/*! 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 72 Consider the following systems i... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 72

Consider the following systems in unknowns \(x\) and \(y:\) (a) \(a x+b y=1\) \(c x+d y=0\) (b) \(a x+b y=0\) \(c x+d y=1\)

Short Answer

Expert verified
In short, for system (a): \(x = \frac{a d}{a (a d - c b)}\) and \(y = \frac{-c}{a d - c b}\). For system (b): \(x = \frac{-b a}{a (a d + c b)}\) and \(y = \frac{a}{a d + c b}\).

Step by step solution

01

Solve the first equation for \(x\)

We can solve the first equation for \(x\) by isolating it on one side: \(x = \frac{1 - b y}{a}\)
02

Substitute the expression for \(x\) into the second equation

Now we substitute the expression for \(x\) we found in step 1 into the second equation: \(c \left(\frac{1 - b y}{a}\right) + d y = 0\)
03

Solve for \(y\)

Now we simplify the equation by multiplying through by \(a\) and then solving for \(y\): \(c (1 - b y) + a d y = 0\) \(c - c b y + a d y = 0\) Now, solve for \(y\): \(y (a d - c b) = -c\) \(y = \frac{-c}{a d - c b}\)
04

Substitute the value of \(y\) back into the expression for \(x\)

Now we substitute the value of \(y\) we found in step 3 into the expression for \(x\): \(x = \frac{1 - b \left(\frac{-c}{a d - c b}\right)}{a}\)
05

Simplify the expression for \(x\)

Finally, we simplify the expression for \(x\): \(x = \frac{a d - c b + b c}{a (a d - c b)}\) \(x = \frac{a d}{a (a d - c b)}\) Now, we'll solve system (b): (b) \(a x + b y = 0\) \(c x + d y = 1\) We can use the same steps as we did for system (a).
06

Solve the first equation for \(x\)

We can solve the first equation for \(x\) by isolating it on one side: \(x = \frac{-b y}{a}\)
07

Substitute the expression for \(x\) into the second equation

Now we substitute the expression for \(x\) we found in step 1 into the second equation: \(c \left(\frac{-b y}{a}\right) + d y = 1\)
08

Solve for \(y\)

Now we simplify the equation by multiplying through by \(a\) and then solving for \(y\): \(-c b y + a d y = a\) \(y (a d + c b) = a\) \(y = \frac{a}{a d + c b}\)
09

Substitute the value of \(y\) back into the expression for \(x\)

Now we substitute the value of \(y\) we found in step 3 into the expression for \(x\): \(x = \frac{-b \left(\frac{a}{a d + c b}\right)}{a}\)
10

Simplify the expression for \(x\)

Finally, we simplify the expression for \(x\): \(x = \frac{-b a}{a (a d + c b)}\)

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.

Substitution Method
The substitution method is a technique used to find solutions for simultaneous equations, particularly linear systems. It involves solving one of the equations for one variable, and then substituting that expression into the other equation. This process allows us to convert a system of two equations into a single equation with one unknown.

Here are the essential steps for using the substitution method:
  • First, solve one of the equations for one variable, usually in terms of the other variable. For example, isolate \( x \) in terms of \( y \) or vice versa.
  • Next, substitute this expression into the other equation. This will transform the system into a single equation.
  • Solve the resulting equation for the remaining variable.
  • Finally, substitute the value back into one of the original equations to find the value of the first variable.
This method is particularly useful when one equation is already isolated or when coefficients make substitution straightforward. It simplifies the system and often helps in understanding how the variables relate to one another.
Linear Systems
A linear system consists of a set of equations where each equation represents a line in a two-dimensional space. In general, a linear system with two unknowns can be expressed in the form:
\[ a_1x + b_1y = c_1 \]
\[ a_2x + b_2y = c_2 \]
Each equation corresponds to a straight line when graphed on a coordinate plane, and the solution to the system is the point (or points) where the lines intersect.

There are a few possibilities for solutions in linear systems:
  • Unique Solution: The lines intersect at exactly one point. This is typical if the system has consistent and independent equations.
  • No Solution: The lines are parallel and never meet. This indicates that the system is inconsistent.
  • Infinite Solutions: The lines coincide completely, meaning the equations represent the same line. This situation means the system is consistent and dependent.
Understanding these possibilities helps in determining the nature of the solutions and in selecting the appropriate method, like substitution, for solving the system.
Solution of Equations
Solving equations is the process of finding the values of the variables that make the equation true. For linear systems, finding the solution involves pinpointing the values of the unknowns that satisfy all the equations simultaneously.

To solve linear equations using methods like substitution, we follow these basic principles:
  • Simplify and Manipulate: Use algebraic operations to simplify equations where possible, making it easier to solve.
  • Isolate Variables: Rearrange equations to isolate one variable in terms of the others. This helps in substituting and reducing complexity.
  • Consistency Check: Verify whether the solution satisfies all original equations to avoid errors.
The objective is clarity - redirecting each system towards simpler equations or expressions. Consistent practice with varied equations boosts skills in transforming and solving equations efficiently. Recognizing patterns in solutions also aids in understanding broader algebraic concepts.

