Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 3: Problem 46
$$ \begin{aligned} &\frac{2 x}{3}-\frac{y}{2}=\frac{1}{6} \\ &\frac{x}{2}-\frac{y}{2}=-1 \end{aligned} $$
Short Answer
Expert verified
The solution to the system of equations \(\frac{2x}{3} - \frac{y}{2} = \frac{1}{6}\) and \(\frac{x}{2} - \frac{y}{2} = -1\) is (x, y) = (-1, 1).
Step by step solution
01
Multiply each equation by a necessary multiple to make the coefficients of y's in each equation equal
To eliminate y, we need to make sure that the coefficients of y's in each equation are equal (or equal but with the opposite sign). In this case, the coefficients of y's are already equal (both are -1/2), so we don't need to multiply any equation by any number.
02
Add or subtract the equations to eliminate y
Since the coefficients of y are equal in both equations, we can add the equations to eliminate y:
\(\frac{2x}{3} + \frac{x}{2}=\frac{1}{6} - 1\)
03
Solve for x
Now, we need to solve the resulting equation for x:
\(\frac{2x + 3x}{6} = \frac{1 - 6}{6}\)
\(5x = -5\)
\(x = -1\)
04
Substitute the value of x in either equation and solve for y
Now that we have the value of x, we can substitute it into either of the original equations and solve for y. Let's use the second equation:
\(-\frac{1}{2} - \frac{y}{2} = -1\)
05
Solve for y
Multiplying both sides by 2, we get:
\(-1 - y = -2\)
\(y = 1\)
06
Write the solution as an ordered pair
The solution to the system of equations is:
(x, y) = (-1, 1)
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Solving Linear Systems Algebraically
Understanding how to solve a system of linear equations algebraically is a cornerstone of algebra. This process involves finding a set of values for the variables that make all the equations true simultaneously. There are two main approaches: the substitution method and the elimination method. The substitution method is where you solve one equation for one variable and then substitute that expression into the other equation. On the other hand, the elimination method involves adding or subtracting the equations to cancel out one of the variables, leading to a solution for the other variable. Mastering these techniques allows for solving complex real-world problems that can be expressed as linear systems.
For the given system of equations, we see two equations with two variables, x and y, which we aim to find by manipulating these equations algebraically.
For the given system of equations, we see two equations with two variables, x and y, which we aim to find by manipulating these equations algebraically.
Elimination Method
The elimination method aims to remove one variable by adding or subtracting the equations. This method is particularly useful when the coefficients of a variable are opposites or can easily be made opposite by multiplication. Here's how it works:
First, if necessary, you multiply one or both of the equations by a number that will allow you to eliminate one of the variables when you add or subtract the equations. In the provided example, the equations were ready for elimination since the y coefficients were the same.
First, if necessary, you multiply one or both of the equations by a number that will allow you to eliminate one of the variables when you add or subtract the equations. In the provided example, the equations were ready for elimination since the y coefficients were the same.
Applying Elimination on Our System
By adding the two equations, we managed to eliminate y, simplifying the system to a single variable equation where we could solve for x. Once you find the value of x, you plug it back into one of the original equations to find y. This two-step process leads you to the ordered pair solution.Substitution Method
In contrast to the elimination method, the substitution method involves solving one of the equations for one variable in terms of the other, and then plugging this back into the second equation. This turns the system into a single equation with a single variable, making it solvable.
How Does Substitution Work?
For instance, if one of the original equations were solved for y, we'd get an equation like y = 2x - 1. We would then take this expression for y and substitute it into the other equation where y appears. Once we solve for x, we substitute this found value back into our 'solved for y' equation to find the corresponding y value, thus arriving at an ordered pair solution for the system. This method is particularly useful when the equations are already solved for one variable, or can be manipulated easily to do so.Ordered Pair Solutions
An ordered pair solution is the final goal in solving systems of linear equations. It represents values of x and y that satisfy all equations within the system simultaneously. An ordered pair is written in the form (x, y).
To validate that you have the correct solution, you can take these x and y values and substitute them back into the original equations. If the substitution balances both equations, you can be confident in your solution. In our case, we found the ordered pair (-1, 1), which when tested, satisfies both of the initial equations, confirming its validity as the solution to the system.
To validate that you have the correct solution, you can take these x and y values and substitute them back into the original equations. If the substitution balances both equations, you can be confident in your solution. In our case, we found the ordered pair (-1, 1), which when tested, satisfies both of the initial equations, confirming its validity as the solution to the system.
