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Chapter 1: Problem 36

Solve the following system for \(x, y,\) and \(z\) $$\begin{aligned}\frac{1}{x}+\frac{2}{y}-\frac{4}{z} &=1 \\\\\frac{2}{x}+\frac{3}{y}+\frac{8}{z} &=0 \\\\-\frac{1}{x}+\frac{9}{y}+\frac{10}{z} &=5 \end{aligned}$$

Short Answer

Expert verified
The solution is \(x = -\frac{13}{7}\), \(y = \frac{91}{54}\), \(z = -\frac{91}{8}\).

Step by step solution

01

Express Each Fraction Separately

Given the system of equations:\[ \frac{1}{x} + \frac{2}{y} - \frac{4}{z} = 1 \]\[ \frac{2}{x} + \frac{3}{y} + \frac{8}{z} = 0 \]\[ -\frac{1}{x} + \frac{9}{y} + \frac{10}{z} = 5 \]Let's denote \( a = \frac{1}{x} \), \( b = \frac{1}{y} \), and \( c = \frac{1}{z} \). The system becomes:\[ a + 2b - 4c = 1 \]\[ 2a + 3b + 8c = 0 \]\[ -a + 9b + 10c = 5 \].
02

Solve the System of Linear Equations

We will solve the linear system in terms of \(a\), \(b\), and \(c\):1. Start with the first two equations: \[ a + 2b - 4c = 1 \] \[ 2a + 3b + 8c = 0 \] Multiply the first equation by 2: \[ 2a + 4b - 8c = 2 \] Subtract the second equation from the modified first equation: \[ (2a + 4b - 8c) - (2a + 3b + 8c) = 2 - 0 \] Simplify to get: \[ b - 16c = 2 \] (Equation 4)2. Use Equations 1 and 3: \[ a + 2b - 4c = 1 \] \[ -a + 9b + 10c = 5 \] Add these equations: \[ 11b + 6c = 6 \] (Equation 5).3. Use Equations 4 and 5 to solve for \(b\) and \(c\): First rearrange Equation 5: \[ b = 16c + 2 \] Substitute into Equation 5: \[ 11(16c + 2) + 6c = 6 \] \[ 176c + 22 + 6c = 6 \] \[ 182c = -16 \] \[ c = -\frac{8}{91} \].
03

Substitute and Solve for Remaining Variables

Substitute \(c = -\frac{8}{91}\) back into \(b = 16c + 2\):\[ b = 16\left(-\frac{8}{91}\right) + 2 \]\[ b = -\frac{128}{91} + 2 \]\[ b = -\frac{128}{91} + \frac{182}{91} \]\[ b = \frac{54}{91} \].Now substitute \(b\) and \(c\) into one of the original equations to find \(a\). Use Equation 1:\[ a + 2\left(\frac{54}{91}\right) - 4\left(-\frac{8}{91}\right) = 1 \]\[ a + \frac{108}{91} + \frac{32}{91} = 1 \]\[ a + \frac{140}{91} = 1 \]\[ a = 1 - \frac{140}{91} \]\[ a = \frac{91}{91} - \frac{140}{91} \]\[ a = -\frac{49}{91} \].
04

Solve for Original Variables \(x\), \(y\), and \(z\)

We have \(a = \frac{1}{x} = -\frac{49}{91}\), \(b = \frac{1}{y} = \frac{54}{91}\), and \(c = \frac{1}{z} = -\frac{8}{91}\).Thus, solve for \(x\), \(y\), and \(z\):- \( x = -\frac{91}{49} = -\frac{13}{7} \)- \( y = \frac{91}{54} \)- \( z = -\frac{91}{8} \).
05

