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Chapter 5: Problem 81

Use the substitution \(u=\frac{1}{x}\) and \(v=\frac{1}{y}\) to rewrite the equations in the system in terms of the variables \(u\) and \(v .\) Solve the system in terms of \(u\) and \(v .\) Then back substitute to determine the solution set to the original system in terms of \(x\) and \(y\). $$ \begin{array}{l} \frac{1}{x}+\frac{2}{y}=1 \\ -\frac{1}{x}+\frac{4}{y}=-7 \end{array} $$

Short Answer

Expert verified
The solution is \( x = \frac{1}{3} \) and \( y = -1 \).

Step by step solution

01

Substitutions

Let the substitutions be: \( u = \frac{1}{x} \) and \( v = \frac{1}{y} \). Rewrite the original system of equations using these substitutions.
02

Rewrite the system

Using the substitutions, the system becomes: \( u + 2v = 1 \) \(-u + 4v = -7 \)
03

Solve for one variable

Add the two equations to eliminate \(u\):\( (u + 2v) + (-u + 4v) = 1 - 7 \) This simplifies to: \( 6v = -6 \) So, \( v = -1 \)
04

Solve for the other variable

Substitute \( v = -1 \) into the first equation: \( u + 2(-1) = 1 \) This simplifies to: \( u - 2 = 1 \) So, \( u = 3 \)
05

Back Substitution

Recall that \( u = \frac{1}{x} \) and \( v = \frac{1}{y} \). Substitute back \( u = 3 \) and \( v = -1 \): \( 3 = \frac{1}{x} \) => \( x = \frac{1}{3} \)\( -1 = \frac{1}{y} \) => \( y = -1 \)
06

Solution set

The solution set in terms of \( x \) and \( y \) is \( x = \frac{1}{3} \) and \( y = -1 \).

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Key Concepts

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

solving systems of equations
When solving systems of equations, the goal is to find values for the variables that make all equations true simultaneously. In this exercise, we deal with two equations involving the variables \(x\) and \(y\). The key is to use algebraic techniques to simplify and solve them step-by-step, ensuring every step gets us closer to the solution. We begin by transforming these variables into simpler forms using substitution, allowing us to solve the system more easily.
variable substitution
Variable substitution is a powerful algebraic technique where we replace a variable with another expression that simplifies our equations. In this problem, we use the substitutions \(u = \frac{1}{x}\) and \(v = \frac{1}{y}\). This helps transform our original complicated equations into simpler ones. This approach restructures the problem, making it more straightforward to solve.
algebraic transformation
An algebraic transformation involves manipulating an equation or system of equations to a different form that is easier to solve. After substituting \(u\) and \(v\), our system becomes:
\(u + 2v = 1 \)
\( -u + 4v = -7 \)
These simplified equations can now be solved using standard algebraic methods. We aimed to eliminate one variable by adding the two equations, leading us to a single equation with one variable.
back substitution
Back substitution is the final step in solving systems of equations using substitution. Once we have found the values of the substituted variables \(u = 3\) and \(v = -1\), we then revert to the original variables by back substituting these values.
We know that \(u = \frac{1}{x}\) so \(3 = \frac{1}{x}\) leading to \(x = \frac{1}{3}\).
Similarly, \(v = \frac{1}{y}\) gives us \(-1 = \frac{1}{y}\) resulting in \(y = -1\).
This back substitution process ensures that we translate our solutions correctly back to the original variables.

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Most popular questions from this chapter

Describe the solution set to the system of inequalities. \(x \geq 0, y \geq 0, x \leq 1, y \leq 1\)

Two or more linear equations taken together are called a _____ of linear equations.

Refer to Section 2.5 for a review of linear cost functions and linear revenue functions. A cleaning company charges \(\$ 100\) for each office it cleans. The fixed monthly cost of \(\$ 480\) for the company includes telephone service and the depreciation on cleaning equipment and a van. The variable cost is \(\$ 52\) per office and includes labor, gasoline, and cleaning supplies. a. Write a linear cost function representing the cost \(C(x)\) (in \(\$$ ) to clean \)x\( offices per month. b. Write a linear revenue function representing the revenue \)R(x)\( (in \$) for cleaning \)x$ offices per month. c. Determine the number of offices to be cleaned per month for the company to break even. d. If 28 offices are cleaned, will the company make money or lose money?

Solve the system using any method. $$ \begin{array}{l} 0.05 x+0.01 y=0.03 \\ x+\frac{y}{5}=\frac{3}{5} \end{array} $$

The attending physician in an emergency room treats an unconscious patient suspected of a drug overdose. The physician does not know the initial concentration \(A_{0}\) of the drug in the bloodstream at the time of injection. However, the physician knows that after \(3 \mathrm{hr}\), the drug concentration in the blood is \(0.69 \mu \mathrm{g} / \mathrm{dL}\) and after \(4 \mathrm{hr}\), the concentration is \(0.655 \mu \mathrm{g} / \mathrm{dL}\). The model \(A(t)=A_{0} e^{-k t}\) represents the drug concentration \(A(t)\) (in \(\mu \mathrm{g} / \mathrm{dL}\) ) in the bloodstream \(t\) hours after injection. The value of \(k\) is a constant related to the rate at which the drug is removed by the body. a. Substitute 0.69 for \(A(t)\) and 3 for \(t\) in the model and write the resulting equation. b. Substitute 0.655 for \(A(t)\) and 4 for \(t\) in the model and write the resulting equation. c. Use the system of equations from parts (a) and (b) to solve for \(k .\) Round to 3 decimal places. d. Use the system of equations from parts (a) and (b) to approximate the initial concentration \(A_{0}\) (in \(\mu \mathrm{g} / \mathrm{dL}\) ) at the time of injection. Round to 2 decimal places. e. Determine the concentration of the drug after \(12 \mathrm{hr}\). Round to 2 decimal places.
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