easier. In slope-intercept form, \(m\) represents the slope, indicating the line's steepness, and \(b\) is the y-intercept, showing where the line crosses the y-axis.
  • Single Solutions and Graphs: When solving systems of linear equations, each solution is where two lines intersect. If the lines are parallel, there is no solution, as they never intersect. If they coincide, every point on the line is a solution, and there are infinitely many solutions.
  • Applications: Linear equations are used in a wide range of real-life situations. They can model anything from budgeting expenses to calculating distances or predicting trends over time.
Understanding linear equations helps build the groundwork for more advanced mathematical concepts, and mastering them equips you with invaluable problem-solving skills.

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Most popular questions from this chapter

In Exercises 13-16, 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. Objective function: \( z = 3x + 2y \) Constraints: \( \hspace{1cm} x \ge 0 \) \( \hspace{1cm} y \ge 0 \) \( 5x + 2y \le 20 \) \( 5x + y \ge 10 \)

Fill in the blanks. If a linear programming problem has a solution, it must occur at a ________ of the set of feasible solutions.

An animal shelter mixes two brands of dog food. Brand \(\mathrm{X}\) costs \(\$ 25\) per bag and contains two units of nutritional element A, two units of element \(\mathrm{B},\) and two units of element \(\mathrm{C}\) , Brand \(\mathrm{Y}\) costs \(\$ 20\) per bag and contains one unit of nutritional element A, nine units of element \(\mathrm{B},\) and three units of element \(\mathrm{C}\) . The minimum required amounts of nutrients \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{Care}\) 12 units, 36 units, and 24 units, respectively. What is the optimal number of bags of each brand that should be mixed? What is the optimal cost?

In Exercises 45-47, determine whether the statement is true or false. Justify your answer. When solving a linear programming problem, if the objective function has a maximum value at more than one vertex, you can assume that there are an infinite number of points that will produce the maximum value.

In Exercises 29-34, the linear programming problem has an unusual characteristic. Sketch a graph of the solution region for the problem and describe the unusual characteristic. Find the minimum and maximum values of the objective function (if possible) and where they occur. Objective function: \( z = 3x + 4y \) Constraints: \( \hspace{1cm} x \ge 0 \) \( \hspace{1cm} y \ge 0 \) \( x + y \le 1 \) \( 2x + y \le 4 \)
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