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Chapter 3: Problem 50

If a system has an infinite number of solutions, use set-builder notation to write the solution set. If a system has no solution, state this. Solve using any algebraic method. $$ \begin{aligned} &3(a-b)=15\\\ &4 a=b+1 \end{aligned} $$

Short Answer

Expert verified
The solution set is \( \left\{ \left( \frac{-4}{3}, \frac{-19}{3} \right) \right\}. \)

Step by step solution

01

- Simplify the First Equation

The first equation is given as: \( 3(a-b)=15 \). To simplify, divide both sides by 3:\[ a - b = 5. \]
02

- Express One Variable in Terms of Another

From the simplified form of the first equation, solve for \(a\):\[ a = b + 5. \]
03

- Substitute into the Second Equation

Substitute \( a = b + 5 \) from Step 2 into the second equation \( 4a = b + 1 \):\[ 4(b + 5) = b + 1. \]
04

- Solve for the Variable

Distribute and solve for \( b \):\[ 4b + 20 = b + 1. \] Subtract \( b \) and 20 from both sides:\[ 3b + 20 - 20 = 1 - 20 \]\[ 3b = -19 \] \[ b = \frac{-19}{3}. \]
05

- Back Substitute to Solve for the Other Variable

Substitute \( b = \frac{-19}{3} \) back into \( a = b + 5 \):\[ a = \frac{-19}{3} + 5 \] \[ a = \frac{-19}{3} + \frac{15}{3} \] \[ a = \frac{-19 + 15}{3} \] \[ a = \frac{-4}{3}. \]
06

- Write the Solution Set

Since we found a unique solution for \( a \) and \( b \), the system does not have an infinite number of solutions. The solution set can be written as:\[ \left\{ \left( \frac{-4}{3}, \frac{-19}{3} \right) \right\}. \]

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Key Concepts

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

Set-builder notation
Set-builder notation is a method of describing a set by stating the properties that its members must satisfy. This is particularly useful in mathematics for specifying solution sets. For example, if a linear system had infinitely many solutions, we could express all possible solutions using set-builder notation. For our system, however, we found unique values for both variables, so this notation isn't necessary here. Instead, our solution is simply a single point.
Unique solutions
A unique solution in the context of a system of linear equations means that there is exactly one set of values for the variables that satisfies all equations simultaneously. In our exercise, after simplifying and substituting, we found distinct values for both variables (ewline a ewline and ewline bewline). This indicates that the system has unique solutions. Systems with unique solutions typically graph as lines that intersect at precisely one point.
Substitution method
The substitution method is a technique used to solve systems of equations where one equation is solved for one variable, which is then substituted into the other equation. In our example, we first solved ewline a - b = 5ewline for ewlineaewline, giving us ewline a = b + 5ewline. Then, we substituted ewline a = b + 5 ewline into the second equation ewline 4a = b + 1 ewline. This allowed us to solve for ewline b ewline. Finally, we back-substituted ewline b ewline into ewline a = b + 5 ewline to find ewline aewline.
Algebraic methods
Algebraic methods for solving systems of linear equations involve manipulating the equations to isolate variables and solve for their values. Key algebrAic methods include:
  • Substitution Method: Solving one equation for one variable and substituting the result into the other equation, as we used in the example.
  • Elimination Method: Adding or subtracting equations to eliminate a variable, making it easier to solve for the remaining variable.
  • Matrix Methods: Using matrices and operations such as row reduction to solve systems, which is particularly useful for larger systems.
In our exercise, we used the substitution method, which is often simpler for systems of two equations with two variables.

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Most popular questions from this chapter

Gold used to make jewelry is often a blend of gold, silver, and copper. The relative amounts of the metals determine the color of the alloy. Red gold is \(75 \%\) gold, \(5 \%\) silver, and \(20 \%\) copper. Yellow gold is \(75 \%\) gold, \(12.5 \%\) silver, and \(12.5 \%\) copper. White gold is \(37.5 \%\) gold and \(62.5 \%\) silver. If 100 g of red gold costs 4177.15,100 dollars g of yellow gold costs 4185.25 dollars, and \(100 \mathrm{g}\) of white gold costs 2153.875 dollars, how much do gold, silver, and copper cost per gram?

A plane flying the 3458 -mi trip from New York City to London has a 50 -mph tailwind. The flight's point of no return is the point at which the flight time required to return to New York is the same as the time required to continue to London. If the speed of the plane in still air is \(360 \mathrm{mph},\) how far is New York from the point of no return?

Investments. A business class divided an imaginary investment of 80,000 dollars among three mutual funds. The first fund grew by \(4 \%\), the second by \(6 \%,\) and the third by \(8 \% .\) Total earnings were 4400 dollars. The earnings from the third fund were 200 dollars more than the earnings from the first. How much was invested in each fund?

Solve each system. If a system's equations are dependent or if there is no solution, state this. $$ \begin{aligned} 2 x-3 y+z &=5 \\ x+3 y+8 z &=22 \\ 3 x-y+2 z &=12 \end{aligned} $$

A magic show's audience of 100 people consists of adults, students, and children. The ticket prices are 10 dollars each for adults, 3 dollars each for students, and 50\(\not {c}\) dollars each for children. A total of 100 dollars is taken in. How many adults, students, and children are in attendance? Does there seem to be some information missing? Do some more careful reasoning.
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