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Chapter 8: Problem 16

Use elimination to solve each system of equations. Check your solution. $$\left\\{\begin{aligned} 5 x+3 y &=-1 \\ -10 x+2 y &=26 \end{aligned}\right.$$

Short Answer

Expert verified
The solution to the system of equations is \(x = -2\) and \(y = 3\).

Step by step solution

01

Preparation

Get rid of one of the variables. To make the coefficients of \(x\) cancel each other out when the two equations are added, multiply the first equation by 2 and the second one by 1. This gives us: \[\begin{{align*}} 10x + 6y &= -2 \ -10x + 2y &= 26 \end{{align*}}\]
02

Adding the two equations

Add the two equations to eliminate \(x\). These manipulations gives us: \[8y = 24\]
03

Solve for \(y\)

To find the value of the variable \(y\), divide both sides of the equation by 8. So, \(y = 24 / 8 = 3\)
04

Substitute \(y\) into one of the original equations

Substitute \(y = 3\) into the first equation \(5x + 3y = -1\): So, \(5x + 3*3 = -1\) or \(5x= -10\)
05

Solve for \(x\)

Divide both sides of the equation by 5 to find the value of \(x\). So, \(x = -10 / 5 = -2\)
06

Check the Solution

Substitute \(x = -2\) and \(y = 3\) into both equations in the original system. If both equations hold true, 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.

Understanding System of Equations
A system of equations is a collection of two or more equations with a common set of unknowns or variables. In our case, the system consists of two equations with the variables \( x \) and \( y \). We aim to find values for these variables that satisfy all equations in the system.
Solving a system can offer valuable insights, especially in real-world scenarios where multiple conditions or relationships exist simultaneously, like in business, science, or engineering.
  • Each equation represents a line when graphed on a coordinate plane.
  • Solving the system involves finding the point or points where these lines intersect, giving us the solutions for \( x \) and \( y \).
  • Methods to solve systems include graphing, substitution, and elimination, each having its own strategic advantage depending on the system's complexity.
All About Linear Equations
Linear equations are equations of the first degree, which means that the variables are not raised to any powers other than one. They are represented commonly in the form \( ax + by = c \). Linear equations graph as straight lines, and when solving a system of such equations, we are essentially finding where these lines intersect.
  • In our example, both equations are linear: \( 5x + 3y = -1 \) and \( -10x + 2y = 26 \).
  • The coefficients (numbers in front of the variables) play a crucial role in the elimination process, as they can be adjusted to facilitate solving.
  • Our goal in solving linear systems is to reduce the problem to finding a solution that satisfies all of these equations simultaneously.
This type of equation is fundamental in mathematics because they serve as the basis for many more complex equations and systems seen in advanced topics.
Using Variable Substitution Effectively
Variable substitution is an effective method for solving systems of equations and involves replacing one variable with another expression to simplify equations. In the context of the elimination method, substitution is often the final step.
In our problem, once we determined \( y = 3 \), we substituted this into one of the original equations to find \( x \):
  • Placed \( y = 3 \) into the first equation: \( 5x + 3y = -1 \).
  • Calculated: \( 5x + 3(3) = -1 \), which simplifies to \(5x = -10\).
  • Finally, solved for \( x \) by substitution: \( x = -10 / 5 = -2 \).
This illustrates the flexibility and utility of substitution, where simplifying an equation can turn a two-variable problem into one that's easier to handle and solve. Always remember to substitute back into the original equations to verify the solution's accuracy.

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Most popular questions from this chapter

A family owns and operates three businesses. On their income-tax return, they have to report the depreciation deductions for the three businesses separately. In \(2004,\) their depreciation deductions consisted of use of a car, plus depreciation on 5 -year equipment (on which onc-fifth of the original value is deductible per year) and 10-year equipment (on which one- tenth of the original valuc is deductible per year). The car use (in miles) for cach business in 2004 is given in the following table, along with the original value of the depreciable 5 - and 10 year equipment used in each business that year. $$\begin{array}{|c|c|c|c|}\hline & \begin{array}{c}\text { Car } \\\\\text { Use }\end{array} & \begin{array}{c}\text { Original } \\\\\text { Value, 5-Year }\end{array} & \begin{array}{c}\text { Original } \\\\\text { Value, 10-Year }\end{array} \\\\\text { Business } & \text { (miles) } & \text { Equipment (S) } & \text { Equipment (S) } \\\\\hline 1 & 3200 & 9850 & 435 \\\2 & 8800 & 12,730 & 980 \\\3 & 6880 & 2240 & 615\\\\\hline\end{array}$$ The depreciation deduction for car use in 2004 was 37.5 cents per mile. Use matrix multiplication to determine the total depreciation deduction for each business in 2004.

A financial advisor offers three specific investment instruments: a stock- based mutual fund, a high-yield bond, and a certificate of deposit (CD). Risk factors for individual instruments can be quantified on a scale of 1 to \(5,\) with 1 being the most risky. The risk factors associated with these particular instruments are summarized in the following table.$$\begin{array}{lc} \text { Type of Investment } & \text { Risk Factor } \\ \text { Stock-based mutual fund } & 3 \\\\\text { High-yield bond } & 1 \\\\\text { CD } & 5\end{array}$$.One of the advisor's clients can tolerate an overall risk level of \(3.5 .\) In addition, the client stipulates that the amount of money invested in the mutual fund must equal the sum of the amounts invested in the high-yield bond and the CD. To satisfy the client's requirements, what percentage of the total investment should be allocated to each instrument?

Consider the following system of equations. $$\left\\{\begin{aligned} x+y &=3 \\\\-2 x-2 y &=-6 \\\\-x-y &=-3 \end{aligned}\right.$$ Use Gauss-Jordan elimination to show that this system Thas infinitely many solutions. Interpret your answer in merms of the graphs of the given equations.

A chemist wishes to make 10 gallons of a \(15 \%\) acid solution by mixing a \(10 \%\) acid solution with a \(25 \%\) acid solution. (a) Let \(x\) and \(y\) denote the total volumes (in gallons) of the \(10 \%\) and \(25 \%\) solutions, respectively. Using the variables \(x\) and \(y,\) write an equation for the total volume of the \(15 \%\) solution (the mixture). (b) Using the variables \(x\) and \(y,\) write an equation for the total volume of acid in the mixture by noting that Volume of acid in \(15 \%\) solution \(=\) volume of acid in \(10 \%\) solution \(+\) volume of acid in \(25 \%\) solution. (c) Solve the system of equations from parts (a) and (b), and interpret your solution. (d) Is it possible to obtain a \(5 \%\) acid solution by mixing a \(10 \%\) solution with a \(25 \%\) solution? Explain without solving any equations.

Find \(\left(A^{2}\right)^{-1}\) and \(\left(A^{-1}\right)^{2},\) where \(A=\left[\begin{array}{rr}1 & -2 \\ -1 & 3\end{array}\right] .\) What do you observe? Use the definition of the inverse of a matrix, together with the fact that \((A B)^{-1}=A^{-1} B^{-1},\) to show that \(\left(A^{2}\right)^{-1}=\left(A^{-1}\right)^{2}\) for every square matrix \(A\)
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