Chapter 8: Problem 28
Solve the system by the method of substitution. Use a graphing utility to verify your results. $$\left\\{\begin{aligned} -\frac{2}{3} x+y &=2 \\ 3 x-\frac{1}{2} y &=4 \end{aligned}\right.$$
Short Answer
Expert verified
The solution for the system of equations is \(x = \frac{15}{8}\) and \(y = \frac{5}{4}\)
Step by step solution
01
Isolate one variable
In the first equation, isolate the variable 'y'. That turns \(-\frac{2}{3} x + y = 2\) into \(y = \frac{2}{3}x + 2\)
02
Substitute in second equation
Substitute \(y\) from first equation into the second equation. So, \(3x - \frac{1}{2} (\frac{2}{3}x + 2) = 4\) turns into \(3x - \frac{1}{3}x -1 = 4\)
03
Solve for x
Simplify and solve for 'x'. So, \(\frac{8}{3} x = 5\) yields \(x = \frac{15}{8}\)
04
Substitute x to find y
Substitute \(x = \frac{15}{8}\) into the \(y\) equation to find 'y'. So, \(y = \frac{2}{3}\) * \(\frac{15}{8} + 2\) turns into \(y = \frac{5}{4}\).
05
Verification with graphing utility
A graphing utility can be utilized to verify the obtained solution. Plotting the equations \(-2/3 x + y = 2\) and \(3x - 1/2 y = 4\), the intersection point should be the point \((15/8, 5/4)\).
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 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Method of Substitution
The method of substitution is a fundamental technique to solve systems of equations, particularly when dealing with two variables.
It involves a step-by-step process to replace one variable with an equivalent expression, allowing us to work with a single equation at a time. Let's break it down:
It involves a step-by-step process to replace one variable with an equivalent expression, allowing us to work with a single equation at a time. Let's break it down:
- First, choose one of the equations and isolate either of the variables, say 'y'. This creates a manageable expression where 'y' is expressed in terms of 'x'. For example, from \(-\frac{2}{3} x + y = 2\), we derive \(y = \frac{2}{3}x + 2\).
- Next, substitute the expression for 'y' into the other equation, turning it into a single-variable equation. This reduces \(3x - \frac{1}{2} y = 4\) to \(3x - \frac{1}{3}x -1 = 4\).
- Finally, solve this equation to find the value of 'x'. In this case, we find \(x = \frac{15}{8}\).
Graphing Utility Verification
Once you have a solution from substitution or another algebraic method, verifying with a graphing utility ensures accuracy.
Graphing utilities, such as Desmos or a calculator capable of graphing, allow us to visualize equations as lines in the coordinate plane.Here's how verification works:
Graphing utilities, such as Desmos or a calculator capable of graphing, allow us to visualize equations as lines in the coordinate plane.Here's how verification works:
- Plot each equation on the same set of axes. They will appear as lines on the graph. In this exercise, we graph \(-\frac{2}{3} x + y = 2\) and \(3x - \frac{1}{2} y = 4\).
- Observe where the two lines intersect. The intersection point represents the solution \((x, y)\) that satisfies both equations simultaneously.
- Verify that the intersection point \(\left(\frac{15}{8}, \frac{5}{4}\right)\) matches the calculated solution.
Linear Equations
Linear equations are equations of the first order that represent straight lines when graphed on a coordinate plane.
They are typically formatted as \(ax + by = c\), where ''a', 'b', and 'c' are constants.Key properties of linear equations include:
They are typically formatted as \(ax + by = c\), where ''a', 'b', and 'c' are constants.Key properties of linear equations include:
- A linear equation in two variables (like 'x' and 'y') describes a line in a two-dimensional space.
- The equation can be manipulated into slope-intercept form \(y = mx + b\), where 'm' is the slope and 'b' is the y-intercept.
- The solutions to a linear equation are all the points on its line; however, for systems, we seek the common solution point.
Simultaneous Equations
Simultaneous equations, or systems of equations, entail solving multiple equations with common variables.
We aim to find a solution that satisfies all given equations.When dealing with simultaneous equations:
We aim to find a solution that satisfies all given equations.When dealing with simultaneous equations:
- Each equation represents a condition that must be true at the same time, forming lines or curves in coordinate space.
- Solutions correspond to intersection points where these lines meet, representing values of variables that are consistent across all equations.
- Methods like substitution and elimination are commonly used to find these points, depending on the problem structure.