Verification

Substitute the values back into the original equations to verify:1. \( \frac{1}{x} + \frac{2}{y} - \frac{4}{z} = 1 \) Produces: \( -\frac{49}{91} + 2(\frac{54}{91}) - 4(-\frac{8}{91}) = \frac{-49+108+32}{91} = 91/91 = 1 \)2. \( \frac{2}{x} + \frac{3}{y} + \frac{8}{z} = 0 \) Produces: \( 2(-\frac{49}{91}) + 3(\frac{54}{91}) + 8(-\frac{8}{91}) = \frac{-98+162-64}{91} = 0 \)3. \( -\frac{1}{x} + \frac{9}{y} + \frac{10}{z} = 5 \) Produces: \( \frac{49}{91} + 9(\frac{54}{91}) + 10(-\frac{8}{91}) = \frac{49+486-80}{91} = 5 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Variable Substitution
When solving a system of linear equations, sometimes terms are complex, like those involving fractions. We can simplify these by using a process called **variable substitution**.
In this method, we substitute complex fractional terms with simpler variables. For example, if you have an equation with sums like \( \frac{1}{x} \), \( \frac{2}{y} \), and \( \frac{4}{z} \), you can set these fractions equal to other variables.
Here's how it's done in our exercise:
  • Set \( a = \frac{1}{x} \)
  • Set \( b = \frac{1}{y} \)
  • Set \( c = \frac{1}{z} \)
Now, instead of dealing with fractions directly, you work with the new variables \( a \), \( b \), and \( c \). This transformation makes the equations linear and easier to handle.
Solution Verification
After calculating the values of variables from a system of equations, it's crucial to verify if they actually solve the original equations. **Solution verification** ensures accuracy and helps catch any potential mistakes in your calculations.
In our problem, we found \( x = -\frac{13}{7}, y = \frac{91}{54}, z = -\frac{91}{8} \). To verify the solution, substitute these values back into the original set of equations:
  • Check the equation \( \frac{1}{x} + \frac{2}{y} - \frac{4}{z} = 1 \). With the values for \( x, y, \) and \( z \), compute each term.
  • Do the same for \( \frac{2}{x} + \frac{3}{y} + \frac{8}{z} = 0 \).
  • Finally, validate \( -\frac{1}{x} + \frac{9}{y} + \frac{10}{z} = 5 \).
Each calculation should yield true statements if the solution is correct. Double-check your math if there's a discrepancy.
Fractional Equations
Sometimes, equations like these involve fractions, making them tricky to solve. **Fractional equations** are equations where variables appear in the denomination. Solving these equations typically requires clearing the fractions to simplify the problem.
Here's how to manage them:
  • Commonly, you multiply every term by a common denominator to eliminate fractions.
  • For example, from \( \frac{1}{x} + \frac{2}{y} - \frac{4}{z} = 1 \), it's easier if expressed without fractions.
  • By using variable substitution, as we did by letting \( \frac{1}{x} = a \), these equations transform into simpler linear equations.
Handling fractional equations this way can prevent errors and streamline solving the system, making complex problems more approachable.

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Most popular questions from this chapter

Let \(A=\left[a_{i j}\right]\) be an \(n \times n\) matrix. Determine whether \(A\) is symmetric. (a) \(a_{i j}=i^{2}+j^{2}\) (b) \(a_{i j}=i^{2}-j^{2}\) (c) \(a_{i j}=2 i+2 j\) (d) \(a_{i j}=2 i^{2}+2 j^{3}\)

In each part, find a \(6 \times 6\) matrix \(\left[a_{i j}\right]\) that satisfies the stated condition. Make your answers as general as possible by using letters rather than specific numbers for the nonzero entries. (a) \(a_{i j}=0\) if \(i \neq j\) (b) \(a_{i j}=0\) if \(i>j\) (c) \(a_{i j}=0\) if \(i1\) Answer: A. $$\left[\begin{array}{cccccc} a_{11} & 0 & 0 & 0 & 0 & 0 \\ 0 & a_{22} & 0 & 0 & 0 & 0 \\ 0 & 0 & a_{33} & 0 & 0 & 0 \\ 0 & 0 & 0 & a_{44} & 0 & 0 \\ 0 & 0 & 0 & 0 & a_{55} & 0 \\ 0 & 0 & 0 & 0 & 0 & a_{66} \end{array}\right]$$ B. $$\left[\begin{array}{cccccc} a_{11} & a_{12} & a_{13} & a_{14} & a_{15} & a_{16} \\ 0 & a_{22} & a_{23} & a_{24} & a_{25} & a_{26} \\ 0 & 0 & a_{33} & a_{34} & a_{35} & a_{36} \\ 0 & 0 & 0 & a_{44} & a_{45} & a_{46} \\ 0 & 0 & 0 & 0 & a_{55} & a_{56} \\ 0 & 0 & 0 & 0 & 0 & a_{66} \end{array}\right]$$ C. $$\left[\begin{array}{llllll} a_{11} & 0 & 0 & 0 & 0 & 0 \\ a_{21} & a_{22} & 0 & 0 & 0 & 0 \\ a_{31} & a_{32} & a_{33} & 0 & 0 & 0 \\ a_{41} & a_{42} & a_{43} & a_{44} & 0 & 0 \\ a_{51} & a_{52} & a_{53} & a_{54} & a_{55} & 0 \\ a_{61} & a_{62} & a_{63} & a_{64} & a_{65} & a_{66} \end{array}\right]$$ D. $$\left[\begin{array}{cccccc} a_{11} & a_{12} & 0 & 0 & 0 & 0 \\ a_{21} & a_{22} & a_{23} & 0 & 0 & 0 \\ 0 & a_{32} & a_{33} & a_{34} & 0 & 0 \\ 0 & 0 & a_{43} & a_{44} & a_{45} & 0 \\ 0 & 0 & 0 & a_{54} & a_{55} & a_{56} \\ 0 & 0 & 0 & 0 & a_{65} & a_{66} \end{array}\right]$$

Show that if $$ A=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & c \end{array}\right] $$ is an elementary matrix, then at least one entry in the third row must be a zero.

Prove that if \(A\) is an invertible matrix and \(B\) is row equivalent to \(A\), then \(B\) is also invertible.

Solve the following system of nonlinear equations for \(x, y,\) and \(z\) $$\begin{aligned} x^{2}+y^{2}+z^{2} &=6 \\\x^{2}-y^{2}+2 z^{2} &=2 \\\2 x^{2}+y^{2}-z^{2} &=3 \end{aligned}$$
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