/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 Solve the system by the method o... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Solve the system by the method of substitution. $$\left\\{\begin{array}{l} 0.5 x+3.2 y=9.0 \\ 0.2 x-1.6 y=-3.6 \end{array}\right.$$

Short Answer

Expert verified
The solutions for the system of equations are \(x = -2.34\) and \(y = 3.21\).

Step by step solution

01

Isolate a Variable

From equation \(0.5x + 3.2y = 9.0\), we can isolate \(x\) as \(x = 18 - 6.4y\).
02

Substitute into the Other Equation

Now substitute \(x = 18 - 6.4y\) into the second equation \(0.2x - 1.6y = -3.6\). This gives us \(0.2(18 - 6.4y) - 1.6y = -3.6\). Simplifying the left side of the equation results in \(3.6 - 2.24y = -3.6\). This simplifies further to give \(2.24y = 7.2\).
03

Solve for the Substitute Variable

Solve the equation \(2.24y = 7.2\) to get the value of \(y\) as \(y = 7.2/2.24 = 3.21\).
04

Substitute y into the Isolated Variable

Using the value of \(y = 3.21\) into \(x = 18 - 6.4y\), we find that \(x = 18 - 6.4*3.21 = -2.34\).
05

Verify the Solution

Plugging \(x = -2.34\) and \(y = 3.21\) into both original equations to verify that both equations are satisfied. We find that both equations are indeed satisfied, thus the solution is correct.

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.

Substitution Method
The substitution method is a technique used to solve simultaneous linear equations. It involves solving one of the equations for one variable and then substituting this expression into the other equation. This strategy effectively reduces a system of equations with multiple variables to a single equation with one variable, making it simpler to solve.

Let's take the provided exercise as an example. With the equation \(0.5x + 3.2y = 9.0\), we can solve for \(x\) to get an expression in terms of \(y\) which is \(x = 18 - 6.4y\). By substituting this into the second equation, we create an equation with a single variable \(y\), allowing us to find its value. Once we have the value of \(y\), we substitute it back into the isolated variable equation to find the value of \(x\). This step-by-step process simplifies the solution and ensures that you work methodically towards the correct answer.
Isolate a Variable
Isolating a variable is crucial in solving equations and is the foundation of the substitution method. To isolate a variable means to manipulate the equation in such a way that you get the targeted variable on one side of the equation and all other terms on the other side.

In the given problem, we start by isolating \(x\) in the first equation. Isolation is achieved by moving all terms involving \(y\) to the right side and keeping \(x\) alone on the left side. This step not only prepares us for substitution but also helps to see the relationship between the variables at a glance. Being comfortable with moving terms across the equals sign through addition, subtraction, multiplication, or division is necessary for this process.
Simultaneous Equations
Simultaneous equations are a set of equations containing multiple variables that are all true at the same time. The goal when solving simultaneous equations is to find the values of the variables that satisfy all the equations in the system at once. Systems can be composed of two or more equations, and the strategies to solve them include substitution, elimination, and graphical methods.

In our problem, we have two simultaneous equations that characterize a linear relationship between \(x\) and \(y\). To solve these equations, we use the substitution method, which is ideal when one equation can be conveniently solved for one variable. This approach often provides a clear path to the algebraic solution for both variables.
Algebraic Solutions
Algebraic solutions involve finding the exact numerical answers to equations using algebraic manipulations. Unlike graphical solutions, which might only give an approximate answer based on where lines intersect on a graph, algebraic solutions provide precise values.

In our exercise, algebraic solution entails precisely determining the values of \(x\) and \(y\) that satisfy both equations simultaneously. After isolating \(x\), substituting it into the second equation, and solving for \(y\), we acquire an exact numerical value, which is then used to compute the exact value of \(x\). Verifying the calculated values by plugging them back into the original equations to ensure they are true solutions is a critical final step in confirming the algebraic solution.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find two systems of linear equations that have the ordered triple as a solution. (There are many correct answers.) $$(-5,-2,1)$$

A fruit grower raises crops \(A\) and \(B\). The yield is 300 bushels per acre for crop \(\mathrm{A}\) and 500 bushels per acre for crop B. Research and available resources indicate the following constraints. "The fruit grower has 150 acres of land for raising the crops. \(\cdot\) It takes 1 day to trim an acre of crop \(A\) and 2 days to trim an acre of crop \(\mathbf{B},\) and there are 240 days per year available for trimming. \(\cdot\) It takes 0.3 day to pick an acre of crop \(\mathrm{A}\) and 0.1 day to pick an acre of crop \(\mathbf{B},\) and there are 30 days per year available for picking. What is the optimal acreage for each fruit? What is the optimal yield?

The linear programming problem has an unusual characteristic. Sketch a graph of the solution region for the problem and describe the unusual characteristic. Find the minimum and maximum values of the objective function (if possible) and where they occur. Objective function: \(z=-x+2 y\) Constraints: $$\begin{aligned}x & \geq 0 \\\y & \geq 0 \\ x & \leq 10 \\\x+y & \leq 7\end{aligned}$$

Use a graphing utility to graph the region determined by the constraints. Then find the minimum and maximum values of the objective function and where they occur, subject to the constraints. Objective function: \(z=3 x+y\) Constraints: $$\begin{aligned}x & \geq 0 \\\y & \geq 0 \\ x+4 y & \leq 60 \\\3 x+2 y & \geq 48\end{aligned}$$

Solve the system of linear equations and check any solutions algebraically. $$\left\\{\begin{array}{l} 3 x-5 y+5 z=1 \\ 5 x-2 y+3 z=0 \\ 7 x-y+3 z=0 \end{array}\right.$$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.