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Chapter 8: Problem 62

Use any method to solve the system. \(\left\\{\begin{array}{r}-x+3 y=17 \\ 4 x+3 y=7\end{array}\right.\)

Short Answer

Expert verified
The solution for the given system of equations is \( x = 10/3 \) and \( y = 61/9 \).

Step by step solution

01

Arrange the System of Equations

The given system of linear equations is: \[\begin{align*} -x + 3y & = 17 \ 4x + 3y & = 7 \end{align*}\] These equations are already arranged in standard form, with the 'x' and 'y' variable terms and constant terms aligned.
02

Apply the Elimination Method

In elimination method, the two equations are added to eliminate one variable. Here, if we add the two equations, the variable 'y' is eliminated. Which yields: \[\begin{align*} -x + 4x & = 17 - 7 \ 3x & = 10 \end{align*}\] The last step yields the equation in one variable 'x'.
03

Solve for 'x'

Now we solve for 'x': \[\begin{align*} x & = 10 / 3 \ x & = 10/3 \end{align*}\]
04

Substitute 'x' in the first equation

Substitute the obtained value of 'x' in the first equation: \[\begin{align*} -10/3 + 3y & = 17 \ 3y & = 17 + 10/3 \ 3y & = 61/3 \ y & = 61/9 \end{align*}\] The above steps result in the solutions for the variables 'x' and 'y'.

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Key Concepts

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

Elimination Method
The elimination method is a powerful technique used to solve systems of linear equations. It focuses on eliminating one variable to make the problem easier to solve. Think of it like balancing a scale. If one side is too heavy, you can take away or add weights to balance it out.

Here's how it works in this exercise:
  • You have two equations that need solving together.
  • By adding or subtracting the equations, you eliminate one of the variables.
In our example, by adding the two equations, the 'y' terms cancel each other out, allowing us to solve for 'x' directly.
Standard Form
Standard form is the format where equations are arranged with variables and constants aligned. This alignment makes it easier to compare and manipulate equations.

The general format is:
  • Ax + By = C
Each equation follows this format, making them ready for methods like elimination. The exercise already presents the equations in standard form, highlighting the importance of organized data. This organization helps in visualizing and executing the steps more efficiently.
Variable Elimination
Variable elimination is the goal of the elimination method. By removing one variable, you simplify the equations into a single-variable problem. In this exercise, the 'y' variable is eliminated:
  • Add the equations to cancel out the 'y' terms.
  • This step reduces the problem to solving a single equation in 'x'.
By focusing on one variable at a time, you streamline the process, making it logical and methodical. It's like peeling an onion layer by layer to get to the core.
Solving for Variables
Once a single variable is isolated, solving it becomes straightforward. For 'x', we ended up with the equation:
  • \(3x = 10\)
To solve for 'x':
  • Divide both sides by 3, giving \(x = \frac{10}{3}\).
With 'x' known, you substitute back into one of the original equations to find 'y'. This substitution involves basic algebra, and it's the final step to reach a full solution. Each variable holds a piece of the overall answer, completing the puzzle of equations.

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Most popular questions from this chapter

Chemistry Thirty liters of a \(40 \%\) acid solution are obtained by mixing a \(25 \%\) solution with a \(50 \%\) solution. (a) Write a system of equations in which one equation represents the amount of final mixture required and the other represents the percent of acid in the final mixture. Let \(x\) and \(y\) represent the amounts of the \(25 \%\) and \(50 \%\) solutions, respectively. (b) Use a graphing utility to graph the two equations in part (a) in the same viewing window. (c) As the amount of the \(25 \%\) solution increases, how does the amount of the \(50 \%\) solution change? (d) How much of each solution is required to obtain the specified concentration of the final mixture?

Write the system of linear equations represented by the augmented matrix. Then use back-substitution to find the solution. (Use the variables \(x, y,\) and \(z,\) if applicable.) $$\left[\begin{array}{rrrrr} 1 & -1 & 4 & \vdots & 0 \\ 0 & 1 & -1 & \vdots & 2 \\ 0 & 0 & 1 & \vdots & -2 \end{array}\right]$$

Use the matrix capabilities of a graphing utility to solve (if possible) the system of linear equations. $$\left\\{\begin{array}{l} 2 x+3 y+5 z=4 \\ 3 x+5 y-9 z=7 \\ 5 x+9 y+17 z=13 \end{array}\right.$$

Sales The projected sales \(S\) (in millions of dollars) of two clothing retailers from 2015 through 2020 can be modeled by \(\left\\{\begin{array}{ll}S-149.9 t=415.5 & \text { Retailer } \mathrm{A} \\\ S-183.1 t=117.3 & \text { Retailer } \mathrm{B}\end{array}\right.\) where \(t\) is the year, with \(t=5\) corresponding to 2015 (a) Solve the system of equations using the method of your choice. Explain why you chose that method. (b) Interpret the meaning of the solution in the context of the problem. (c) Interpret the meaning of the coefficient of the \(t\) -term in each model. (d) Suppose the coefficients of \(t\) were equal and the models remained the same otherwise. How would this affect your answers in parts (a) and (b)?

Four test plots were used to explore the relationship between wheat yield \(y\) (in bushels per acre) and amount of fertilizer applied \(x\) (in hundreds of pounds per acre). The results are given by the ordered pairs \((1.0,32),(1.5,41),(2.0,48),\) and (2.5,53) (a) Find the least squares regression line \(y=a x+b\) for the data by solving the system for \(a\) and \(b\) \(\left\\{\begin{array}{l}4 b+7.0 a=174 \\ 7 b+13.5 a=322\end{array}\right.\) (b) Use the regression feature of a graphing utility to confirm the result in part (a). (c) Use the graphing utility to plot the data and graph the linear model from part (a) in the same viewing window. (d) Use the linear model from part (a) to predict the yield for a fertilizer application of 160 pounds per acre.
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