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

Solve each of the following systems: \(\begin{array}{rrr}x+2 y-4 z=-4 & x+2 y-3 z=-1 & x+2 y-3 z=1 \\ 2 x+5 y-9 z=-10 & -3 x+y-2 z=-7 & 2 x+5 y-8 z=4 \\ 3 x-2 y+3 z=11 & 5 x+3 y-4 z=2 & 3 x+8 y-13 z=7 \\ \text { (a) } & \text { (b) } \\ \begin{array}{rrr} & & \end{array}\end{array}\) Reduce each system to triangular or echelon form using Gaussian elimination: a) Apply "Replace \(L_{2}\) by \(-2 L_{1}+L_{2}^{\prime \prime}\) and "Replace \(L_{3}\) by \(-3 L_{1}+L_{3}^{n}\) to eliminate \(x\) from the second and third equations, and then apply "Replace \(L_{3}\) by \(8 L_{2}+L_{3}\) " to eliminate \(y\) from the third equation. These operations yield $$ \begin{array}{rlrl} x+2 y-4 z & =-4 & x+2 y-4 z & =-4 \\ y-z & =-2 & \text { and then } & y-z & =-2 \\ -8 y+15 z & =23 & 7 z & =7 \end{array} $$ The system is in triangular form. Solve by back-substitution to obtain the unique solution \(u=(2,-1,1)\). b) Eliminate \(x\) from the second and third equations by the operations "Replace \(L_{2}\) by \(3 L_{1}+L_{2} "\) and "Replace \(L_{3}\) by \(-5 L_{1}+L_{3} . "\) This gives the equivalent system $$ \begin{array}{r} x+2 y-3 z=-1 \\ 7 y-11 z=-10 \\ -7 y+11 z=7 \end{array} $$ The operation "Replace \(L_{3}\) by \(L_{2}+L_{3}\) " yields the following degenerate equation with a nonzero constant: $$ 0 x+0 y+0 z=-3 $$ This equation and hence the system have no solution. c) Eliminate \(x\) from the second and third equations by the operations "Replace \(L_{2}\) by \(-2 L_{1}+L_{2} "\) "and "Replace \(L_{3}\) by \(-3 L_{1}+L_{3}\)." This yields the new system $$ \begin{aligned} x+2 y-3 z &=1 \\ y-2 z &=2 \\ 2 y-4 z &=4 \end{aligned} \quad \text { or } \quad \begin{aligned} x+2 y-3 z=1 \\ y-2 z=2 \end{aligned} $$ (The third equation is deleted, because it is a multiple of the second equation.) The system is in echelon form with pivot variables \(x\) and \(y\) and free variable \(z\). To find the parametric form of the general solution, set \(z=a\) and solve for \(x\) and \(y\) by backsubstitution. Substitute \(z=a\) in the second equation to get \(y=2+2 a\). Then substitute \(z=a\) and \(y=2+2 a\) in the first equation to get $$ x+2(2+2 a)-3 a=1 \quad \text { or } \quad x+4+a=1 \quad \text { or } \quad x=-3-a $$ Thus, the general solution is $$ x=-3-a, \quad y=2+2 a, \quad z=a \quad \text { or } \quad u=(-3-a, \quad 2+2 a, \quad a) $$ where \(a\) is a parameter.

Prove Theorem \(3.15 .\) Let \(v_{0}\) be a particular solution of \(A X=B,\) and let \(W\) be the general solution of \(A X=0 .\) Then \(U=v_{0}+W=\left\\{v_{0}+w: w \in W\right\\}\) is the general solution of \(A X=B\) Let \(w\) be a solution of \(A X=0 .\) Then

Prove Theorem 3.4: Suppose a system \(\mathscr{H}\) of linear equations is obtained from a system \(\mathscr{L}\) by a sequence of elementary operations. Then \(\mathscr{H}\) and \(\mathscr{L}\) have the same solutions. Each step of the sequence does not change the solution set (Problem 3.44). Thus, the original system \(\mathscr{L}\) and the final system \(\mathscr{M}\) (and any system in between) have the same solutions.

Describe the Gauss-Jordan elimination algorithm, which also row reduces an arbitrary matrix \(A\) to its row canonical form. The Gauss-Jordan algorithm is similar in some ways to the Gaussian elimination algorithm, except that here each pivot is used to place 0 's both below and above the pivot, not just below the pivot, before working with the next pivot. Also, one variation of the algorithm first normalizes each row-that is, obtains a unit pivot-before it is used to produce 0's in the other rows, rather than normalizing the rows at the end of the algorithm.

Suppose \(A B\) is defined. Prove (a) Suppose \(A\) has a zero row. Then \(A B\) has a zero row. (b) Suppose \(B\) has a zero column. Then \(A B\) has a zero column.
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

Read Explanation

Probability and Statistics

Read Explanation
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.