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Chapter 3: Problem 53

For Exercises 52 and \(53,\) let u represent \(1 / x,\) v represent \(1 / y,\) and \(w\) represent \(1 / z .\) Solve for \(u, v,\) and \(w,\) and then solve for \(x, y,\) and \(z\). $$ \begin{aligned} &\frac{2}{x}-\frac{1}{y}-\frac{3}{z}=-1\\\ &\frac{2}{x}-\frac{1}{y}+\frac{1}{z}=-9\\\ &\frac{1}{x}+\frac{2}{y}-\frac{4}{z}=17 \end{aligned} $$

Short Answer

Expert verified
u = -1, v = 5, w = -2; x = -1, y = \( \frac{1}{5} \), z = -\( \frac{1}{2} \)

Step by step solution

01

Substitution of Variables

Start by letting \(u = \frac{1}{x}\), \(v = \frac{1}{y}\), and \(w = \frac{1}{z}\). Substitute these expressions into the given system of equations to rewrite them in terms of \(u\), \(v\), and \(w\). The system becomes:\[ \begin{aligned} 2u - v - 3w &= -1 \ 2u - v + w &= -9 \ u + 2v - 4w &= 17 \end{aligned} \]
02

Solve for One Variable

Considering the first equation: \(2u - v - 3w = -1\), solve for \(v\) in terms of \(u\) and \(w\):\[ v = 2u - 3w + 1 \]
03

Substitute in the Second Equation

Substitute the expression for \(v\) from Step 2 into the second equation: \(2u - v + w = -9\):\[ 2u - (2u - 3w + 1) + w = -9 \]Simplify and solve for \(w\):\[ 2u - 2u + 3w - 1 + w = -9 \] \[ 4w - 1 = -9 \]\[ 4w = -8 \]\[ w = -2 \]
04

Find \(v\) and Substitute in the Third Equation

Use \(w = -2\) in the expression found in Step 2:\[ v = 2u - 3(-2) + 1 \]\[ v = 2u + 6 + 1 \]\[ v = 2u + 7 \] Next, substitute \(v = 2u + 7\) and \(w = -2\) into the third equation: \(u + 2v - 4w = 17\):\[ u + 2(2u + 7) - 4(-2) = 17 \]\[ u + 4u + 14 + 8 = 17 \]\[ 5u + 22 = 17 \]\[ 5u = -5 \]\[ u = -1 \]
05

Solve for \(v\)

Substitute \(u = -1\) back into the expression for \(v\):\[ v = 2(-1) + 7 \]\[ v = -2 + 7 \]\[ v = 5 \]
06

Solve for Original Variables

Now that \(u, v,\) and \(w\) are known: \(u = -1\), \(v = 5\), and \(w = -2\) convert back to the original variables \(x, y,\) and \(z\):\[ \frac{1}{x} = u = -1 \implies x = -1 \]\[ \frac{1}{y} = v = 5 \implies y = \frac{1}{5} \]\[ \frac{1}{z} = w = -2 \implies z = -\frac{1}{2} \]

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Key Concepts

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

Substitution Method
The substitution method is a common way to solve systems of equations. It involves replacing one variable with an expression obtained from another equation. This turns a system of equations into an easier-to-manage situation.
In this exercise, we start by letting three new variables represent the reciprocals of the original ones: let \(u = \frac{1}{x}\), \(v = \frac{1}{y}\), and \(w = \frac{1}{z}\).
By substituting these new variables, the initial complex problem transforms into a simpler linear system:
  • \(2u - v - 3w = -1\)
  • \(2u - v + w = -9\)
  • \(u + 2v - 4w = 17\)
With these substitutions, solving for one variable becomes straightforward, and we follow this strategy all the way through to the solution.
Solving for Variables
Solving for variables step-by-step ensures accuracy. We first solve one equation for one variable in terms of the others and then substitute this into the remaining equations.
Consider our first equation: \(2u - v - 3w = -1\).
To isolate \(v\), we rearrange the equation:
  • \(v = 2u - 3w + 1\)
This reveals \(v\) as a function of \(u\) and \(w\).
Next, we substitute this expression for \(v\) into the second equation. This reduces the number of variables, making the system easier to solve:
  • \(2u - (2u - 3w + 1) + w = -9\)
Simplifying this gives us \(w = -2\).
We continue this process until all variables are found.
Inverse Relationships
Inverse relationships play a key role in this problem. The variables \(x, y, z\) are the inverses of \(u, v, w\), respectively:
  • \(u = \frac{1}{x}\)
  • \(v = \frac{1}{y}\)
  • \(w = \frac{1}{z}\)
Once the new system of linear equations is solved, we revert \(u, v, w\) back to \(x, y, z\).
For instance, if we know \(u = -1\), then solving for \(x\) involves calculating \(x = \frac{1}{u} = -1\).
This principle simplifies converting the solved new variables back into the original variables.
Linear Equations
A linear equation is an equation where each term is either a constant or the product of a constant and a single variable. In the context of this exercise, we manipulate and solve these linear equations.
Linear equations are fairly simple and follow specific patterns, such as:
  • \(ax + by + cz = d\)
In our exercise, after substituting the variables, we manipulate equations such as:
  • \(2u - v - 3w = -1\)
  • \(2u - v + w = -9\)
  • \(u + 2v - 4w = 17\)
If we solve them step-by-step while keeping track of our substitutions and arithmetical operations, we can find their solutions efficiently.

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Most popular questions from this chapter

Solve each system. If a system's equations are dependent or if there is no solution, state this. $$ \begin{aligned} x-y+z &=4 \\ 5 x+2 y-3 z &=2 \\ 4 x+3 y-4 z &=-2 \end{aligned} $$

Automotive Maintenance. The radiator in Natalie's car contains \(6.3 \mathrm{L}\) of antifreeze and water. This mixture is \(30 \%\) antifreeze. How much of this mixture should she drain and replace with pure antifreeze so that there will be a mixture of \(50 \%\) antifreeze?

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Simplify. \(\frac{1.2 \times 10^{3}}{2.4 \times 10^{-17}}\)
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