Chapter 3: Problem 39
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 the elimination method. $$ \begin{aligned} &\frac{1}{2} x-\frac{1}{6} y=10\\\ &\frac{2}{5} x+\frac{1}{2} y=8 \end{aligned} $$
Short Answer
Expert verified
The solution is \( x = 20 \) and \( y = 0 \).
Step by step solution
01
Eliminate the Fractions
First, clear the fractions in both equations by finding a common multiple for each denominator. Multiply the first equation by 6 and the second equation by 10.
02
Simplify the Equations
After multiplying, the equations become: \[ 3x - y = 60 \] and \[ 4x + 5y = 80 \]
03
Align Coefficients for Elimination
Decide which variable to eliminate. Here, eliminate y by making the coefficients of y in both equations equal. Multiply the first equation by 5
04
Rewriting the Equations
The rewritten system of equations is: \[ 15x - 5y = 300 \] and \[ 4x + 5y = 80 \]
05
Perform Elimination
Add the two equations to eliminate y: \[ 15x - 5y + 4x + 5y = 300 + 80 \]
06
Simplify and Solve for x
Combine like terms: \[ 19x = 380 \] Then solve for x: \[ x = \frac{380}{19} = 20 \]
07
Substitute to Find y
Substitute x = 20 back into one of the original equations. Use \[ 3x - y = 60 \]: \[ 3(20) - y = 60 \] \[ 60 - y = 60 \] Solve for y: \[ y = 0 \]
08
State the Solution
The solution to the system is \( x = 20 \) and \( y = 0 \). Since there is a unique solution, the system has one solution.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Linear Equations
Let's start by understanding **linear equations**. A linear equation is an algebraic equation of the form \( ax + by = c \), where \( a \), \( b \), and \( c \) are constants. The graph of a linear equation forms a straight line. In our exercise, the original equations are linear equations: \( \frac{1}{2} x - \frac{1}{6} y = 10 \) and \( \frac{2}{5} x + \frac{1}{2} y = 8 \).
The coefficients of \( x \) and \( y \) ((\frac{1}{2}, \frac{1}{6}) and (\frac{2}{5}, \frac{1}{2})) determine the steepness and direction of the lines. By eliminating fractions and solving for variables, these types of equations can be simplified and solved methodically.
The coefficients of \( x \) and \( y \) ((\frac{1}{2}, \frac{1}{6}) and (\frac{2}{5}, \frac{1}{2})) determine the steepness and direction of the lines. By eliminating fractions and solving for variables, these types of equations can be simplified and solved methodically.
System of Equations
A **system of equations** consists of two or more equations with the same set of variables. In our case, we have two equations and two variables, \( x \) and \( y \). The goal in solving a system of equations is to find the values of the variables that satisfy all equations simultaneously. There are three possible outcomes when solving a system:
Once one variable is found, we substitute it back into one of the equations to find the other variable.
- One unique solution
- No solution
- An infinite number of solutions
Once one variable is found, we substitute it back into one of the equations to find the other variable.
Set-Builder Notation
When describing the solution set of a system of equations, especially when it has infinite solutions, we use **set-builder notation**. This notation allows us to specify a set of numbers that satisfy a certain condition. It is expressed as:
\( \{ x \, | \, \text{condition} \} \)
For example, if the solution set of a system is all pairs \( (x, y) \) that satisfy a particular linear equation, we would write something like:
\( \{ (x, y) \, | \, y = 2x + 3 \} \).
In our exercise, we found a unique solution. Hence, we didn't need set-builder notation. But, if the system had infinite solutions, we would use this notation to describe all possible solutions.
\( \{ x \, | \, \text{condition} \} \)
For example, if the solution set of a system is all pairs \( (x, y) \) that satisfy a particular linear equation, we would write something like:
\( \{ (x, y) \, | \, y = 2x + 3 \} \).
In our exercise, we found a unique solution. Hence, we didn't need set-builder notation. But, if the system had infinite solutions, we would use this notation to describe all possible solutions.
Solving Algebraic Equations
Finally, let's break down the steps involved in **solving algebraic equations** using the elimination method. This process helps isolate variables and find solutions:
- First, eliminate fractions by finding common denominators and multiplying through to simplify the equations.
- Next, align the coefficients of one of the variables by multiplying the entire equation by appropriate constants.
- Subtract or add the equations to eliminate one variable, resulting in a single-variable equation.
- Solve this equation to find the value of the remaining variable.
- Substitute this value back into one of the original equations to solve for the other variable.